ap e unit 8 - acids & bases

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Acids & BasesISPS Chemistry Feb 2021 page 1

Unit 8 - Acids & Bases

8.1 Introduction to Acids and Bases 8.2 pH and pOH of Strong Acids and Bases 8.3 Weak Acid and Base Equilibria 8.4 Acid-Base Reactions and Buffers 8.5 Acid-Base Titrations 8.6 Molecular Structure of Acids and Bases 8.7 pH and pKa 8.8 Properties of Buffers 8.9 Henderson-Hasselbalch Equation 8.10 Buffer Capacity

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8.1 Introduction to Acids and Bases

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Original Definitions of Acids & Bases - Arrhenius Definition

In earlier courses, you were probably introduced to the idea that Acids are substances that release H+ ions into water, for example,

HCl(g) HCl(aq) H+(aq) + Cl—

(aq)

hydrogen chloride gas hydrogen chloride solution hydrochloric acid

H2SO4(l) H2SO4(aq) 2 H+(aq) + SO4

2—(aq)

hydrogen sulfate liquid hydrogen sulfate solution sulfuric acid

CO2(g) CO2(aq) H2CO3(aq) 2 H+(aq) + CO3

2—(aq)

carbon dioxide gas carbon dioxide solution hydrogen carbonate solution carbonic acid

All of our acids start as covalent molecules that ionise when dissolved in water. The significance of is that we would consider the acid has ionised 100% and we would define the acid as a

strong acid.

The significance of is that we would consider that the acid has only partially ionised and we would define the acid as a weak acid.

Again, our earliest definition was that a Base is a substance that releases OH`— ions into water, for example,

NaOH(s) Na+(aq) + OH—

(aq)

sodium hydroxide sodium hydroxide solution

Al(OH)3(s) Al3+(aq) + 3 OH—

(aq) Ksp = 1.8 x 10-5

aluminium hydroxide aluminium hydroxide solution

NH3(g) NH3(aq) NH4+

(aq) + OH—(aq)

ammonia gas ammonia solution ammonium hydroxide solution

Apart from ammonia, most bases are already ionic so the significance of is that we would consider the base has dissolved 100% and we would define the base as a strong base.

The significance of is that we would consider that the base has only partially dissolved or that the base has only partially ionised and we would define the acid as a weak base.

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Role of Water

Though water has a major role as the solvent, it plays even more crucial roles in acid-base chemistry.

Hydrogen ions are too small (basically they are the positive nucleus of the original hydrogen atom) to exist as stable particles.

Instead they form a coordinate covalent bond, using one of the lone pairs of electrons on a water molecule, and form a number of complex ions such as the hydroxonium or hydronium ion.

With strong bases, water is usually just a solvent but with weak bases, water gets involved in the ionisation and is the source of the OH— ions.

This time a H+ ion forms a coordinate covalent bond with the base, using one of the lone pairs of electrons on the NH3 molecule to form an ammonium ion,NH4

+.

However, water has an even more pervasive influence on acid-base chemistry due to the autoionisation of water:

H2O(l) + H2O(l) ⇌ H3O+

(aq) + OH—(aq)

Water Constant, Kw

The significance of the autoionisation of water is that there will be H+ ions and OH — ions present in all solutions, and their relative concentrations will be linked by their equilibrium constant.

K = [H3O+][OH—] but measurements show that (at 25°C) [H+] = [OH—] = 10-7 mol L-1,

[H2O]2 whereas in a Litre of water, [H2O] would be 1000/18 ≃ 55 mol L-1

Therefore, any change in [H2O] will be negligible and we can incorporate [H2O] into our constant. This is consistent with the rule met in Unit 7 - Equilibrium, where pure liquids and pure solids are treated as having an 'activity' = 1 which means 'effective concentration' or [H2O] = 1

K [H2O]2 = [H3O+][OH—] becomes Kw = [H3O

+][OH—]

or Kw = [H+][OH—] = 1.0 x 10-14 at 25° C

⇌

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Effect of Temperature on Kw H2O(l) ⇌ H+(aq) + OH—

(aq)

The ionisation of water is endothermic (bond breaking) so will be favoured by an increase in temperature, as shown in the graph opposite.

At 25° C, neutral solutions will have:

pKw = 14.0 so Kw = 1.0 x 10-14 mol2 L-2

Kw = [H+][OH—] = 1.0 x 10-14 mol2 L-2

[H+] = [OH—] = 1.0 x 10-7 mol L-1

and pH = pOH = 7

However, neutral solutions will have different pH's at different temperatures.

At 250° C, neutral solutions will have: pKw = 11.0 so Kw = 1.0 x 10-11 mol2 L-2

Kw = [H+][OH—] = 1.0 x 10-11 mol2 L-2

[H+] = [OH—] = √1.0 x 10-11 = 3.16 x 10-6 mol L-1 and pH = pOH = 5.5

At ~0° C, neutral solutions will have: pKw = 14.8 so Kw = 2.0 x 10-15 mol2 L-2

Kw = [H+][OH—] = 2.0 x 10-15 mol2 L-2

[H+] = [OH—] = √2.0 x 10-15 = 3.98 x 10-8 mol L-1 and pH = pOH = 7.4

While the vast majority of our data will be measures at 25°C (s.t.p.), you need to be aware that values such as Kw and hence, pH and pOH , are temperature dependent.

But at 25°C

As acid is added: [H+]⬀ so [OH—]⬂

[H+] > 10-7 mol L-1 pH < 7

In neutral: [H+] = [OH—]

[H+] = 10-7 mol L-1 pH = 7

As base is added: [OH—]⬀ so [H+]⬂

[H+] < 10-7 mol L-1 pH > 7

Kw = [H3O+][OH—] = 1.0 x 10-14

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The 'potent' Scale

In an attempt to simplify working with very small numbers, Søren Sørensen, a Swedish chemist, suggested, in 1909, a scale based on the 'powers of ten' ('potent' is German for power).

This was first applied to [H+]'s so [H+] = 1.0 x 10-7 became pH = 7

Mathematically, pH = - log [H+] and inversely [H+] = 10 -pH

similarly, pOH = - log [OH—] and inversely [OH—] = 10 -pOH

and, pKa = - log Ka and inversely Ka = 10 -pKa

As well as : Kw = [H3O+][OH—] = 1.0 x 10-14

Other useful relationships include: pKw = pH + pOH = 14

Due to the logarithmic nature of thepH scale:

a change of one pH unit represents a 10 fold increase or decrease in

ion concentration

1M HCl, [H+] = 100, pH = 010M HCl, [H+] = 101, pH = -1

but 100M HCl not possible so pH scale has a 'minimum' between -1 and -2.

1M NaOH, [H+] = 10-14, pH = 1410M NaOH, [H+] = 10-15, pH = 15

but 100M NaOH not possible so pH scale has a 'maximum' between 15 and 16.

Most strong acids and strong bases form saturated solutions in the range 20 - 30 M.

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8.1 Practice Problems 1. In pure water, some of the molecules ionize according to the equation H2O ⇌ H+ + OH−

The extent of the ionization increases with temperature. A student heats pure water and records the measured pH at 50°C as 6.6.

Based on this information, which of the following mathematical relationships gives the pOH of pure water at 50°C ?

A pOH = pH B pOH = 1/pH C pOH = 14−pH D pOH = (1×10−14)/pH

2. In pure water, some of the molecules ionize according to the equation

H2O ⇌ H+ + OH−

The extent of the ionization increases with temperature.

The graph opposite shows the pH of pure water at different temperatures.

Which of the following represents the variations in the pOH of pure water under the same conditions?

A B

C D

O

O

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3. H2O(l) + H2O(l) ⇋ H3O+

(aq) + OH—(aq)

At 5.0°C , the value of Kw or the equilibrium shown above is 1.9 × 10−15 and the value of pKw is 14.73.

Based on this information, which of the following is correct for pure water at this temperature?

A

B

C

D

4. Which of the following accounts for the observation that the pH of pure water at 37°C is 6.8?

A At 37 ° C water is naturally acidic.

B At 37 ° C the autoionization constant for water, Kw, is larger than it is at 25 ° C.

C At 37 ° C water has a lower density than it does at 25 ° C; therefore, [H+] is greater.

D At 37 ° C water ionizes to a lesser extent than it does at 25 ° C

5. 2 H2O(l) ⇋ H3O+

(aq) + OH—(aq)

The autoionization of water is represented by the equation above. Values of pKw at various temperatures are listed in the table below.

Based on the information above, which of the following statements is true?

A The dissociation of water is an exothermic process.

B The pH of pure water is 7.00 at any temperature.

C As the temperature increases, the pH of pure water increases.

D As the temperature increases, the pH of pure water decreases.

O

O

O

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6. 5 H2O2(aq) + 2 MnO4—

(aq) + 6 H+(aq) → 2 Mn2+

(aq) + 8 H2O(l) + 5 O2(g)

In a titration experiment, H2O2(aq) reacts with aqueous MnO4—

(aq) as represented by the equation above. The dark purple KMnO4 solution is added from a buret to a colorless, acidified solution of H2O2(aq) in an Erlenmeyer flask. (Note: At the end point of the titration, the solution is a pale pink color.)

Which of the following best describes what happens to the pH of the H2O2 solution as the titration proceeds?

A The +2 charge on the manganese ions maintains the acidity of the solution.

B The production of water dilutes the solution, making it basic.

C As H+ ions are consumed, the solution becomes less acidic and the pH increases.

D As H+ ions are consumed, the solution becomes less acidic and the pH decreases.

7. At 25°C, aqueous solutions with a pH of 8 have a hydroxide ion concentration, [OH—], of

A 1 x 10-14 M B 1 x 10-8 M C 1 x 10-6 M D 1 M E 8 M

8. A student is given a solution of phenol red of unknown concentration. Solutions of phenol red are bright pink under basic conditions.

The student uses a pH probe and determines that the pH of the solution is 11.20. Calculate the hydrogen ion concentration, [H+], in the solution of phenol red.

The response gives a calculation equivalent to the following.

[H+] = 10 -pH = 10 -11.20 = 6.3 x 10-12 M

O

O

AP Chemistry


8.2 pH and pOH of Strong Acids and Bases

pH = - log [H+]pOH = - log [OH—]

Kw = [H3O+][OH—] = 10-14

pKw = pH + pOH = 14

a change of one pH unit represents a

10 fold increase or decrease in ion concentration

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Strong AcidsFor strong acids, since we can assume 100% ionisation, there is a quick and easy route to calculating pH and, if necessary, pOH.

0.20 M HCl [HCl] = 0.20 M [H+] = 0.20 M

pH = - log [H+] = -log(0.20) = 0.70 pOH = 14 - pH = 14 - 0.70 = 13.30

0.20 M H2SO4 [H2SO4] = 0.20 M [H+] = 0.40 M

pH = - log [H+] = -log(0.40) = 0.40 pOH = 14 - pH = 14 - 0.40 = 13.60

0.02 M H2SO4 [H2SO4] = 0.02 M [H+] = 0.04 M

pH = - log [H+] = -log(0.04) = 1.40 pOH = 14 - pH = 14 - 1.40 = 12.62

Notice that diluting the H2SO4 by a factor of 10 (0.20M → 0.02M) causes pH and pOH to change by 1 unit.

Strong BasesFor strong bases, since we can again assume 100% ionisation, there is a similar quick and easy route to calculating pOH though we will normally want to convert this to pH.

0.015 M NaOH [NaOH] = 0.015 M [OH—] = 0.015 M

pOH = - log [OH—] = -log(0.015) = 1.82 pH = 14 - pOH = 14 - 1.82 = 12.18

0.015 M Ca(OH)2 [Ca(OH)2] = 0.015 M [OH—] = 0.030 M


0.15 M Ca(OH)2 [Ca(OH)2] = 0.15 M [OH—] = 0.30 M


Notice that increasing the [Ca(OH)2] by a factor of 10 (0.015M → 0.15M) causes pH and pOH to change by 1 unit.

Effect of TemperatureIn the previous lesson we saw that Kw is temperature dependent. At ~0° C, pKw = 14.8

Changing temperature, however, has no effect on stoichiometry and our formulae for calculating pH and pOH do not change. So,

0.20 M HCl [HCl] = 0.20 M [H+] = 0.20 M

pH = - log [H+] = -log(0.20) = 0.70

However, pOH = 14.8 - pH = 14.8 - 0.70 = 14.1

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8.2 Practice Problems1. Which of the following is the correct mathematical expression to use to calculate the pH of a 0.10 M aqueous Ba(OH)2 solution at 25°C ?

A pH = −log (0.10)

B pH = −log (0.20)

C pH = 14.00 − log (0.10)

D pH = 14.00 + log (0.20)

2. Which of the following statements about the pH of 0.010 M HClO4 is correct?

A pH = 2.00 , because [H+] = 1.0 × 10−2 M .

B pH = 2.00 , because [H+] = 2.0 × 10−2 M .

C pH > 2.00 , because HClO4 is a strong acid.

D pH < 2.00 , because HClO4 is a weak acid.

3. Which of the following pairs of mathematical expressions can be used to correctly calculate the pH and pOH of a 0.0015 M KOH(aq) solution at 25°C ?

A

B

C

D

4. How can 100. mL of sodium hydroxide solution with a pH of 13.00 be converted to a sodium hydroxide solution with a pH of 12.00 ?

A By diluting the solution with distilled water to a total volume of 108 mL

B By diluting the solution with distilled water to a total volume of 200 mL

C By diluting the solution with distilled water to a total volume of 1.00 L

D By adding 100. mL of 0.10 M HCl

E By adding 100. mL of 0.10 M NaOH

O

O

O

O

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5. Which of the following is a strong electrolyte in aqueous solution

A NH3(g) B BH3(g) C H2(g) D H2S(g) E HBr(g)

6. Refer to the following.

The pH of solutions of four acids prepared at various concentrations were measured and recorded in the table above.

The four acids are, in no particular order, chlorous, hydrochloric, lactic, and propanoic.

Which of the four acids listed in the table is hydrochloric acid?

A Acid 1 B Acid 2 C Acid 3 D Acid 4

7. The pH of a solution prepared by the addition of 10. mL of 0.002 M KOH(aq) to 10. mL of distilled water is closest to

A 12 B 11 C 10 D 4 E 3

O

O

O

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8.3 Weak Acid and Base Equilibria

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AP Chemistry


Brønsted–Lowry Definition of Acids and Bases

This concept is based on the fact that

acid–base reactions involve the transfer of H+ ions from one substance to another.

Conjugate pairs, that differ by a H+ ion, will exist with one molecule being theacid while the other is the base.

One consequence of this definition is that water, H2O, will often fulfill the role of an acid or base in many reactions.

An ACID is a substance (molecule or ion) that donates a proton (H+)

A BASE is a substance (molecule or ion) that accepts a proton (H+)

Weak Acids

Most acids do not ionise 100%, they are described as weak acids and are represented by the following equation: HA(aq) + H2O(l) ⇋ H3O

+(aq) + A—

(aq)

The equilibrium constant is

K = [H3O+] [A—]

[HA] [H2O]

but, as usual, [H2O] can be assumed to stay unchanged so, we re-define the equilibrium constant as the acid-dissociation constant, Ka

Ka = [H+] [A—] [HA]

HCl ⇋ Cl— NH3 ⇋ NH4+

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As always, an equilibrium constant provides useful information about the ratio of products and reactants and helps us compare the relative strengths of different weak acids.

Calculating pH of Weak Acids

We can use the acid-dissociation constant to calculate [H+] ( and then pH = -log[H+]) but only if we make an important assumption about the value of [HA]equilibrium.

In essence, we have to assume the acid is so weak and such a small amount of ionisation will take place that: [HA]equilibrium = [HA]initial = C (referred to as the analytical concentration)

In other words, if we start with a 0.01 M ethanoic acid solution, we will assume [CH3COOH]equ is still 0.01 M.

So Ka = [H+] [A—] becomes Ka = [H+] [A—] We also know that [HA]equilibrium [HA]initial [H+] = [A—]

so Ka = [H+]2 and [H+] = √(Ka[HA]initial) [HA]initial then pH = -log[H+]

AP Chemistry


Example 1: Calculate the pH of a 0.01 M solution of ethanoic (acetic) acid given that Ka = 1.8 x 10-5

Method 1

step 1: [H+] = √(Ka [HA]initial ) = √(1.8 x 10-5 x 0.01) = 4.2 x 10-4 M step 2: pH = -log [H+] = -log (4.2 x 10-4) = 3.37

A Second Method would be to use the ICE table, met in the Equilibrium Unit, to calculate [H+] at equilibrium and then use pH = -log[H+].

CH3COOH(aq) ⇋ H+(aq) + CH3COO—

(aq)

Initial 0.01 0.00 0.00

Change - x + x + x

Equilibrium 0.01 - x x x

Ka = [H+] [CH3COO—] = x x x = 1.8 x 10-5

[CH3COOH] 0.01 - x

At this point, we are faced with using the quadratic expression to solve for x. Instead, we use our previous assumption, that [CH3COOH]equilibrium ≃ [CH3COOH]initial = 0.01M, to make life easier.

x2 = 1.8 x 10-5 x 0.01 so x = √(1.8 x 10-7) = 4.2 x 10-4

so [H+] = 4.2 x 10-4 M and pH = -log(4.2 x 10-4) = 3.37

Assumptions There are two major assumptions made in these calculations:-

① that we can ignore the [H+] being produced by the auto-ionisation of water.

Given that this is normally ≃ 10-7 M, this will only be an issue if our weak acid is at very low concentration.

② that we can assume [HA]equilibrium ≃ [HA]initial

This is justifiable as long as the % of acid molecules that ionise is <5%. However, % ionisation increases at lower concentrations so ......

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Example 2: Calculate the ionisation constant, Ka, of a 0.1 M solution of a weak monoprotic acid that has a pH equal to 4.0. Method

step 1: [H+] = 10-pH = 1.0 x 10-4 M step 2: Ka = [H+]2 / [HA]initial = (1.0 x 10-4)2 / 0.1 = 1 x 10-7

Calculating Percent Ionisation of a Weak Acid

We have seen that the magnitude of Ka indicates the strength of an acid. Another measure of the strength of an acid is its percent ionisation, which is defined as

percent ionisation = ionised acid concentration at equilibrium x 100 % initial concentration of acid

The stronger the acid, the greater the percent ionisation. For a monoprotic acid HA, the concentration of the acid that undergoes ionisation is equal to the concentration of the H+ ions or the concentration of the A— ions at equilibrium.

Therefore, we can write the percent ionisation as

percent ionisation = [H+] x 100 % [HA]

Example: The pH of a 0.036 M nitrous acid (HNO2) solution is measured as 2.4.

HNO2(aq) ⇋ H+(aq) + NO2

—(aq)

Calculate the percent ionisation for nitrous acid.

step 1: [H+] = 10-pH = 10-2.4 = 3.98 x 10-3 M

step 2: percent ionisation = [H+] / [HA] x 100 % = 3.98 x 10-3 / 0.036 x 100 %

= 11.1 %

So about 1 in 9 molecules of nitrous acid would ionise which is consistent with a weak acid.

Earlier it was mentioned that one of our main assumptions for pH, pKa calculations relies on ionisation being < 5%, so we would be unable to rely on these calculations for 0.036 M HNO2 .

AP Chemistry


Calculating pH of Weak Bases

The ionisation of weak bases is treated in the same way as the ionisation of weak acids. When ammonia dissolves in water, it undergoes the reaction

NH3(aq) + H2O(l) ⇋ NH4+

(aq) + OH—(aq)

The equilibrium constant is given by K = [NH4

+][OH—] [NH3][H2O]

Compared with the total concentration of water, very few water molecules are consumed by this reaction, so we can treat [H2O] as a constant. Thus, we can write the base ionisation constant (Kb), which is the equilibrium constant for the ionisation reaction, as

K[H2O] = Kb = [NH4+][OH—] = [BH+][OH—]

[NH3] [B]

Similarly, the magnitude of Kb indicates the strength of a base. Notice also that the stronger the base, the weaker is the conjugate acid. (Ka x Kb = 1.0 x 10-14 for a conjugate pair)

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We can use the base-dissociation constant to calculate [OH—] ( and then, if necessary, convert to [H+] to allow pH = -log[H+]). But again we need to make an important assumption about the value of [NH3]equilibrium.

In essence, we have to assume the base is so weak and such a small amount of ionisation will take place that: [NH3]equilibrium = [NH3]initial

In other words, if we start with 0.01 M ammonia solution, we will assume [NH3]equ is still 0.01 M.

So Kb = [NH4+][OH—] becomes Kb = [NH4

+][OH—] We also know that [NH3]equilibrium [NH3]initial [NH4

+] = [OH—]

so Kb = [OH—]2 from here there are a number of different routes that can be taken [NH3]initial

Example 1: What is the pH of a 0.40 M ammonia solution? Kb = 1.8 x 10-5

step 1: [OH—] = √(Kb C) = √(1.8 x 10-5 x 0.4) = 2.7 x 10-3 M

step 2: [H+] = 1.0 x 10-14 /[OH—] Kw = [H+][OH—] = 1.0 x 10-14

= 1.0 x 10-14 / 2.7 x 10-3

= 3.7 x 10-12

step 3: pH = -log [H+] = -log (3.7 x 10-12) = 11.4

Example 1 again: What is the pH of a 0.40 M ammonia solution? Kb = 1.8 x 10-5

step 1: [OH—] = √(Kb C) = √(1.8 x 10-5 x 0.4) = 2.7 x 10-3 M

step 2: pOH = -log[OH—] = -log(2.7 x 10-3)

= 2.6

step 3: pH = 14 - pOH pKw = pH + pOH = 14 = 14 - 2.6 = 11.4

Again, we are assuming that OH— ions from auto-ionisation of water will be insignificant and that the weak base does not ionise enough (< 5%) to ensure [base]equilibrium = [base]initial.

AP Chemistry


Formulae for Calculations

We have an almost bewildering number of formulae that we can either derive as needed or attempt to memorise.

Lets start, however, with the formulae provided in the 'AP® Chemistry Equations & Constants' sheet.

Acid Equations

① ② ③

Base Equations

④ ⑤ ⑥

General Equations ⑦

⑧ ⑨only applies to conjugate pairs

The following could be very useful and can then be derived or memorised.

From ①we can derive From ④we can derive

From ⑨we can derive pKw = 14 = pKa + pKb only applies to conjugate pairs

You should also be able to do the inverse of ② , ③ , ⑤and ⑥

[H+] = 10-pH [OH—] = 10-pOH Ka = 10-pKa Kb = 10-pKb

Over the rest of this unit you should try solving questions using the minimum number of memorised formulae - wherever possible, use the formulae provided by 'AP® Chemistry Equations & Constants' sheet and derive any others needed.

Kb = [OH—]2

[B]initial

AP Chemistry


8.3 Practice Problems1. The equilibrium reaction shown above represents the partial ionization of the weak acid HCN(aq). A 0.200 M HCN(aq) solution has a pH≈4.95. If 0.05g (0.010mol) of NaCN(s) is added to 100mL of 0.200 M HCN(aq), which of the following explains how and why the pH of the solution changes?

A The pH will be higher than 4.95 because adding CN− will disrupt the equilibrium, resulting in an increased production of HCN that decreases the concentration of H3O

+.

B The pH will be lower than 4.95 because adding CN− will disrupt the equilibrium, resulting in an increased production of HCN that decreases the concentration of H3O

+.

C The pH will be higher than 4.95 because CN− is a strong base that can neutralize HCN.

D The pH will remain close to 4.95 because the Ka is so small that hardly any products form.

2. (CH3)2NH(aq) + H2O(l) ⇋ (CH3)2NH2+

(aq) + OH−(aq) Kb=5.4 × 10−4 at 25°C

The equilibrium for the reaction between (CH3)2NH, a weak base, and water is represented by the equationabove.

The table shows the pH of three solutions of (CH3)2NH(aq) at 25°C.

Based on the information given, which of the following is true?

A Solutions with a higher concentration of (CH3)2NH have a higher pOH because to reach equilibrium a smaller amount of the conjugate acid (CH3)2NH2

+ is produced.

B Solutions with a higher concentration of (CH3)2NH have a higher pOH because to reach equilibrium more OH− is produced.

C Solutions with a higher concentration of (CH3)2NH have a higher pH because to reach equilibrium a smaller amount of the conjugate acid (CH3)2NH2

+ is produced.

D Solutions with a higher concentration of (CH3)2NH have a higher pH because to reach equilibrium more OH− is produced.

O

O

AP Chemistry


3.

NH3 is a weak base that reacts with water according to the chemical equilibrium represented above. The table provides some information for two NH3(aq) solutions of different concentration at 25°C . Which of the following is true about the more concentrated 0.30 M NH3(aq), and why?

A [OH−] = 3.2 ×10−3 M and pOH < 2.78 , because a higher [OH−] corresponds to a lower pOH for 0.30 M NH3(aq) compared to 0.15 M NH3(aq).

B [OH−] = 3.2 ×10−3 M and pOH < 2.78 , because a higher [OH−] corresponds to a higher pOH for 0.30 M NH3(aq) compared to 0.15 M NH3(aq).

C [OH−] = 2.3 ×10−3 M and pOH < 2.78 , because a higher [OH−] corresponds to a lower pOH for 0.30 M NH3(aq) compared to 0.15 M NH3(aq).

D [OH−] = 2.3 ×10−3 M and pOH < 2.78 , because a higher [OH−] corresponds to a higher pOH for 0.30 M NH3(aq) compared to 0.15 M NH3(aq).

4. Caffeine (C8H10N4O2) is a weak base with a Kb value of 4 x 10-4. The pH of a 0.01 M solution of caffeine is in the range of

A 2 - 3 B 5 - 6 C 7 - 8 D 11 - 12

5. A 0.10 M solution of a weak monoprotic acid has a pH equal to 4.0. The ionization constant, Ka , of the acid is

A 1 x 10-3 B 1 x 10-4 C 1 x 10-7 D 1 x 10-8 E 1 x 10-9

6. The pH of a 0.01 M HNO2(aq) solution is in which of the following ranges? (For HNO2(aq), Ka = 4.0 x 10-4)

A Between 1 and 2 B Between 2 and 3

C Between 4 and 5 D Between 6 and 7

7. The pH of 0.1 M ammonia solution is approximately

A 1 B 4 C 7 D 11 E 14

O

O

O

O

O

AP Chemistry


The pH of solutions of four acids prepared at various concentrations were measured and recorded in the table below.


8. Of the following species, which has the greatest concentration in a 1.0 M solution of acid 1 at equilibrium?

A OH— B H3O+ C Acid 1 D Conjugate base of Acid 1

9. For which acid is the value of the acid-dissociation constant, Ka , the smallest?

A Acid 1 B Acid 2 C Acid 3 D Acid 4

10. What is the pH of a 1.0 x 10–2-molar solution of HCN? (For HCN, Ka = 4.0 x 10–10.)

A 10 B 7 - 10 C 7 D 4 - 7 E 4

11. A 0.10 M solution of a weak monoprotic acid has a pH equal to 4.0. The ionization constant, Ka , of the acid is

A 1 x 10-3 B 1 x 10-4 C 1 x 10-7 D 1 x 10-8 E 1 x 10-9

12. The ionization constant for acetic acid is 1.8 x 10-5; that for hydrocyanic acid is 4 x 10-10. In 0.1 M solutions of sodium acetate and sodium cyanide, it is true that

A [H+] equals [OH—] in each solution

B [H+] excedes [OH—] in each solution

C [H+] of the sodium acetate solution is less than that of the sodium cyanide solution

D [OH—] of the sodium acetate solution is less than that of the sodium cyanide solution

E [OH—] for the two solutions is the same

O

O

O

O

O

AP Chemistry


13. Which of the following correctly ranks the three monoprotic acids listed in the table opposite from the weakest to the strongest?

A X < Y < Z B X < Z < Y C Y < Z < X D Z < Y < X

14. The following chemical equation represents the reaction that occurs when methylamine dissolves in water to form a basic solution.

CH3NH2(aq) + H2O(l) ⇋ CH3NH3+

(aq) + OH−(aq)

The pH of 2.65 M CH3NH2(aq) is 12.54. Determine the value of Kb for methylamine.

The response meets all of the following three criteria:

1 point - The correct calculation of [OH—]:

pH = 12.54 pOH = 14 - 12.54 = 1.46

[OH—] = 10-1.46 = 3.47 x 10-2 M

1 point - The use of the correct expression for Kb :

Kb = [CH3NH3+] [OH—] / [CH3NH2]

1 point - The correct substitution of the numbers and calculation of the value:

[OH—] = 3.47 x 10-2 M [CH3NH3

+] = [OH—] = 3.47 x 10-2 M [CH3NH2] = 2.65 (it will decrease by 3.47 x 10-2 but this can be considered negligible)

Kb = [CH3NH3+] [OH—] / [CH3NH2] = (3.47 x 10-2)2 / 2.65 = 4.54 x 10-4

O

AP Chemistry


8.4 Acid-Base Reactions and Buffers

AP Chemistry


AP Chemistry


Strong Acid–Strong Base Neutralisation Reaction

The reaction between a strong acid (say, HCl) and a strong base (say, NaOH) can be represented by: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)or more generally: OH—

(aq) + H+(aq) H2O(l) net ionic equation

Changes in pH can be monitored by setting the reaction up as a titration and using a pH meter to measure the pH throughout :-

A 0.100 M NaOH solution is added from a buret to 25.0 mL of a 0.100 M HCl solution in an Erlenmeyer flask. This curve is sometimes referred to as a titration curve.

The equivalence point represents the stoichiometric balance when [H+] = [OH-]. This can be different from the 'end-point' of a titration - the point at which an indicator changes colour as a slight excess of H+ ions or OH— ions may be needed to complete a colour change.

Only for strong acid / strong base titrations will pH = 7 at equivalence point.

Reactions involving weak acids or weak bases will produce acidic salts or basic salts , so pH ≠ 7

Changes in pH can be calculated using our 'traditional' and 'new' formulae:

C = n / L n = m / M pH = -log [H+] [H+] = 10-pH

Kw = 10-14 = [H+] [OH—] or 14 = pH + pOH pOH = -log [OH—]

AP Chemistry


Example 1: A 25.0 mL sample of H2SO4 requires 46.23 mL of a standard 0.203 M NaOH solution to reach the equivalence point. Calculate the [H2SO4].

step1: calculate number of moles of OH— ions added,

n = C x L n = 0.203 x 0.04623 = 0.00938 moles of OH— ions

step2: use the stoichiometry of the balanced equation to convert to moles of H2SO4 ,

2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(l)

2 moles ≣ 1 mole

0.0938 moles of OH— ions ≣ 0.00938/2 = 0.00469 moles of H2SO4

step3: calculate [H2SO4],

C = n /L C = 0.00469 / 0.0250 = 0.188 M

Example 2: A 25.0 mL sample of 0.188 M H2SO4 has 50 mL of a standard 0.203 M NaOH solution added. Calculate the pH of the resulting solution.

step 1: calculate number of moles of H+ ions present at start,

n = C x L n = 0.188 x 0.025 = 0.00470 moles of H2SO4 = 0.00940 moles of H+

step 2: calculate number of moles of OH— ions added,


step 3: subtract to determine moles of excess ions,

0.01015 moles of OH— ions - 0.00940 moles of H+ = 0.00075 moles of excess OH—

step 4: calculate [OH—], (don't forget to combine volumes for overall volume of solution)

C = n /L C = 0.00075 / 0.075 = 0.01 M [OH—] = 10-2

step 5: convert to [H+],

Kw = 10-14 = [H+] [OH—] [H+] = 10-14 / 10-2 = 10-12 M

step 6: calculate pH,

pH = -log[H+] pH = -log(10-12) = 12

AP Chemistry


Weak Acid–Strong Base Neutralisation Reaction

The calculations in the previous sections are made 'easier' by the fact that everything - acids, bases and salts - ionise 100%. Next 'easiest' is where we have one weak / one strong.

The reaction between a weak acid (say, CH3COOH) and a strong base (say, NaOH) can be represented by: NaOH(aq) + CH3COOH(aq) ⇋ NaCH3COO(aq) + H2O(l)or more generally: OH—

(aq) + HA(aq) ⇋ A—(aq) + H2O(l)

Changes in pH can be monitored by setting the reaction up as a titration and using a pH meter to measure the pH throughout :-

A 0.100 M NaOH solution is added from a buret to 25.0 mL of a 0.100 M CH3COOH solution in an Erlenmeyer flask.

The equivalence point represents the stoichiometric balance when moles of CH3COOH present at start = moles of NaOH added.

Only for strong acid / strong base titrations will pH = 7 at equivalence point. When a weak acid / strong base is used the salt formed will be basic so pH > 7.

Changes in pH can still be calculated using our 'traditional' and 'new' formulae but we may need to allow for the partial ionisation of the acid so extra equations needed:

C = n / L n = m / M pH = -log [H+] [H+] = 10-pH

Kw = 10-14 = [H+] [OH—] or 14 = pH + pOH pOH = -log [OH—] Ka = [H+] [A—] / HA or Ka = [H+]2 / C etc.

AP Chemistry


NB. When excess base is used, the weak acid molecule can be treated as ionising 100% as the equilibrium is driven over to the right until all the molecules are consumed. So,

NaOH(aq) + CH3COOH(aq) NaCH3COO(aq) + H2O(l)

Similarly, we can assume 100% removal of acid molecules by OH— ions, but any remaining acid molecules will be only partially ionising depending on

Ka = [H+] [A—] / HA but [A—] and [HA] will have changed as will [H+] and pH

Example 1: Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid by sodium hydroxide after the addition to the acid solution of a) 10.0 mL of 0.100 M NaOH, b) 25.0 mL of 0.100 M NaOH, c) 35.0 mL of 0.100 M NaOH. Ka = 1.8 x 10-5.

a) step 1: calculate number of moles of OH— ions added,


step 2: calculate the number of moles of CH3COOH present at start

n = C x L n = 0.100 x 0.025 = 0.0025 moles of CH3COOH

step 3: calculate the number of moles of CH3COOH remaining and the number of moles of CH3COO— formed in the new equilibrium mixture,


CH3COOH = 0.0025 - 0.0010 = 0.0015 moles CH3COO— = 0.0010 moles

step 4: calculate [H+],

Ka = [H+] [A—] / HA so [H+] = Ka [HA] / [A—]

we are about to substitute values in moles for what should be Molar concentrations. We could calculate concentrations using new overall volumes but this would not change the ratio [HA] / [A—] so we can get the [H+] quicker and easier.

[H+] = Ka [HA] / [A—] = (1.8 x 10-5) (0.0015)/(0.0010) = 2.7 x 10-5 M


pH = -log[H+] so pH = -log (2.7 x 10-5) = 4.57

A solution containing a reasonably high concentration of acid molecules (CH3COOH) and high concentration of salt ions (CH3COO—) will produce a buffering effect. This explains why pH changes 'slow down' in the previous pH curve, between about 5 and 20 ml added. Buffers will be covered in more detail later.

AP Chemistry


Example 1: Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid by sodium hydroxide after the addition to the acid solution of a) 10.0 mL of 0.100 M NaOH, b) 25.0 mL of 0.100 M NaOH, c) 35.0 mL of 0.100 M NaOH. Ka = 1.8 x 10-5.

b) This would be the equivalence point - all CH3COOH molecules would have reacted and there would be no excess OH— ions.

The pH would be determined by the salt ions (CH3COO—). Later in this unit, you will learn how to calculate the pH of salt solutions. c) step 1: calculate number of moles of OH— ions added,


step 2: calculate the number of moles of CH3COOH present at start

n = C x L n = 0.100 x 0.025 = 0.0025 moles of CH3COOH

step 3: calculate the number of moles excess OH— ions


OH— = 0.0035 - 0.0025 = 0.0010 moles

we are going to assume that the contribution due to the salt ions (CH3COO—) will be negligible compared to the excess OH— ions.

step 4: calculate [OH—],(don't forget to combine volumes for overall volume of solution)

C = n /L C = 0.0010 / 0.060 = 0.0167 M [OH—] = 1.67 x 10-2

step 5: convert to [H+],

Kw = 10-14 = [H+] [OH—] [H+] = 10-14 / (1.67 x 10-2) = 5.99 x 10-13 M


pH = -log[H+] pH = -log(5.99 x 10-13) = 12.22

In the previous example, we had a significant amount of our original acid molecule (HA) and the conjugate base (A-) so we had to use our equilibrium constant, Ka , to calculate pH - in effect we were factoring in the buffering effect of the A - ion.

[H+] = Ka [HA] / [A—]

Later you will be introduced to the Henderson- Hasselbalch equation that provides a direct route from relative amounts of salt (A-) and acid (HA) to pH:

pH = pKa + log( [A—] / [HA])

AP Chemistry


8.4 Practice Problems1.

Nitrous acid reacts with ammonia according to the balanced chemical equation shown above. If 50.mL of 0.20 M HNO2(aq) and 50.mL of 0.20 M NH3(aq) are mixed and allowed to reach equilibrium at 25°C , what is the approximate [NH3] at equilibrium?

A 0.00010 M B 0.0010 M C 0.010 M D 0.10 M

2. A student mixes 40.mL of 0.10 M HBr(aq) with 60.mL of 0.10 M KOH(aq) at 25°C. What is the [OH−] of the resulting solution?

A [OH−] = 0.060 M B [OH−] = 0.033 M

C [OH−] = 0.020 M D [OH−] = 0.00000010 M

3.

The equilibrium for the acid ionization of HC2HO3 is represented by the equation above. If 10.0mL of 0.20 M HC2HO3 react with 5.0mL of 0.10 M NaOH, which of the following could be used to calculate the correct pH of the resulting solution?

A

B

C

D

4. A solution containing HCl and the weak acid HClO2 has a pH of 2.4. Enough KOH(aq) is added to the solution to increase the pH to 10.5. The amount of which of the following species increases as the KOH(aq) is added?

A Cl—(aq) B H+

(aq) C ClO2—

(aq) D HClO2(aq)

O

O

O

O

AP Chemistry


5. Acid-dissociation constants of two acids are listed in the table above. A 20. mL sample of a 0.10 M solution of each acid is titrated to the equivalence point with 20. mL of 0.10 M NaOH.

Which of the following is a true statement about the pH of the solutions at the equivalence point?

A Solution 1 has a higher pH at the equivalence point because CH3CO2H is the stronger acid.

B Solution 1 has a higher pH at the equivalence point because CH3CO2H has the stronger conjugate base.

C Solution 1 has a lower pH at the equivalence point because CH3CO2H is the stronger acid.

D Solution 1 has a lower pH at the equivalence point because CH3CO2H has the stronger conjugate base.

6. The net ionic equation for the reaction that occurs during the titration of nitrous acid with sodium hydroxide is

A HNO2 + Na+ + OH— → NaNO2 + H2O

B HNO2 + NaOH → Na+ + NO2— + H2O

C H+ + OH— → H2O

D HNO2 + H2O → NO2— + H3O

+

E HNO2 + OH— → NO2— + H2O

7. Addition of sulfurous acid (a weak acid) to barium hydroxide (a strong base) results in the formation of a precipitate. The net ionic equation for this reaction is

A 2 H+(aq) + 2 OH-

(aq) ⇄ 2 H2O(l)

B H2SO3(aq) + Ba2+(aq) + 2 OH-

(aq) ⇄ BaSO3(s) + 2 H2O(l)

C 2 H+(aq) + SO3

2-(aq) + Ba2+

(aq) + 2 OH-(aq) ⇄ BaSO3(s) + 2 H2O(l)

D H2SO3(aq) + Ba2+(aq) + 2 OH-

(aq) ⇄ Ba2+(aq) + SO3

2-(aq) + 2 H2O(l)

E H2SO3(aq) + Ba(OH)2(aq) ⇄ BaSO3(s) + 2 H2O(l)

O

O

O

AP Chemistry


8. A solution prepared by mixing 10 mL of 1 M HCl and 10 mL of 1.2 M NaOH has a pH of

A 0 B 1 C 7 D 13 D 14

9. When 200. mL of 2.0 M NaOH(aq) is added to 500. mL of 1.0 M HCl(aq), the pH of the resulting mixture is closest to

A 1.0 B 3.0

C 7.0 D 13.0

10.

A student prepares three solutions, X, Y, and Z, as described in the table above. The values of Ka for the acidic species in the solutions are given in the table opposite. a) Using the information above, write the letters of the solutions in the boxes below to rank the solutions in order of increasing pH. Explain your reasoning for the ranking.

1 point is earned for the correct ordering. 1 point is earned for a valid explanation Solution Y is a strong acid solution with a very low pH. Solution Z is a buffer solution with pKa = 4.74 = pH. Solution X is a neutral solution created from equimolar amounts of a strong acid and a strong base that react in a 1:1 ratio.

b) Does the pH of solution Y increase, decrease, or remain the same when 100 mL of water is added? Justify your answer.

1 point is earned for the correct choice and a valid justification.

The pH of the solution increases.

The addition of water will decrease [H+]; therefore, the pH will increase.

O

O

Y Z X

AP Chemistry


Q10 cont. c) The student adds 0.0010 mol of NaOH(s) to solution Y, and adds 0.0010 mol of NaOH(s) to solution Z. Assume that the volume of each solution does not change when the NaOH(s) is added. The pH of solution Y changes much more than the pH of solution Z changes. Explain this observation.

1 point is earned for a valid explanation.

Solution Z is a buffer system (composed of a weak acid and its conjugate base), whereas solution Y is not a buffer.

11. C6H5COOH(s) ⇋ C6H5COO–(aq) + H+

(aq) Ka = 6.46 x 10–5

Benzoic acid, C6H5COOH, dissociates in water as shown in the equation above.

A 25.0 mL sample of an aqueous solution of pure benzoic acid is titrated using standardised 0.150 M NaOH.

a) After addition of 15.0 mL of the 0.150 M NaOH, the pH of the resulting solution is 4.37. Calculate each of the following.

i) [H+] in the solution

1 point is earned for the correct answer.

[H+] = 10−pH [H+] = 10−4.37 M = 4.3 x 10−5 M

ii) [OH—] in the solution


[OH-] = Kw / [H+] = 10-14 / 4.3 x 10−5 = 2.3 x 10-10 M

iii) The number of moles of NaOH added


mol OH− = 0.0150 L x 0.150 mol L−1 = 2.25 x 10−3 mol

iv) The number of moles of C6H5COO–(aq) in the solution


mol OH− added = mol C6H5COO−(aq) generated,

thus mol C6H5COO−(aq) in solution = 2.25 x 10−3 mol

AP Chemistry


Q11 contd. C6H5COOH(s) ⇋ C6H5COO–(aq) + H+

(aq) Ka = 6.46 x 10–5

a) v) The number of moles of C6H5COOH in the solution

1 point is earned for the correct molarity.

Ka = [C6H5COO-] [H+] / [C6H5COOH] so [C6H5COOH] = [C6H5COO-] [H+] / Ka

[C6H5COOH] = [2.25 x 10-3 mol / 0.040 L] [4.3 x 10-5 M] / 6.46 x 10-5

= 3.7 x 10-2 M


n = M x L thus, mol C6H5COOH = ( 3.7 x 10−2 M)(0.040 L) = 1.5 x 10−3 mol

b) State whether the solution at the equivalence point of the titration is acidic, basic, or neutral. Explain your reasoning.

1 point is earned for the prediction and the explanation.

At the equivalence point the solution is basic due to the presence of C6H5COO− (the conjugate base of the weak acid) that hydrolyzes to produce a basic solution as represented below.

C6H5COO− + H2O ⇄ C6H5COOH + OH−

c) In a different titration, a 0.7529 g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved in water and titrated with 0.150 M NaOH.

The equivalence point is reached when 24.78 mL of the base solution is added. Calculate each of the following

i) The mass, in grams, of benzoic acid in the solid sample

1 point is earned for the correct moles n = M x L

thus, mol OH- added = ( 0.150 M)(0.02478 L) = 3.72 x 10−3 mol = mol C6H5COOH

1 point is earned for the correct mass m = n x M

thus, mass C6H5COOH = ( 3.72 x 10-3)(122 g) = 0.453 g

ii) The mass percentage of benzoic acid in the solid sample

1 point is earned for the correct answer

% C6H5COOH = ( 0.453 g)(0.7529 g) x 100 = 60.2 %

AP Chemistry


12. CH3CH2COOH(aq) + H2O(l) ⇋ CH3CH2COO-(aq) + H3O

+(aq)

Propanoic acid, CH3CH2COOH, is a carboxylic acid that reacts with water according to the equation above. At 25°C the pH of a 50.0 mL sample of 0.20 M CH3CH2COOH is 2.79.

a) Identify a Brønsted-Lowry conjugate acid-base pair in the reaction. Clearly label which is the acid and which is the base.

1 point is earned for writing (or naming) either of the Brønsted-Lowry conjugate acid-base pairs with a clear indication of which is the acid and which is the base.

CH3CH2COOH and CH3CH2COO-

acid base OR H3O

+ and H2O acid base

b) Determine the value of Ka for propanoic acid at 25°C.

1 point is earned for correctly solving for [H3O+].

[H3O+] = 10-pH = 10-2.79 = 1.6 x 10-3 M

1 point is earned for the Ka expression for propanoic acid

Ka = [CH3CH2COO-][H3O+]/ [CH3CH2COOH]

1 point is earned for the Ka expression for propanoic acid

Ka = (1.6 x 10-3)2 / [0.20] = 1.3 x 10-5

c) For each of the following statements, determine whether the statement is true or false. In each case, explain the reasoning that supports your answer.

i) The pH of a solution prepared by mixing the 50.0 mL sample of 0.20 M CH3CH2COOH with a sample 50.0 mL 0.20 M NaOH is 7.00.

1 point is earned for noting that the statement is false AND providing a supporting explanation.

False. The conjugate base of a weak acid undergoes hydrolysis at equivalence to form a solution with a pH > 7.

ii) If the pH of a hydrochloric acid solution is the same as the pH of a propanoic acid solution, then the molar concentration of the hydrochloric acid solution must be less than the molar concentration of the propanoic acid solution.

1 point is earned for noting that the statement is true AND providing a supporting explanation.

True. HCl is a strong acid that ionizes completely. Fewer moles of HCl are needed to produce the same [H3O

+] as the propanoic acid solution, which only partially ionizes.

AP Chemistry


Q2 contd CH3CH2COOH(aq) + H2O(l) ⇋ CH3CH2COO-(aq) + H3O

+(aq)

A student is given the task of determining the concentration of a propanoic acid solution of unknown concentration.

A 0.173 M NaOH solution is available to use as the titrant. The student uses a 25.00 mL volumetric pipet to deliver the propanoic acid solution to a clean, dry flask. After adding an appropriate indicator to the flask, the student titrates the solution with the 0.173M NaOH, reaching the end point after 20.52 mL of the base solution has been added.

d) Calculate the molarity of the propanoic acid solution.

1 point is earned for correctly calculating the number of moles of acid that reacted at the equivalence point.

n = M x L thus, mol OH- = ( 0.173 M)(0.02052 L) = 3.55 x 10-3 mol = mol propanoic acid

1 point is earned for the correct molarity of acid.

M = n / L thus, molarity CH3CH2COOH- = (3.55 x 10-3 mol) / (0.02500 L) = 0.142 M

e) The student is asked to redesign the experiment to determine the concentration of a butanoic acid solution instead of a propanoic acid solution.

For butanoic acid the value of pKa is 4.83. The student claims that a different indicator will be required to determine the equivalence point of the titration accurately.

Based on your response to part b), do you agree with the student’s claim? Justify your answer.

1 point is earned for disagreeing with the student's claim and making a valid justification using pKa, Ka, pH arguments.

1 point is earned for numerically comparing either: the two pKa values, the two Ka values, or the two pH values at the equivalence point.

Disagree with the student's claim

From part b) above, pKa for propanoic acid is log(1.3 x 10-5) = 4.89.

Because 4.83 is so close to 4.89, the pH at the equivalence point in the titration of butanoic acid should be close enough to the pH in the titration of propanoic acid to make the original indicator appropriate for the titration of butanoic acid.

AP Chemistry


8.5 Acid-Base Titrations

AP Chemistry


AP Chemistry


pH Titrations - Strong Acid / Strong Base

Though titrations are usually done with indicators, we learn much more by monitoring pH throughout the course of a titration as shown in the arrangement below.

The features of these Titration Curves are as follows:

Volume added(mL)

Adding NaOH to 0.1 M HCl Adding HCl to 0.1 M NaOH

HCl(aq) + NaOH(aq) ⇨ NaCl(aq) + H2O(l) NaOH(aq) + HCl(aq) ⇨ NaCl(aq) + H2O(l)

0

Acid is 100% ionised so H+(aq) and Cl—

(aq) ions present in large quantities. [H+]=[Cl—]

pH = -log (0.1) = 1.0Small number of OH—

(aq) ions (10-13) present due to autoionisation of water.

Base is 100% ionised so Na+(aq) and OH—

(aq) ions present in large quantities. [OH—]=[Na+]

pOH = -log (0.1) = 1, pH = 13.0Small number of H+

(aq) ions (10-13) present due to autoionisation of water.

0 - 20[H+] drops to half - 0.05M (replaced by Na+

(aq))

pH = -log (0.05) = 1.30pH changes very little.

[OH—] drops to half - 0.05M (replaced by Cl—(aq))

pOH = -log (0.05) = 1.30, pH = 12.70pH changes very little.

36(90% completed)

[H+] drops to 10% - 0.01M (replaced by Na+(aq))

pH = -log (0.01) = 2.0Tenfold dilution cause pH to change by one.

[OH—] drops to 10% - 0.01M (replaced by Cl—(aq))

pOH = -log (0.01) = 2.0, pH = 12.0Tenfold dilution cause pH to change by one.

39.6(99% completed)

[H+] drops to 1% - 0.001M (replaced by Na+(aq))

pH = -log (0.001) = 3.0Tenfold dilution cause pH to change by one.

[OH—] drops to 1% - 0.001M (replaced by Cl—(aq))

pOH = -log (0.001) = 3.0, pH = 11.0Tenfold dilution cause pH to change by one.

39.96 (99.9% completed, 1 drop (0.05 mL)

away from completion

[H+] drops to 0.1% - 0.0001MpH = -log (0.0001) = 4.0

Tenfold dilution cause pH to change by one.

[OH—] drops to 0.1% - 0.0001MpOH = -log (10-4) = 4.0, pH = 10.0

Tenfold dilution cause pH to change by one.

AP Chemistry


Volume added(mL)

Adding NaOH to 0.1 M HCl Adding HCl to 0.1 M NaOH

HCl(aq) + NaOH(aq) ⇨ NaCl(aq) + H2O(l) NaOH(aq) + HCl(aq) ⇨ NaCl(aq) + H2O(l)

39.96 - 40.0(100% completed)

[H+] drops to 10-7 M (same as water)

pH = -log (10-7) = 7.0During last 0.04 mL (about 1 drop) the pH will change by at least 3 units - sometimes more, as slight excess OH— in last drop.

[OH—] drops to 10-7 M (same as water)

pOH = -log (10-7) = 7.0, pH = 7.0During last 0.04 mL (about 1 drop) the pH will change by at least 3 units - sometimes more, as slight excess H+ in last drop.

40(equivalence

point)

moles NaOH added = moles HCl originally

Concentration of acid solution could be calculated (if unknown) or checked (if known)

moles HCl added = moles NaOH originally

Concentration of base solution could be calculated (if unknown) or checked (if known)

> 40

Effectively adding NaOH to water so [OH—] will increase and pH will continue to rise.

Each addition of NaOH is having smaller effect than at start of titration due to larger volume of solution.

Tenfold increase in [OH—] cause pH to change by one.

Effectively adding HCl to water so [H+] will increase and pH will continue to fall.

Each addition of HCl is having smaller effect than at start of titration due to larger volume of solution.

Tenfold increase in [H+] cause pH to change by one.

Understanding the shape and behaviour of a titration curve is crucial to selecting the correct indicator for a particular titration.

The large pH changes around the equivalence point in a strong acid/strong base titration make it relatively easy to find suitable indicators.

You will learn about indicators in more detail in a later section, but the crucial thing is that it takes a pH change of 2 to ensure a complete change in colour.

For example, phenolpthalein is colourless at all pH's <8, but will be strong pink at pH > 10.

Phenolpthalein could be successfully used in a titration as described above but, to ensure a strong pink colour (pH 10), a slight excess of NaOH would be needed so the end-point detected could be slightly different from the true equivalence point. The difference, however, should be no more than a single drop from the burette, so may be acceptable.

AP Chemistry


Similarly, methyl red would complete it's colour change (4.5 - 6.5, yellow - red) slightly before the equivalence point. However, as previously shown, the last drop added from the burette will cause a massive pH change so end-point and equivalence point will probably still coincide.

Bromothymol blue would be ideal indicator as end-point and equivalence point should be same.

Rule of thumb: the pH of mid-point of the colour change and the pH of the equivalence point should be close as possible.

Remember that it is only for strong acid / strong base titrations that pH of equivalence point = 7.

pH Titrations involving Weak Acids and Weak Bases

There are a number of general differences between the curves involving a weak acid compared to a strong acid, and between a weak base compared to a strong base.

Strong Acid Weak Acid Weak Base Strong Base

e.g 0.1M HCl(aq) e.g 0.1M CH3COOH(aq) eg. 0.1M NH3(aq) e.g 0.1M NaOH(aq)

starts at low pH = 1, due to 100% ionisation

starts at higher pH ≃ 3, due to lower ionisation <1%

starts at lower pH ≃ 11, due to lower ionisation <1%

starts at high pH = 13, due to 100% ionisation

pH changes slowly until very close to equivalence

pH changes rapidly at first but then more slowly until very close to equivalence

pH changes rapidly at first but then more slowly until very close to equivalence


large region of pH change around equivalence - about 8 pH units

smaller region of pH change around equivalence - about 4 pH units



At equivalence, pH = 7 At equivalence, pH > 7 At equivalence, pH < 7 At equivalence, pH = 7

AP Chemistry


To avoid confusion, let's concentrate on explaining these observations for a Weak Acid only - but the same explanations would apply equally to a weak base.

Strong Acid Weak Acid

e.g 0.1M HCl(aq) e.g 0.1M CH3COOH(aq) , Ka = 1.8 x 10-5 , pKa = 4.74

starts at low pH = 1, due to 100% ionisation

pH = -log [H+] = -log(0.1) = 1

starts at higher pH ≃ 3, due to lower ionisation <1%

[H+] = √(KaC) = √(1.8 x 10-6) = 0.0013M, pH = 2.87


Tenfold dilution cause pH to change by one, so to raise pH from 1 to 2:

90% of 0.0025 moles of H+ (assuming 25 mL volume) would have to react - so 0.00225 moles of OH— needed to cause that change.

[OH—] = 0.1M so 22.5 ml needed to cause that change

(V = n/C, 0.00225/0.1 = 22.5 ml)

pH changes rapidly at first .......

This can be very hard to demonstrate mathematically -

need to allow for the fact that as the initial H+ ions present are consumed, more molecules will ionise to replace these H+ ions,

also need to allow for the effect of the CH3COO— ions (our conjugate base) forming.

However, starting with lower levels of H+ ions means that relatively smaller changes are needed to have the same impact on pH.

From graph, only about 3 ml of NaOH needed to cause similar change - an increase of 1 pH unit.

..... but then more slowly until very close to equivalence

Once titration has progressed sufficiently, the [CH3COO—] becomes very significant as a buffering effect is set up.

For a while, the calculation of pH is determined by the Henderson-Hasselbach equation (covered in detail later):

In Titration curves, the mid-point (at 12.5 mL in curve above) is extremely significant as half our acid has reacted, to be replaced by our conjugate base, so

[CH3COOH] = [CH3COO—]

The pH at this point will tell us the pKa of the weak acid

since [base] / [acid] = 1 and log 1 = 0



At equivalence, pH = 7

Our salt has no effect on the water equilibrium so

[H+] = [OH-] = 10-7 M and pH = 7

At equivalence, pH > 7

Our salt is a conjugate base, so it removes H+ ions from the water equilbrium so

[H+] < [OH-] < 10-7 M and pH > 7

AP Chemistry


Working with polyprotic acids (molecules that ionise in stages to release more than one H+ ion)can be difficult, so detailed calculations will not be required.

However, titration curves can still be useful and we can extract the following information:-

① the number of ionisation steps,

in this case two steps, so H2A initially

② the pKa values for each step

pKa1 = 2.34 for H2A ⟹ HA—

pKa1 = 9.68 for HA— ⟹ A—

③ the pH at the equivalence points

pH = 6.02 and pH = 11.85

AP Chemistry


8.5 Practice Problems1. B(aq) + HCl(aq) ⇋ HB+

(aq) + Cl—(aq)

The graph opposite shows the titration curve for an aqueous solution of a weak base, B , with HCl as the titrant. Based on the graph, which of the following best estimates the pKa of HB+ ?

A 12.0 B 10.8

C 6.0 D 1.8

2. B(aq) + HCl(aq) ⇋ HB+(aq) + Cl—

(aq)

The reaction of a weak base, B , with HCl is represented by the equation above. The graph above (Q1) shows the titration curve for 25.0mL of an aqueous solution of B titrated with 0.100MHCl.

Based on the graph, which of the following best estimates the initial concentration of the solution of the weak base?

A 0.20 M B 0.10 M C 0.067M D 0.048 M

3. A 40.0mLsample of the weak base C5H11N was titrated with 1.00 M HCl at 25°C.

Based on the resulting titration curve shown opposite, which of the following pairs provide the best estimates for the pKb and Kb of C5H11N ?

A pKb≈ 3.0 and Kb≈ 1× 10−3 B pKb≈5.8 and Kb ≈ 2 × 10−6

C pKb≈ 11.0 and Kb≈ 1× 10−11 D pKb≈ 12.5 and Kb≈ 3× 10−2

O

O

O

AP Chemistry


A solution of a weak monoprotic acid is titrated with a solution of a strong base, KOH.

For Q 4-6 that follow, choose from the points labeled (A) through (E) on the titration curve.

4. The point at which the concentrations of the weak acid and its conjugate base are approximately equal A A B B C C

D D E E

5. The point at which the moles of the added strong base are equal to the moles of the weak acid initially present A A B B C C D D E E

6. The point at which the pH is closest to that of the strong base being added A A B B C C D D E E

7. Data collected during the titration of a 20.0 mL sample of a 0.10 M solution of a monoprotic acid with a solution of NaOH of unknown concentration are plotted in the graph opposite.

Based on the data, which of the following are the approximate pKa of the acid and the molar concentration of the NaOH?

A pKa= 4.7 and [NaOH] = 0.050 M B pKa= 4.7 and [NaOH] = 0.10 M

C pKa= 9.3 and [NaOH] = 0.050 M D pKa= 9.3 and [NaOH] = 0.10 M

O

O

O

O

AP Chemistry


A 50.0 mL sample of an acid, HA, of unknown molarity is titrated, and the pH of the resulting solution is measured with a meter and graphed as a function of the volume of 0.100 M NaOH added.

For some of the questions that follow, choose from the points labeled (Q) through (U) on the titration curve.

8. At point R in the titration, which of the following species has the highest concentration? A HA B A— C H3O

+ D OH—

9. A student carries out the same titration, but uses an indicator instead of a pH meter. If the indicator changes color slightly past the equivalence point, what will the student obtain for the calculated concentration of the acid? A Slightly less than 0.0800 M B Slightly more than 0.0800 M

C Slightly less than 0.125 M D Slightly more than 0.125 M

10. Which of the following is the best particulate representation of the species (other than H2O) that are present in significant concentrations in the solution at point U in the titration?

A B C D

11. At which point on the titration curve is [A-] closest to twice that of [HA]? A R B S C T D U

O

O

O

O

AP Chemistry


12. An unknown acid is dissolved in 25 mL of water and titrated with 0.100 M NaOH.

The results are shown in the titration curve below.

Which of the following could be the unknown acid?

A Fluoroacetic acid, pKa = 2.6

B Glycolic acid, pKa = 3.8

C Propanoic acid, pKa = 4.9

D Hypochlorous acid, pKa = 7.5

E Boric acid, pKa = 9.3

13. A 0.35 g sample of Li(s) is placed in an Erlenmeyer flask containing 100 mL of water at 25°C.

A balloon is placed over the mouth of the flask to collect the hydrogen gas that is generated.

After all of the Li(s) has reacted with H2O(l), the solution in the flask is added to a clean, dry buret and used to titrate an aqueous solution of a monoprotic acid.

The pH curve for this titration is shown in the diagram above. On the basis of the pH curve, the pKa value of the acid is closest to A 4 B 5 C 8 D 12

O

O

AP Chemistry


14. To determine the concentration of a NaOH(aq) solution, a student titrated a 50. mL sample with 0.10 M HCl(aq).

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

The reaction is represented by the equation above.

The titration is monitored using a pH meter, and the experimental results are plotted in the graph opposite.

At the point labeled R on the pH curve, which of the following ions are present in the reaction mixture at a concentration greater than 0.01 M ?

A Na+ and Cl− only B Na+ , Cl− and H+ only

C Na+ , Cl− and OH− only D Na+ , Cl− , H+ and OH−

15. The graph opposite shows the titration curve that resulted when a sample of 0.1 M monoprotic acid was titrated with a solution of NaOH.

Based on the graph, the pKa of the acid is closest to

A 3.0 B 4.0 C 6.0

D 8.0 E 12.0

16. When a strong acid is titrated with a strong base using phenolphthalein as an indicator, the color changes abruptly at the endpoint of the titration and can be switched back and forth by the addition of only one drop of acid or base.

The reason for the abruptness of this color change is that

A a large change in pH occurs near the endpoint of the titration

B a buffer solution exists at the endpoint of the titration

C phenolphthalein is a strong proton donor

D the pH of water is very resistant to change

E phenolphthalein is much more sensitive to the pH of a solution than most other indicators

O

O

O

AP Chemistry


17. A student performs an acid-base titration and plots the experimental results in the graph opposite.

Which of the following statements best explains the experimental findings?

A A strong acid was titrated with a strong base, as evidenced by the equivalence point at pH = 7.

B A strong acid was titrated with a strong base, as evidenced by the equivalence point at pH > 7.

C A weak acid was titrated with a strong base, as evidenced by the equivalence point at pH > 7.

D A weak acid was titrated with a weak base, as evidenced by the equivalence point at pH approximately 7.

18. Refer to the following.



If equal volumes of the four acids at a concentration of 0.50 M are each titrated with a strong base, which will require the greatest volume of base to reach the equivalence point?

A Acid 1 B Acid 2 C Acid 3

D All the acids will require the same volume of base to reach the equivalence point.

O

O

AP Chemistry


8.5 Quick Check FRQ1. Methylamine is a weak base with the formula CH3NH2.

a) In the following box, complete the Lewis electron-dot diagram for a molecule of methylamine.

Show all bonding and nonbonding valence electrons.

The student’s drawing includes exactly 14 electrons; one pair of electrons between every two adjacent atoms plus one lone pair of electrons on the nitrogen atom.

Electron pairs can be represented either by dots, or line segments.

The following chemical equation represents the reaction that occurs when methylamine dissolves in water to form a basic solution.

CH3NH2(aq) + H2O(l) → CH3NH3+

(aq) + OH−(aq)

b) The pH of 2.65 M CH3NH2(aq) is 12.54. Determine the value of Kb for methylamine.

The response meets all of the following three criteria:

The correct calculation of [OH—]:

pH = 12.54 pOH = 14 - 12.54 = 1.46 [OH—]= 10-1.46 = 3.47 x 10-2 M

The use of the correct expression for Kb :

Since [CH3NH3+] = [OH—] and [CH3NH2] ≈ 2.65 M since % ionisation is low

Kb = [CH3NH3+][OH—] / [CH3NH2] = [OH—]2 / 2.65

The correct substitution of the numbers and calculation of the value:

Kb = (3.47 x 10-2)2 / 2.65 = 4.54 x 10-4

H H | |H—C—N—H | H

..

AP Chemistry


Q1 contd. A 25.00 mL sample of a CH3NH2(aq) solution of unknown concentration is titrated with 1.84 M HCl(aq).

Opposite is a graph that shows pH versus the volume of 1.84 M HCl(aq) added during the titration.

c) If 28.25 mL of 1.84 M HCl(aq) was required to reach the equivalence point, calculate the concentration of the CH3NH2(aq) solution of unknown concentration.

The response includes both of the following criteria:

The correct calculation of moles of HCl(aq) added:

n = C x L n = 1.84 x 0.02825 = 0.05198 mol HCl(aq)

The correct calculation of the concentration of the CH3NH2(aq) solution:

0.05198 mol HCl(aq) ⇒ 0.05198 mol CH3NH2(aq)

C = n / L C = 0.05198 / 0.02500 = 2.08 M

d) Using the symbols in the legend opposite, draw particles in the following beaker to represent the relative amounts of the two species, CH3NH2 and CH3NH3

+ , in the solution after the first 5.00 mL of titrant had been added to the CH3NH2(aq) solution.

The response shows more CH3NH2 particles than CH3NH3

+ particles

e) Explain why the titration curve shows only a small change in pH per volume of acid added when the total amount of acid added is about 14.0 mL.

Include a balanced chemical equation as part of your answer.


An explanation that, near the half equivalence point, the concentrations of the base and of its conjugate acid are approximately equal (the buffer zone). The effect on of adding more acid to this system is partially counteracted because the added acid reacts with the base in the solution.

The balanced equation: H3O+ + CH3NH2 ⇋ CH3NH3

+ + H2O

AP Chemistry


2. HC2H3O2(aq) + H2O(l) ⇋ H3O+

(aq) + C2H3O2−

(aq)

The dissociation of ethanoic acid, HC2H3O2(aq), is represented above. A student is given the task of determining the value of Ka for HC2H3O2(aq) using two different experimental procedures.

a) The student is first asked to prepare 100.0 mL of 0.115 M HC2H3O2 (aq) using a 2.000 M standard solution.

i) Calculate the volume, in mL, of 2.000 M HC2H3O2(aq) the student needs to prepare 100.0 mL of 0.115 M HC2H3O2(aq).

1 point is earned for the correct volume.

Mi Vi = Mf Vf Vi = (0.115 M)(100.0 mL)/ 2.000 M = 5.75 mL

ii) Describe the procedure the student should use to prepare 100.mL of 0.115 M HC2H3O2(aq) using appropriate equipment selected from the list below.

Assume that the student uses appropriate safety equipment.

100 mL beaker 100 mL graduated cylinder 100 mL volumetric flask

Eye dropper 500 mL wash bottle filled with distilled water

2.000 M HC2H3O2(aq) in a 50 mL burette

1 point is earned for dispensing from the burette.

Use the burette to deliver 5.75 mL of 2.000 M HC2H3O2 to the 100 mL volumetric flask.

1 point is earned for diluting the solution to the calibration mark of the volumetric flask.

Then add distilled water from the wash bottle to the flask (adding the last few drops with an eyedropper) until the volume of liquid in the flask is at the calibration mark

AP Chemistry


Q2 contd b) Using a pH probe, the student determines that the pH of 0.115 M HC2H3O2(aq) is 2.92.

i) Using the pH value, calculate the value of Ka for HC2H3O2(aq).

1 point is earned for correct conversion of pH to [H3O+].

pH = 2.92 ⇒ [H3O+] = 10-2.92 = 0.0012 M

1 point is earned for a value of Ka consistent with the student’s value of [H3O

+].

Ka = [ H3O+][C2H3O2

-]/ [HC2H3O2]

Since [H3O+] = [C2H3O2

-], then

Ka = (0.0012)(0.0012)/(0.115−0.0012) = (0.0012)2/(0.114) =1.3 × 10−5

ii) Calculate the percent dissociation of ethanoic acid in 0.115 M HC2H3O2(aq).

1 point is earned for the correct percent dissociation.

Percent dissociation = [C2H3O2−]/[HC2H3O2] x 100

= (0.0012 / 0.115) ×100 = 1.0%

AP Chemistry


Q2 contd

In a separate experimental procedure, the student titrates 10.0 mL of the 2.000 M HC2H3O2(aq)with an NaOH(aq) solution of unknown concentration.

The student monitors the pH during the titration. The titration curve opposite was created using the experimental data presented in the table.

c) Write the balanced net ionic equation for the reaction that occurs when HC2H3O2(aq) and NaOH(aq) are combined.

1 point is earned for the correct equation.

HC2H3O2(aq) + OH-(aq) → C2H3O2

-(aq) + H2O(l)

d) Calculate the molar concentration of the NaOH(aq) solution.

1 point is earned for determining the moles of acid.

From the pH curve, the equivalence point occurs at 14.0 mL.

10 mL× 2.000 mol HC2H3O2/1000 mL =0.0200 mol HC2H3O2(aq)

1 point is earned for determining the molar concentration of the base.

0.0200 mol HC2H3O2(aq) × 1 mol NaOH/1 mol HC2H3O2 =0.0200 mol NaOH

0.0200 mol NaOH/ 0.0140 L solution =1.43 M NaOH(aq)

e) Explain how the student can estimate the value of Ka for HC2H3O2(aq) using the titration curve.

1 point is earned for a correct explanation (numerical explanation not required).

At the half-equivalence point (~7.0 mL) the pH of the solution is equal to the pKa of the acid. The antilog of the negative pH is equal to the value of Ka.

AP Chemistry


8.6 Molecular Structure of Acids and Bases

AP Chemistry


Warning: There is no easy 'single explanation' for the effect that molecular structure has on the relative strength of acids and bases. There are however, a number of 'guiding principles' that should be considered.

'Suitable' Protons

In general, only hydrogen's attached to atomswith a large electronegativity are going to havea tendency to ionise when dissolved in water.

Hδ+—Aδ- ⇋ H+(aq) + A—

(aq)

Stability of Conjugate

Both ethanol and ethanoic acid have a very polarH—O bond, but their acid properties are extremelydifferent - in fact, ethanol is not usually considered an acid.

The difference is due to the ability ofthe ethanoate ion to be stabilised dueto resonance structures which caneffectively share the charge over a groupof atoms.

However, the weakness of ethanol as an acid means that the conjugate base, the ethoxide ion, is considered a strong base. Here, the concentrated nature of the negative charge is an advantage.

The presence of nearby electronegative atoms can also help to 'share' the negative charge, which increases the stability of the conjugate ion and makes it easier for the molecule to release the H+ ion.

trichloroacetic acid dichloroacetic acid monochloroacetic acid Ka = 2.2 x 10-1 Ka = 4.5 x 10-2 Ka = 1.3 x 10-3

This is referred to as the inductive effect - where an electronegative atom draws electrical charge towards it.

Whilst it can help make weak acids stronger, it will have the opposite effect on bases, such as NH3, which rely on their lone pair of electrons to help accept H+ions.

AP Chemistry


Ethanoic acid benefits from the extra stability that comes from the neighbouring oxygen in the C=O group and the ability to set up resonance structures.

Stronger acids benefit from having a central atom that is more electronegative than a carbon atom. However, the ability to set upresonance structures remains crucial.

This can be demonstrated by looking at the various oxyacids formed with chlorine.

The more resonance structures - the more the conjugate base can be stabilised - the stronger the acid.

Nitrous acidKa = 7.2 x 10-4

AP Chemistry


Similarly, the strength of bases can be effected by the stability of the conjugate acid that forms. For ammonia and amines, the positive charge on the conjugate acid must be stabilised.

Kb = 1.8 × 10−5 Kb = 4.4 × 10−4 Kb=5.4×10−4 Kb = 6.3 × 10−5

Groups, such as alkyl groups, that attract electrons less strongly than hydrogen are said to have + I (positive inductive) effect (electron repelling or electron releasing).

This helps stabilise the positive charge on the ammonium ions.

the longer the alkyl chain the more electron releasing the more alkyl chains the more electron releasing

AP Chemistry



AcidHIOHBrOHClO

Each particle diagram shown is a representation of an aqueous solution of one of the acids listed in the table. The molarity of the acids in the solutions is the same. Based on the information, which particle diagram best corresponds to the indicated acid?

A Diagram 1 corresponds to HClO B Diagram 2 corresponds to HClO

C Diagram 2 corresponds to HIO. D Diagram 3 corresponds to HBrO

2. Which of the following dissolves in water to form an acidic solution

A CH3CH2CH2CH3 B CH3CH2CH2CH2OH

C CH3COCH3 D CH3COOH E CH3CH2CH2NH2

3.Base CH3NH2 NH2NH2 HONH2 ClNH2

Conjugate Acid

The table above includes the Lewis diagrams of the conjugate acids of some weak nitrogenous bases. Based on the relative stability of the conjugate acids, which of the following bases has the largest value for Kb ?

A CH3NH2 B NH2NH2 C HONH2 D ClNH2

O

O

O

AP Chemistry


4.

Structure

Kb 5.7 × 10−10 1.5 × 10−11 7.9 × 10−12

The table above provides the chemical structures for weak bases and their ionization constants, Kb. Based on the data, which of the following provides the best reason for the trend in base strengths?

A The number of hydrogen atoms B The number of resonance structures

C The different electronegativities of D The different molar masses H, I, and Br

5. The Lewis electron-dot diagrams of the HClO3 molecule and the HClO2 molecule are shown opposite at the left and right, respectively.

Which of the following statements identifies the stronger acid and correctly identifies a factor that contributes to its being the stronger acid?

A HClO3(aq) is the stronger acid because its molecules experience stronger London dispersion forces.

B HClO3(aq) is the stronger acid because the additional electronegative oxygen atom on the chlorine atom stabilizes the conjugate base.

C HClO2(aq) is the stronger acid because its molecules experience weaker London dispersion forces.

D HClO2(aq) is the stronger acid because the lone pairs of electrons on the chlorine atom stabilize the conjugate base.

6. Which of the following is a weak acid in aqueous solution?

A HCl B HClO4 C HNO3 D H2S E H2SO4

O

O

O

AP Chemistry


8.6 Quick Check FRQ1. Answer the following questions that relate to the chemistry of halogen oxoacids.

a) Use the information in the table below to answer part i).

i) Which of the two acids is stronger, HOCl or HOBr ?

Justify your answer in terms of Ka .

HOCl is the stronger acid because its Ka value is greater than the Ka value of HOBr.

ii) Draw a complete Lewis electron-dot diagram for the acid that you identified in part a) i).

1 point is earned for a correct diagram.

iii) Hypoiodous acid has the formula HOI. Predict whether HOI is a stronger acid or a weaker acid than the acid that you identified in part a) i).

Justify your prediction in terms of chemical bonding.

1 point is earned for predicting that HOI is a weaker acid than HOCl and stating that iodine has a lower electronegativity than chlorine and EITHER

stating that this results in a stronger O–H bond in HOI OR stating that this decreases the stability of the OI– ion in solution OR The conjugate base OCl— is more stable than OI— because Cl, being more electronegative, is better able to accommodate the negative charge.

b) Write the equation for the reaction that occurs between hypochlorous acid and water.

HOCl + H2O ⇄OCl− + H3O+ OR HOCl ⇄ OCl− + H+

H O Cl.. .. ... .. .

. .. .

AP Chemistry


8.7 pH and pKa

AP Chemistry


Relationships Between pH and pKa

Consider a weak acid,

HA(aq) + H2O(l) ⇋ H3O+

(aq) + A—(aq)

The equilibrium constant is simplified tothe acid-dissociation constant, Ka ,

Ka = [H+] [A—] [HA]

To help see the relationship between pH andpKa we need to take logs of both sides and 'tidy up', so: pH = pKa + log ([A—]/ [HA]) or pH = pKa + log ([base]/ [acid])

When examining the pH curve for a weak acid, the 'mid point' had special significance as it identified the point at which [acid] = [base] and pH = pKa

It is relatively simple to establish other relationships;

If pH < pKa then log ([base]/ [acid]) must be negative so [acid] > [base]

If pH = pKa then log ([base]/ [acid]) must be 0 so [acid] = [base]

If pH > pKa then log ([base]/ [acid]) must be positive so [base] < [acid]

Quick 'rule of thumb' ;

lower pH favours acid so [acid] > [base]

same pH favours neither so [acid] = [base]

higher pH favours base so [base] > [acid]

Or, from first principles; HA(aq) + H2O(l) ⇋ H3O+

(aq) + A—(aq)

lower pH means [H3O+] ⬀ so reverse reaction favoured so [acid] ⬀

higher pH means [H3O+] ⬂ so forward reaction favoured so [base] ⬀

knowing pKa , we can predict [acid]: [base] relationships at any pH

mid point

AP Chemistry


pH = pKa + log ([A—]/ [HA]) or pH = pKa + log ([base]/ [acid])

Finally, it can be worthwhile putting some more detail onto some of this.

For example,

pH = pKa + 1 means [base]/ [acid] = 101 10 times more base than acid ~ 91% base, ~9% acid in mixture

pH = pKa + 2 means [base]/ [acid] = 102 100 times more base than acid ~ 99% base, ~1% acid in mixturesimilarly,

pH = pKa - 1 means [base]/ [acid] = 10-1 10 times less base than acid ~ 9% base, ~99% acid in mixture

pH = pKa - 2 means [base]/ [acid] = 10-2 100 times less base than acid ~ 1% base, ~99% acid in mixture

relatively small changes in pH can dramatically effect equilibrium mixture

Understanding Indicators

Indicators are mixtures containing a conjugate pair of molecules - molecules whose structures differ by a H+ ion.

One molecule is a weak acid (HA) while the other molecule is the conjugate base (normally A—).

So, HA(aq) + H2O(l) ⇋ H3O

+(aq) + A—

(aq) and pH = pKa + log ([A—]/ [HA])

However, the structure of indicator molecules are more complex and the changes that occur when donating a H+ ion can be more complicated. For example, methyl red

In— HIn yellow red

has an acid form (HIn - red) that can donate a H+ ion , and a base form (In— - yellow) that can accept a H+ ion. Though the base form is not actually a negative ion, we treat indicators as identical to weak acids, but change our labels.

So, HIn(aq) + H2O(l) ⇋ H3O

+(aq) + In—

(aq) and pH = pKIn + log ([In—]/ [HIn])

AP Chemistry


Phenolpthalein is even more complicated but we can simplify things if we consider it to be a mixture of the main two forms and ignore the fact that there are 2 H+ ions involved.

HIn(aq) ⇋ H+

(aq) + In—(aq)

colourless pink

and

pH = pKIn + log ([In—]/ [HIn])

Methyl red has a pKIn = 5.1.

At low pH, the acid form (HIn) will dominate and we will see a red colour. HIn - red At pH = 5.1 (the mid point), [HIn] = [In—] so red and yellow will be equal and we see an orange colour.

At high pH, the base form (In—) will dominate and we will see a In— - yellow yellow colour.

Studies have shown that, in a mixture, if one colour is ten times stronger than the other colour then our eyes will only detect the dominant colour.

So either, [In-] / [HIn] = 10 or [In-] / [HIn] = 0.1 log ([In-] / [HIn]) = 1 or log ([In-] / [HIn]) = -1 pH = pKIn + 1 or pH = pKIn - 1

yellow red indicators completely change colour over 2 pH units, pKIn ± 1

AP Chemistry


Notice that it is not strictly true to say that an indictor, for example - methyl red, is 'red in acid' and 'yellow in base' since both colours are seen in solutions with pH < 7.

It is safer to say 'red at low pH's' and 'yellow at higher pH's'.

Choosing Indicators

Indicators should be chosen to change colouras close as possible to the equivalence point.

With a strong acid/strong base titration, theequivalence point is pH = 7.

With a pKIn = 7.1, bromothymol blue is theobvious indicator as it will change colour inthe pH range 6 - 8.

However, in a strong acid/strong base titration, the final drop will cause a change of about 6 pHunits from about pH = 3 to pH = 10.

This means that indicators such as methyl red (pKIn = 5.1, pH range 4 - 6) and phenolpthalein (pKIn = 9.3, pH range 8 - 10) could also be suitable, as they would change colour during this final drop. The phenolpthalein may not have completed the change to pink but, in practice, the end -point is determined as the point where the first sign of pink appears - no matter how faint.

However, the slight excess of NaOH needed may mean that end-point > equivalence point.

Methyl red would be considered better, as it completes it's colour change only slightly before the equivalence point.

AP Chemistry


Changing the direction of a titration, can alter the choiceof indicator, even though the strong base/strong acid titration, has the same equivalence point of pH = 7.

With a pKIn = 7.1, bromothymol blue remains the obvious indicator as it will change colour in the pH range 8 - 6.

However, phenolpthalein (pKIn = 9.3, pH range 10 - 8) wouldnow be the second best choice as it would complete its colourchange (end-point) just before the equivalence point.

An indicator should have a pKIn as close as possible to the pH of the equivalence point or should complete its colour change just before the equivalence point.

For the weak base/strong acid titration aboveleft, the equivalence point is at pH < 7 (~5.5).

methyl red (pKIn = 5.1, pH range 6 - 4) is the ideal indicator.

For the weak acid/strong base titration aboveright,the equivalence point is at pH > 7 (~8.5).

phenolpthalein (pKIn = 9.3, pH range 8 - 10) isthe ideal indicator.

As acids (or bases) get weaker, the pH changearound the equivalence point becomes smallerand it will become impossible to find a suitable indicator.

AP Chemistry


8.7 Practice Problems1. A student pours a 10.0mL sample of a solution containing HC2H3O2 (pKa = 4.8) and NaC2H3O2 into a test tube. The student adds a few drops of bromocresol green to the test tube and observes a yellow color, which indicates that the pH of the solution is less than 3.8.

Based on this result, which of the following is true about the relative concentrations of HC2H3O2 and NaC2H3O2 in the original solution?

A [HC2H3O2] > [NaC2H3O2 ]

B [HC2H3O2] = [NaC2H3O2 ]

C [HC2H3O2] <[NaC2H3O2 ]

D The relative concentrations cannot be determined without knowing the value of pKb for C2H3O2

−.

2. Benzoic acid, HC7H5O2 , has a pKa of 4.20 and a molar solubility of 0.0278M.

Sodium benzoate, NaC7H5O2 , has a molar solubility of 4.16M.

Several 50.0mL samples of 2.00 M NaC7H5O2(aq) are treated with 3.00M HCl, the pH is recorded, and any solid crystals are filtered, dried, and weighed. The data from the experiment are given in the table above. Which of the following best explains the experimental results?

A When pH < pKa , [HC7H5O2] > [NaC7H5O2] and HC7H5O2(s) precipitates from the solution.

B When pH < pKa , [NaC7H5O2] > [HC7H5O2] and NaC7H5O2(s) precipitates from the solution.

C When pH > pKa , [HC7H5O2] > [NaC7H5O2] and HC7H5O2(s) precipitates from the solution.

D When pH > pKa , [NaC7H5O2] > [HC7H5O2] and NaC7H5O2(s) precipitates from the solution.

O

O

AP Chemistry


3. H3N+CH2COOH(aq) + H2O(l) ⇋ H3N

+CH2COO—(aq) + H3O

+(aq)

H3N+CH2COO—

(aq) + H2O(l) ⇋ H2NCH2COO—(aq) + H3O

+(aq)

The stepwise dissociation of the amino acid glycine is represented by the chemical equations above. A student titrates a sample of glycine dissolved in dilute acid with 0.100 M NaOH(aq) . The data are plotted on the following graph.

Based on the data, which of the following species has the highest concentration in an aqueous solution of glycine with a pH of 7?

A H3N+CH2COOH(aq) B H3N

+CH2COO—(aq)

C H2NCH2COO—(aq) D H2NCH2COOH(aq)

O

AP Chemistry


4. The graph below shows the titration curve that results when 100. mL of 0.0250 M acetic acid is titrated with 0.100 M NaOH.

Which of the following indicators is the best choice for this titration?

Indicator pH Range of Color Change

A Methyl orange 3.2 - 4.4

B Methyl red 3.2 - 4.4

C Bromothymol 6.1 - 7.6

D Phenolphthalein 8.2 - 10.0

E Alizarin 11.0 - 12.4O

AP Chemistry


8.7 Quick Check FRQ1. The indicator HIn is a weak acid with a pKa value of 5.0. It reacts with water as represented in the equation below.

Consider the two beakers opposite. Each beaker has a layer of colorless oil (a nonpolar solvent) on top of a layer of aqueous buffer solution.

In beaker X the pH of the buffer solution is 3, and in beaker Y the pH of the buffer solution is 7.

A small amount of HIn is placed in both beakers. The mixtures are stirred well, and the oil and water layers are allowed to separate.

a) What is the predominant form of HIn in the aqueous buffer in beaker Y, the acid form or the conjugate base form? Explain your reasoning.

1 point is earned for correctly identifying In–(aq) as the predominant form in the

aqueous layer of beaker Y because the solution is not acidic (may be implicit).

1 point is earned for stating that pH > pKa and that this causes the equilibrium to favor products.

The conjugate base form, In–(aq), is the predominant form of the indicator in the

aqueous pH 7 buffer in beaker Y. This is because the pH is greater than the pKa of HIn, causing the equilibrium to form a significant amount of products, In–

(aq) and H3O

+(aq).

b) In beaker X the oil layer is yellow, whereas in beaker Y the oil layer is colorless. Explain these observations in terms of both acid-base equilibria and interparticle forces.

1 point is earned for explaining the yellow color in the oil layer of beaker X in terms of acid-base equilibrium and interparticle forces between HIn molecules and oil molecules.

At pH 3 the acid form, HIn(aq), predominates in the aqueous layer of beaker X because pH < pKa. Since HIn(aq) is a neutral molecule, some of it can dissolve in the oil layer of beaker X because of Londondispersion interactions with the oil, causing the oil layer to be yellow.

1 point is earned for explaining the colorless oil layer of beaker Y in terms of interparticle forces between In– ions and water molecules.

Since In–(aq) is charged, it will preferentially dissolve in the aqueous layer of beaker Y

because of ion dipole interactions with the water, leaving the oil layer colorless.

AP Chemistry


8.8 Properties of Buffers

Weak Acids as Buffers

Normally, a solution of a weak acid is mainly acid molecules with very few of the conjugate base ions present in the mixture. For example, ethanoic acid, HC2H3O2

HC2H3O2(aq) + H2O(l) ⇋ H3O+

(aq) + C2H3O2—

(aq) Ka = 1.8 x 10-5

The addition of extra conjugate base ions (eg sodium ethanoate, NaC2H3O2) or enough NaOH to react about half of the acid (mid-point) transforms the mixture into a buffer solution.

The large number of acid molecules will be able to respond to the addition of more OH— ions:-

HC2H3O2(aq) + OH—(aq) ⟹ H2O(l) + C2H3O2

—(aq)

The large number of conjugate base ions will be able to respond to the addition of more H+ ions:-

C2H3O2—

(aq) + H3O+

(aq) ⟹ H2O(l) + HC2H3O2(aq)

A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

AP Chemistry


Weak Bases as Buffers

Normally, a solution of a weak base is mainly base molecules with very few of the conjugate acid ions present in the mixture. For example, dimethyl amine, (CH3)2NH

(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH2+

(aq) + OH−(aq) Kb = 5.4 × 10−4 at 25°C

The addition of extra conjugate acid ions (eg dimethyl ammonium chloride, (CH3)2NH2Cl) or enough HCl to react about half of the base (mid-point) transforms the mixture into a buffer solution.

The large number of base molecules will be able to respond to the addition of more H+ ions:-

(CH3)2NH(aq) + H+(aq) ⟹ (CH3)2NH2

+(aq)

The large number of conjugate acid ions will be able to respond to the addition of more OH— ions:-

(CH3)2NH2+

(aq)+ OH—(aq) ⟹ H2O(l) + (CH3)2NH(aq)


Polyprotic Buffers - Amphoteric Buffers

H2PO4— is the buffer so this Ka for calculations

Polyprotic acids such as phosphoric acid, H3PO4, ionise to form the conjugate base H2PO4—.

However, this ion is also an acid and will go on to form its own conjugate base, HPO42-.

A substance that can act as both an acid and a base is described as amphoteric. Water is another amphoteric molecule. (Al2O3 and some transition metal oxides are amphoteric - react with both acids and bases).

A mixture with large quantities of an amphoteric molecule and conjugate can also act as a buffer.

The large number of H2PO4— ions will be able to

respond to the addition of more H+ ions:- H2PO4—

(aq) + H+(aq) ⟹ H3PO4(aq)

The large number of H2PO4— ions will be able

to respond to the addition of more OH— ions:- H2PO4—

(aq)+ OH—(aq) ⟹ H2O(l) + HPO4

2-(aq)


AP Chemistry



A buffer solution is made up of acetic acid (CH3COOH) and sodium acetate (NaCH3COO). The major equilibria in the buffer system are represented above.

Which of the following equilibria could be used to support the claim that the addition of a small amount of NaOH to the buffer will result in only a very small change in pH ?

A CH3COOH(aq) + CH3COO−(aq) ⇄ CH3COO−

(aq) + CH3COOH(aq)

B CH3COOH(aq) + OH−(aq) ⇄ CH3COO−

(aq) + H2O(l)

C H3O+

(aq) + OH−(aq) ⇄2H2O(l)

D CH3COOH(aq) + H2O(l) ⇄ H3O+

(aq) + CH3COO−(aq)

2. HF(aq) + H2O(l) ⇄ H3O

+(aq) + F−

(aq) Ka = 6.3 ×10−4 at 25°C

The acid ionization equilibrium for HF is represented by the chemical equation above.

A student claims that the pH of a solution that contains 0.100 M HF(aq) and 0.100 MNaF(aq) will change only slightly when small amounts of acids or bases are added.

Which of the following pairs of equations can the student use to justify the claim?

A HF(aq) + OH−(aq) → H2O(l) + F−

(aq) and OH−(aq) + H3O

+(aq) → 2 H2O(l)

B H3O+

(aq) + OH−(aq) → 2 H2O(l) and F−

(aq) + H3O+

(aq) → HF(aq)+H2O(l)

C HF(aq) + OH−(aq) → H2O(l) + F−

(aq) and F−(aq) + H2O(l) → HF(aq) + OH−

(aq)

D HF(aq) + OH−(aq) → H2O(l) + F−

(aq) and F−(aq) + H3O

+(aq) → HF(aq)+ H2O(l)

O

O

AP Chemistry


3. The equilibrium for the reaction between (CH3)2NH, a weak base, and water is represented by the equation below. The table shows the pH of three solutions of (CH3)2NH(aq) at 25°C .

(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH2+

(aq) + OH−(aq) Kb = 5.4 × 10−4 at 25°C

A student mixes 100.mL of 0.200 M (CH3)2NH(aq) with 100.mL of 0.200 M (CH3)2NH2Cl(aq) and claims that if a small amount of strong base is added to the mixture, then the resulting change in pH of the mixture will be smaller than the change in pH that would result from adding the same amount of strong base to 200.mL of 0.200 M (CH3)2NH(aq) .

Which of the following best explains whether or not the student’s claim is correct?

A The claim is correct because the mixture contains only half of the (CH3)2NH(aq) that the 0.200 M (CH3)2NH(aq) contains. The lesser amount of base in the mixture makes the pH of the mixture lower than the pH of the 0.200 M (CH3)2NH(aq) ; therefore, when a small amount of strong base is added, the change in the pH of the mixture will be smaller than the change in the pH of the 0.200 M (CH3)2NH(aq).

B The claim is correct because the mixture contains a significant amount of the acid (CH3)2NH2

+(aq), which can react with and partially neutralize the added strong

base, thereby reducing the change in pH . In 0.200 M (CH3)2NH(aq), the concentration of the acid (CH3)2NH2

+(aq) is very small; therefore, the added strong

base is not neutralized and the change in pH will be larger.

C The claim is incorrect because the mixture contains only half of the (CH3)2NH(aq) that the 0.200 M (CH3)2NH(aq) contains. The lesser amount of base in the mixture makes the pHof the mixture lower than the pH of the 0.200 M (CH3)2NH(aq); therefore, when a small amount of strong base is added, the change in the pH of the mixture will be larger than the change in the pH of the 0.200 M (CH3)2NH(aq).

D The claim is incorrect because the change in pH resulting from the addition of a small amount of strong base to the mixture and to the 0.200 M (CH3)2NH(aq)

will be the same given that the same amount of strong base is being added to the same volume of solution.

O

AP Chemistry


4.



A 25 mL sample of a 1.0 M solution of acid 1 is mixed with 25 mL of 0.50 M NaOH. Which of the following best explains what happens to the pH of the mixture when a few drops of 1.0 M HNO3 are added?

A The pH of the mixture increases sharply, because HNO3 is a strong acid.

B The pH of the mixture decreases sharply, because H3O+ ions were added.

C The pH of the mixture stays about the same, because the conjugate base of acid 1 reacts with the added H3O

+ ions.

D The pH of the mixture stays about the same, because the OH− ions in the solution react with the added H3O

+ ions.

5. A solution is prepared by adding 100 mL of 1.0 M HC2H3O2(aq) to 100 mL of 1.0 M NaC2H3O2(aq). The solution is stirred and its pH is measured to be 4.73.

After 3 drops of 1.0 M HCl are added to the solution, the pH of the solution is measured and is still 4.73. Which of the following equations represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH?

A H3O+

(aq) + OH—(aq) → 2 H2O(l)

B H3O+

(aq) + Cl—(aq) → HCl(g) + H2O(l)

C H3O+

(aq) + C2H3O2—

(aq) → HC2H3O2(aq) + H2O(l)

D H3O+

(aq) + HC2H3O2(aq) → H2C2H3O2+

(aq) + H2O(l)

O

O

AP Chemistry


6. Mixtures that would be considered buffers include which of the following?

I. 0.10 M HCl + 0.10 M NaCl II. 0.10 M HF + 0.10 M NaF

III. 0.10 M HBr + 0.10 M NaBr

A I only B II only C III only D I & II only E II & III only

7. The graph below shows the titration curve that results when 100. mL of 0.0250 M acetic acid is titrated with 0.100 M NaOH.

What part of the curve corresponds to the optimum buffer action for the acetic acid/acetate ion pair?

A point V

B point X

C point Z

D along all of section WY

E along all of section YZ

O

O

AP Chemistry


8.8 Quick Check FRQ1. Methylamine is a weak base with the formula CH3NH2 .

a) In the box opposite, complete the Lewis electron-dot diagram for a molecule of methylamine.

Show all bonding and nonbonding valence electrons.

The following chemical equation represents the reaction that occurs when methylamine dissolves in water to form a basic solution.

CH3NH2(aq) + H2O(l) → CH3NH3+

(aq) + OH−(aq)

b) The pH of 2.65 M CH3NH2(aq) is 12.54. Determine the value of Kb for methylamine.

1 point: correct calculation of [OH—], pH = 12.54 so pH = 1.46, [OH—] = 10-1.46 = 3.47 x 10-2 M

1 point: use of the correct expression for Kb Kb = [CH3NH3+][OH—] / [CH3NH2]

1 point: correct substitution and calculation Kb = [OH—]2 / 2.65 = [3.47 x 10-2]2 / 2.65 = 4.54 x 10-4

A 25.00mL sample of a CH3NH2(aq) solution of unknown concentration is titrated with 1.84 M HCl(aq) . Following is a graph that shows pH versus the volume of 1.84 M HCl(aq) added during the titration.

|

| |— — —..

AP Chemistry


Q1 contd.

c) If 28.25mL of 1.84 M HCl(aq) was required to reach the equivalence point, calculate the concentration of the CH3NH2(aq) solution of unknown concentration.

1 point: The correct calculation of moles of added: n = C x L = 1.84 x 0.0285 = 0.05198 moles

1 point: The correct calculation of concentration : equivalence, so 0.05198 mol HCl = 0.01598 mol CH3NH2

C = n / L = 0.01598 / 0.025 = 2.08 M

d)

Using the symbols in the legend above, draw particles in the beaker opposite to represent the relative amounts of the two species, CH3NH2 and CH3NH+

3, in the solution after the first 5.00mL of titrant had been added to the CH3NH2(aq) solution.

e) Explain why the titration curve shows only a small change in pH per volume of acid added when the total amount of acid added is about 14.0mL. Include a balanced chemical equation as part of your answer.


An explanation that, near the half equivalence point, the concentrations of the base and of its conjugate acid are approximately equal (the buffer zone). The effect on of adding more acid to this system is partially counteracted because the added acid reacts with the base in the solution.

The balanced equation H3O+ + CH3NH2 → H2O + CH3NH3

+

The response shows more CH3NH2 particles than

CH3NH3+ particles.

AP Chemistry


8.9 Henderson-Hasselbalch Equation

AP Chemistry


Calculating the pH of a Buffer

This is where we will utilise the Henderson-Hasellbach equation:

pH = pKa + log ( [base] [acid] )

'Traditionally', a buffer is often prepared sothat [acid] = [base] (mid-point)

With this mixture pH = pKa + log(1) = pKa

If [acid] > [base] then pH < pKa

If [base] > [acid] then pH > pKa

Eventually, if [acid] >> [base] or [base] >> [acid]the mixture will lose it's ability to function as a buffer.

This usually corresponds with a tenfold difference will mean that the Buffering region is effectively pKa ± 1

Example 1: What is the pH of a buffer that is 0.12 M in lactic acid [CH3CH(OH)COOH, or HC3H5O3] and 0.10 M in sodium lactate [CH3CH(OH)COONa, or NaC3H5O3]? For lactic acid, Ka = 1.4 x 10-4

Step1: Before we can use the Henderson-Hasellbach equation, we need to convert Ka to pKa

pKa = -log Ka = -log (1.4 x 10-4) = 3.85

(BTW. In log calculations the first number sets the 'decimal place' or 'power'. The numbers coming after are the 'true value' and would be to the same sig figs as the original number. So 3.85 is the correct number, even though it looks like we have gained a significant figure.)

Step2: Substitute values into the Henderson-Hasellbach equation.

pH = 3.85 + log (0.1 / 0.12) = 3.77

A lot of the time, questions require a more qualitative approach, in that

add extra base, pH > calculated pH add extra acid, pH < calculated pH

and initial calculation is for [base] = [acid] so 'calculated pH' = pKa

AP Chemistry


Calculating the pH of a Weak Base Buffer

The Henderson-Hasellbach equation can be easily modified to work for a buffer formed from a weak base and its conjugate acid, but, this time, we would be calculating pOH using pKb.

For example, B(aq) + H2O(l) ⇄BH+(aq) + OH—

(aq)

pOH = pKb + log ( [base] notice that the original ratio, [base]:[acid] remains [acid] )

Once pOH has been calculated, pH = 14 - pOH

Example 2: (CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH2

+(aq) + OH−

(aq) Kb = 5.4 × 10−4 at 25°C

Calculate the pH of a solution at 25°C that is 0.100 M (CH3)2NH(aq) and 0.100 M (CH3)2NH2Cl(aq) ?

Step1: Before we can use the modified Henderson-Hasellbach equation, we need to convert Kb to pKb

pKb = -log Kb = -log (5.4 x 10-4) = 3.27

Step2: Substitute values into the modified Henderson-Hasellbach equation.

pOH = 3.27 + log (0.1 / 0.1) = 3.27

Step3: Convert to pH

pH = 14 - pOH = 14 - 3.27 = 11.73

Showing the effect of additions to a Buffer

The diagram opposite illustrates how effective a buffer can be.

Normally, the addition of 0.01 mol of OH — ions (to 1L water) would raise the pH from 7 to12 - a rise of 5 pH units.

Similarly, the addition of 0.01 mol of H + ions (to 1 L water) should lower the pH from 7 to 2 - a fall of 5 pH units.

Instead, the pH of the buffer only changes by about 0.1 pH units.

AP Chemistry


We can show these effects mathematically. Consider the same buffer mixture

HC2H3O2(aq) + H2O(l) ⇋ H3O+

(aq) + C2H3O2—

(aq) Ka = 1.8 x 10-5

Assume that it was made with equal quantities of acid (HC2H3O2) and base (C2H3O2—).

For example, a 1L solution that is 0.10 M HC2H3O2 and 0.10 M NaC2H3O2. (1L makes it easy)

Starting pH: Use Henderson-Hasselbach:

pH = pKa + log ([base] = -log (1.8 x 10-5) + log (0.10/0.10) = 4.76 [acid])

This matches with starting pH in the previous diagram.

after addition Step1: We could do an ICE calculation but there is very little error involved inof 0.01 mol H+: simplifying our calculation to assume:

0.01 mol of H+ ions will react to reduce [base] from 0.10 to 0.09 M

C2H3O2—

(aq) + H+(aq) ⟹ HC2H3O2(aq)

At the same time, [acid] will increase from 0.10 to 0.11 M

Step 2: Use Henderson-Hasselbach again

pH = pKa + log ([base] = 4.76 + log (0.09/0.11) = 4.76 - 0.09 = 4.67 a fall of ~0.1 [acid])

after addition Step1: We could do an ICE calculation but there is very little error involved inof 0.01 mol OH— : simplifying our calculation to assume:

0.01 mol of OH_ ions will react to reduce [acid] from 0.10 to 0.09 M

HC2H3O2(aq) + OH—(aq) ⟹ C2H3O2

—(aq) + H2O(l)

At the same time, [base] will increase from 0.10 to 0.11 M

Step 2: Use Henderson-Hasselbach again

pH = pKa + log ([base] = 4.76 + log (0.11/0.09) = 4.76 + 0.09 = 4.85 a rise of ~0.1 [acid])

There is, of course, a limit to how much a buffer can cope with - the buffer capacity - and that will be the focus of the final lesson.

AP Chemistry


8.9 Practice Problems1. HC4H7O2(aq) + H2O(l) ⇄ H3O

+(aq) + C4H7O

2−(aq)

The chemical equation above represents the acid ionization equilibrium for HC4H7O2 for which pKa = 4.8 . Which of the following is the best estimate for the pH of a buffer prepared by mixing 100.mL of 0.20 M HC4H7O2 with 100.mL of 0.10 M NaC4H7O2 ?

A 1.0 B 4.5 C 4.8 D 7.0

2. A student prepares a lactic acid-sodium lactate buffer solution by mixing 40.mL of 0.50 M HC3H5O3(aq) with 200.mL of 1.0 M NaC3H5O3(aq) . The pKa of HC3H5O3 is 3.08. What is the pH of the resulting solution?

A 2.08 B 3.08 C 3.38 D 4.08

3. The equilibrium for the reaction between (CH3)2NH , a weak base, and water is represented by the equation below. The table shows the pH of three solutions of (CH3)2NH(aq) at 25°C .

(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH2+

(aq) + OH−(aq) Kb = 5.4 × 10−4 at 25°C

[(CH3)2NH] pH at25°C

0.050 11.69

0.10 11.85

0.20 12.01

Which of the following equations can be used to correctly calculate the pH of a solution at 25°C that is 0.100 M (CH3)2NH(aq) and 0.100 M (CH3)2NH2Cl(aq) ?

A

B

C should be pH = 14.00 — (-log 5.4 x 10-4) ?

D

O

O

OpH = pKw - pOH

pH = -log(Kw /[OH]

pOH = pKb + log ([base] / [acid]) pH = 14 - pOH

AP Chemistry


4.

The table above shows the values of Ka for four weak acids. Which of the following pairs of chemical species, when combined in equimolar amounts, results in a buffer with a pH closest to 7.5 ?

A HNO2 and OH−

B HC3H5O2 and C3H5O2−

C HClO and ClO−

D C6H5OH and C6H5O−

5.

The acid-dissociation constants of HC3H5O3(aq) and CH3NH3+

(aq) are given in the table above. Which of the following mixtures is a buffer with a pH of approximately 3?

A A mixture of 100. mL of 0.1 M CH3NH3Cl and 50. mL of 0.1 M NaOH

B A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of 0.1 M NaOH

C A mixture of 100. mL of 0.1 M NaC3H5O3 and 100. mL of 0.1 M NaOH

D A mixture of 100. mL of 0.1 M CH3NH3Cl and 100. mL of 0.1 M CH3NH2

O

O

AP Chemistry


6. Acid Dissociation Constant, Ka

H3PO4 7 x 10–3

H2PO4– 8 x 10–8

HPO42– 5 x 10–3

On the basis of the information above, a buffer with a pH = 9 can best be made by using

A pure NaH2PO4 B H3PO4 + H2PO4– C H2PO4

– + PO42–

D H2PO4– + HPO4

2– E HPO42– + PO4

3–

7. A 50.0 mL sample of an acid, HA, of unknown molarity is titrated, and the pH of the resulting solution is measured with a meter and graphed as a function of the volume of 0.100 M NaOH added.

At which point on the titration curve is [A-] closest to twice that of [HA]?

A R B S

C T D U

8. An acetate buffer solution is prepared by combining 50. mL of 0.20 M acetic acid, HC2H3O2(aq), and 50. mL of 0.20 M sodium acetate, NaC2H3O2(aq). A 5.0 mL sample of 0.10 M NaOH(aq) is added to the buffer solution.

Which of the following is a correct pairing of the acetate species present in greater concentration and of the pH of the solution after the NaOH(aq) is added? (The pKa of acetic acid is 4.7.)

Acetate species pH

A HC2H3O2 pH < 4.7

B HC2H3O2 pH > 4.7

C C2H3O2— pH < 4.7

D C2H3O2— pH > 4.7

O

O

O

AP Chemistry


8.9 Quick Check FRQParts a) and b) were done during 8.6. They are shown just to provide context for remaining sections.

1. Answer the following questions that relate to the chemistry of halogen oxoacids.

a) Use the information in the table below to answer part i).

i) Which of the two acids is stronger, HOCl or HOBr ? Justify your answer in terms of Ka .

ii) Draw a complete Lewis electron-dot diagram for the acid that you identified in part a) i).

iii) Hypoiodous acid has the formula HOI. Predict whether HOI is a stronger acid or a weaker acid than the acid that you identified in part a) i). Justify your prediction in terms of chemical bonding.

b) Write the equation for the reaction that occurs between hypochlorous acid and water.

c) A 1.2 M NaOCl solution is prepared by dissolving solid NaOCl in distilled water at 298 K. The hydrolysis reaction OCl−

(aq) + H2O(l) ⇄HOCl(aq) + OH−(aq) occurs.

i) Write the equilibrium-constant expression for the hydrolysis reaction that occurs between OCl-

(aq) and H2O(l) .

Kb = [HOCl] [OH—] / [OCl—]

ii) Calculate the value of the equilibrium constant at 298 K for the hydrolysis reaction.

Kw = Ka x Kb Kb = Kw / Ka = 1.0 x 10-14 / 2.9 x 10-8 = 3.4 x 10-7

iii) Calculate the value of [OH-] in the 1.2 M NaOCl solution at 298 K .

College board do this by ICE, but then let you assume [OCl—] = 1.2 so becomes Kb = 3.4 x 10-7 = [OH—][HOCl] / [OCl—] = [OH-]2 / 1.2

[OH-] = √(1.2 x 3.4 x 10-7) = 6.4 x 10-4 M

AP Chemistry


Q1 contd. Answer the following questions that relate to the chemistry of halogen oxoacids.

d) A buffer solution is prepared by dissolving some solid NaOCl in a solution of HOCl at 298 K. The pH of the buffer solution is determined to be 6.48.

i) Calculate the value of [H3O+] in the buffer solution.

[H+] = 10-pH = 10-6.48 = 3.3 x 10-7 M

ii) Indicate which of HOCl(aq) or OCl−(aq) is present at the higher concentration in

the buffer solution.

Support your answer with a calculation.

College board method: Ka = [H+][OCl—] / [HOCl]

2.9 x 10-8 = (3.3 x 10-7) [OCl—] / [HOCl]

[OCl—] / [HOCl] = 2.9 x 10-8 / 3.3 x 10-7 = 0.88

Therefore, [OCl—] < [HOCl]

AP Chemistry


8.10 Buffer Capacity

Changing the Concentrations of the Buffer Mixture

Buffer solutions are usually prepared from a weak acid and a salt of that acid or from a weak base and a salt of that base. Two important characteristics of a buffered solution are its buffer capacity and its pH range.

The optimal pH of a buffer is equal to pKa (or pKb ) of the acid (or base) used to prepare the buffer.

The same pH (4.76) would be obtained from any ethanoic acid/ethanoate mixture where [acid] = [base]. Keeping within the 'tenfold' rule would mean that the two buffers could have the same pH range.

However, with lower concentrations, the ability to respond to addition of H+(aq) or

OH—(aq) , would be limited compared to a mixture with higher concentrations - it would have a

higher buffer capacity. increasing concentration of both acid and base equally means same pH, same pH range but increases buffer capacity.

AP Chemistry


Changing the Composition of the Buffer Mixture

Different pH's (3.76 - 5.76) would be obtained from any ethanoic acid/ethanoate mixture where [acid] and[base] are not the same and their pH ranges would shift accordingly, though remain± 1 of the new pH of the buffer.

However, if [acid] > [base] then buffer has an increased capacity to deal with the addition of OH—

(aq) ions.

increasing concentration of acid over base means lower pH, lower pH range but increases buffer capacity to deal with OH—

(aq)

Similarly, if [base] > [acid] then buffer has an increased capacity to deal with the addition of H+

(aq) ions.

increasing concentration of base over acid means higher pH, higher pH range but increases buffer capacity to deal with H+

(aq)

This might be useful in circumstances where it is known that the most likely 'contaminant to be resisted' is, for example, an acid.

The kidneys and the lungs work together to help maintain a blood pH of 7.4 by affecting the components of the buffers in the blood.

The pH of blood is mainly dependent on the ratio of the amount of CO2 present (causes H+, so our acid) to the amount of HCO3

- (bicarbonate ion, H+ acceptor, so our base) present in the blood.

The most likely (and most dangerous) problem is an increase in CO2 which can lead to Acidosis.

The optimum pH for blood is 7.4, but the pKa for H2CO3 / HCO3— is 6.1. Maintaining high levels of

HCO3— is crucial as it:

1. raises pH of buffer by having [base] > [acid]

2. increases buffer capacity to deal with H+ ions by having [base] high.

As these notes are being written, doctors are seeing patients with the coronavirus requiring ventilation to help with O2 in and CO2 out.

Many patients then also developkidney problems requiring dialysis- possibly due to the strain put onthe kidneys in their role of homeostasis?

AP Chemistry


8.10 Practice Problems1. HF(aq) + H2O(l) ⇄ H3O

+(aq) + F−

(aq) pKa = 3.20 at 25°C

The acid ionization equilibrium for the weak acid HF is represented by the equation above.

To prepare a buffer with a pH = 3.50 , a student needs to mix 250.mL of 0.100 M HF and 250.mL of 0.100 M KF. If the student mistakenly mixes 250.mL of 0.0500 M HF and 250.mL of 0.0500 M KF , which of the following is the result of this error?

A The buffer will have a lower capacity because of the smaller number of moles of HF and F− available to react if an acid or base is added.

B The buffer will have a lower capacity because the smaller amount of HF and F− will lower the pH of the buffer, and buffers of lower pH have a lower buffer capacity.

C The buffer will have a higher capacity because a larger proportion of HF and F−

will ionize at lower concentrations, resulting in the neutralization of any added acid or base.

D The buffer will have the same capacity because the large volume of the buffer solution dilutes any added acid or base.

2. HF(aq) + H2O(l) ⇄ H3O+

(aq) + F−(aq)

The equation above represents the acid ionization equilibrium for HF. To prepare a buffer with pH≈3.50, 4.20g of NaF(s) should be added to 500.0mL of 0.100MHF(aq).

The buffer is accidentally prepared using 90% pure NaF(s) instead of 99% pure NaF(s).

Assume that the impurities in the NaF(s) samples are inert. Which of the following explains how the error affects the pH and capacity of the buffer?

A The pH is slightly higher than 3.50 and it has a lower capacity for the addition of acids because less than 4.20g of NaF(s) was added.

B The pH is slightly higher than 3.50 and it has a higher capacity for the addition of acids because more than 4.20g of NaF(s) was added.

C The pH is slightly lower than 3.50 and it has a lower capacity for the addition of acids because less than 4.20g of NaF(s) was added.

D The pH is slightly lower than3.50 and it has a higher capacity for the addition of acids because more than 4.20g of NaF(s) was added.

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AP Chemistry


3. HC2H3O2(aq) + H2O(l) ⇄ H3O+

(aq) + C2H3O2−

(aq) pKa = 4.76

The equilibrium for the acid ionization of HC2H3O2 is represented by the equation above.

A student wants to prepare a buffer with a pH of 4.76 by combining 25.00mL of 0.30 M HC2H3O2 with 75.00mL of 0.10 M NaC2H3O2 .

While preparing the buffer, the student incorrectly measures the volume of NaC2H3O2 so that the actual volume used is 76.00mL instead of 75.00mL.

Based on the error, which of the following is true about the buffer prepared by the student?

A The pH of the buffer will be slightly lower than 4.76 because the total volume of the buffer is 101.00mL instead of 100.00mL , and the HC2H3O2 was diluted.

B The pH of the buffer will be slightly lower than 4.76 because the amount of C2H3O2

− added was higher than the amount of HC2H3O2 added. C The buffer solution will have a slightly higher capacity for the addition of bases than for the addition of acids because the total volume of the buffer is 101.00 mL instead of 100.00mL , and the HC2H3O2 was diluted.

D The buffer solution will have a slightly higher capacity for the addition of acids than for the addition of bases because the amount of C2H3O2

− added was higher than the amount of HC2H3O2 added.

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ap e unit 8 - acids & bases

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