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Acids and Bases Arrhenius Acids and Bases Definition An Arrhenius acid is a substance that when added to water increases the concentration of H1+ ions present. An Arrhenius base is a substance that when added to water increases the concentration of OH1- ions present. HCl is an example of an Arrhenius acid and NaOH is an example of an Arrhenius base.

Bronsted-Lowry acids and bases A Bronsted-Lowry (BL) acid is defined as any substance that can donate a hydrogen ion (proton) and a Bronsted-Lowry base is any substance that can accept a hydrogen ion (proton). BL acids and bases must come in what is called conjugate pairs. For example, consider acetic acid dissolved in water:

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What are the conjugate acid/base pairs? Similarly when ammonia is dissolved in water, one has

What are the conjugate acid/base pairs Another important advantage of the BL definition is that we are not limited to water as the solvent. Consider the reaction that occurs when HCl is dissolved in ammonia:

Here, HCl acts as a BL acid with Cl as its conjugate base. Also, NH acts as a BL base with NH as its conjugate acid. Some species can act either as a BL acid or a BL base. Such beasts are called amphoteric. An example

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is the hydrogen carbonate ion, HCO . When dissolved in water, two posible reaction can occur:

or

H3O+ is a hydronium ion. A water molecules has accepted a proton. In the first of these, HCO acts as a BL acid with CO

as its conjugate base, while In the second it acts as a BL base with H CO as its conjugate acid. Acid base reactions are reversible and therefore equilibrium expressions can be written Lewis Acid and bases A Lewis acid is an electron pair acceptor A Lewis base is an electron pair donor

HCl + :OH2 [H3O]+ + Cl–

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The Brønsted description of this reaction says:

• Hydrogen chloride, HCl, is the proton donor [Brønsted] acid and water, :OH2, is the proton-accepting [Brønsted] base.

• The oxonium ion, [H3O]+, is the conjugate [Brønsted] acid and the chloride ion, Cl–, is the conjugate [Brønsted] base.

The Lewis description of this reaction says:

• The chloride ion, Cl–, and water, :OH2, are both Lewis bases and they compete with each other to complex the proton Lewis acid, H+.

• The water Lewis base "wins" and the proton transfers from chloride ion to water. Thus, hydrogen chloride is an H+/Cl– complex that transfers H+ to water to give the oxonium ion, [H3O]+.

• The oxonium ion is an H+/water complex, H+/:OH2.

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6

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Identify the following as a Lewis acid base, Bronsted acid or base and/or an Arrenius acids or base BH3 NH3 H2SO4 H2O Ba(OH)2

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9

10

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A strong acid is one in which the equilibrium lies far to the right. A strong acid produces a weak conjugate base. A weak acid is one in which the equilibrium lies far to the left. Not much acid has dissociated. A weak acid produces a stronger conjugate base.

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Table of Acid and Base Strength Acid Base

Name Formula Formula Name Large Perchloric acid HClO4 ClO4

- Perchlorate ion 3.2 * 109 Hydroiodic acid HI I- Iodide 1.0 * 109 Hydrobromic acid HBr Br- Bromide 1.3 * 106 Hydrochloric acid HCl Cl- Chloride 1.0 * 103 Sulfuric acid H2SO4 HSO4 - Hydrogen sulfate ion 2.4 * 101 Nitric acid HNO3 NO3 - Nitrate ion -------- Hydronium ion H3O+ H2O Water

5.4 * 10-2 Oxalic acid HO2C2O2H HO2C2O2- Hydrogen oxalate ion

1.3 * 10-2 Sulfurous acid H2SO3 HSO3 - Hydrogen sulfite ion 1.0 * 10-2 Hydrogen sulfate ion HSO4 - SO4 2- Sulfate ion 7.1 * 10-3 Phosphoric acid H3PO4 H2PO4 - Dihydrogen phosphate

ion 7.2 * 10-4 Nitrous acid HNO2 NO3

- Nitrite ion 6.6 * 10-4 Hydrofluoric acid HF F - Fluoride ion 1.8 * 10-4 Methanoic acid HCO2H HCO2

- Methanoate ion 6.3 * 10-5 Benzoic acid C6H5COOH C6H5COO- Benzoate ion 5.4 * 10-5 Hydrogen oxalate ion HO2C2O2- O2C2O2 2- Oxalate ion 1.8 * 10-5 Ethanoic acid CH3COOH CH3COO Ethanoate (acetate)

ion 4.4 * 10-7 Carbonic acid CO3 2- HCO3 - Hydrogen carbonate

ion 1.1 * 10-7 Hydrosulfuric acid H2S HS- Hydrogen sulfide ion 6.3 * 10-8 Dihydrogen phosphate

ion H2PO4 - HPO4 2- Hydrogen phosphate

ion 6.2 * 10-8 Hydrogen sulfite ion HS- S2- Sulfite ion 2.9 * 10-8 Hypochlorous acid HClO ClO- Hypochlorite ion 6.2 * 10-10 Hydrocyanic acid HCN CN- Cyanide ion 5.8 * 10-10 Ammonium ion NH4 + NH3 Ammonia 5.8 * 10-10 Boric acid H3BO3 H2BO3 - Dihydrogen carbonate

ion 4.7 * 10-11 Hydrogen carbonate ion HCO3 - CO3 2- Carbonate ion 4.2 * 10-13 Hydrogen phosphate

ion HPO4 2- PO4 3- Phosphate ion

1.8 * 10-13 Dihydrogen borate ion H2BO3- HBO3 2- Hydrogen borate ion

1.3 * 10-13 Hydrogen sulfide ion HS- S 2- Sulfide ion 1.6 * 10-14 Hydrogen borate ion HBO3 2- BO3 3- Borate ion

--------- water H2O OH- Hydroxide

1. Strong acid are listed at the top left hand corner of the table and have Ka values >1 2. Acid with values less than one are considered weak. 3. The strong bases are listed at the bottom right of the table and get weaker as we move to the top of the table.

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Acidity of Ions anions of strong acids and cations of strong bases are neutral examples: NO3

- of HNO3, Cl- of HCl, Na+ of NaOH, Li+ of LiOH, Ca+2 of Ca(OH)2

Therefore ionic compounds containing these ions are basic. examples: NaCl, LiNO3 produce basic solutions Neutral aqueous anions NO3

-, ClO4-, ClO3

-, Cl-, Br-, I- Neutral aqueous cations Li+, Na+, K+, Rb+, Cs+, Ca2+

, Sr2+, Ba2+

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The Autoionization of water Water is an amphoteric substance. It can act as an acid or a base. water ionizes as following 2H2O === H3O+ + OH- giving the following equilibrium expression Kw = [H3O+] [OH-] = [H+] [OH-] where Kw is the dissociation constant for water. At 25 oC Kw = 1.0 x 10-14 M Kw = [H+] [OH-] = 1.0 x 10-7 M

Solving Equilibrium Problems Involving Weak Acids Example: Consider the process by which we would calculate the H 3O+, OAc-, and HOAc concentrations at equilibrium in an 0.10 M solution of acetic acid in water.

We start this calculation by building a representation of what we know about the reaction. HOAc(aq) + H2O(l) H3O+(aq) + OAc- Ka

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(aq) = 1.8 x 10-5

Initial: 0.10 M 0 0 Equilibrium: ? ? ? We then compare the initial reaction quotient (Qa) with the equilibrium constant (Ka) for the reaction and reach the obvious conclusion that the reaction must shift to the right to reach equilibrium.

Recognizing that we get one H3O+ ion and one OAc- ion each time an HOAc molecule dissociates allows us to write equations for the equilibrium concentrations of the three components of the reaction.

HOAc(aq) + H2O(l) H3O+(aq) + OAc-

(aq)

Ka = 1.8 x 10-5

Initial: 0.10 M 0 0 Equilibrium: 0.10 - C C C Substituting what we know about the system at equilibrium into the Ka expression gives the following equation.

Although we could rearrange this equation and solve it with the quadratic formula, it is tempting to test the assumption

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that C is small compared with the initial concentration of acetic acid.

We then solve this approximate equation for the value of C.

C 0.0013 C is small enough to be ignored in this problem because it

is less than 5% of the initial concentration of acetic acid.

We can therefore use this value of C to calculate the equilibrium concentrations of H3O+, OAc-, and HOAc.

[HOAc] = 0.10 - C 0.10 M [H3O+] = [OAc-] = C 0.0013 M

We can confirm the validity of these results by substituting these concentrations into the expression for Ka.

Our calculation must be valid because the ratio of these concentrations agrees with the value of Ka for acetic acid, within experimental error.

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Calculate the H3O+ concentration in an 0.0001 M solution of hydrocyanic acid (HCN). HCN(aq) + H2O(l) à H3O+(aq) + CN-(aq) Ka = 6 x 10-10 Calculate the pH of an 0.023 M solution of saccharin (HSc), if Ka is 2.1 x 10-12 for this artificial sweetener. In a solution of acetic acid, the equilibrium concentrations are found to be [CH3COOH] = 1.000; [CH3COO-] = 0.0042. Evaluate the pH of this solution and the equilibrium constant of ionization of acetic acid.

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Guided Questions • A weak acid is a compound that

a. is completely ionized in solution, b. is not completely ionized in solution, c. gives a high pH in a solution, d. gives a low pH in its solution,

• The acidity constant, Ka, for a strong acid is

a. infinity, b. very large, c. very small, d. zero. Household vinegar is usually 5% acetic acid by volume. Calculate the molarity of this solution. Assume density of solution to be 1 g/mL.

• Acetic acid is a typical and familiar compound that provides a good example for numerical problems. Its acidity constant Ka is 1.85x 10-5. What is the pH of a concentrated vinegar, which is a 1.0 M acetic acid solution?

• If you dilute the vinegar 100 times in a soup that

you are cooking, the concentration of your soup is

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0.010 M in acetic acid. Other ingredients are ignored. What is the pH of this solution?

• What is the pH of a 0.010 M HCl solution?

• What is the pH of a 1.0x 10-4 acetic acid solution

(Ka = 1.85x 10-5)?

• What is the pH of a 1.0 x 10-3 M chloroacetic acid solution (Ka = 1.4 10-3)?

• What is the pH of a 1.0 x 10-6 M chloroacetic acid solution (Ka = 1.4 x 10-3)?

• What is the pH of a 1.0 x 10-7 M chloroacetic acid solution (Ka = 1.4 x 10-3)?

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Relationship between Ka and Kb Kw = Ka x Kb In strong bases OH- can be found using the molarity for the base.

Calculations involving weak bases

Solving Equilibrium Problems Involving Bases

With minor modifications, the techniques applied to equilibrium calculations for acids are

valid for solutions of bases in water.

Consider the calculation of the pH of an 0.10 M NH3 solution. We can start by writing an

equation for the reaction between ammonia and water.

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Strict adherence to the rules for writing equilibrium constant expressions leads to the

following equation for this reaction.

But, taking a lesson from our experience with acid-dissociation equilibria, we can build

the [H2O] term into the value of the equilibrium constant. Reactions between a base and

water are therefore described in terms of a base-ionization equilibrium constant, Kb.

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For NH3, Kb is 1.8 x 10-5.

We can organize what we know about this equilibrium with the format we used for

equilibria involving acids.

NH3(aq) + H2O(l) NH4+(aq) + OH-

(aq) Kb = 1.8 x 10-5

Initial: 0.10 M 0 0

Equilibrium: 0.10 - C C C

Substituting this information into the equilibrium constant expression gives the following

equation.

Kb for ammonia is small enough to allow us to consider the assumption that C is small

compared with the initial concentration of the base.

Solving this approximate equation gives the following result.

C 1.3 x 10-3

This value of is small enough compared with the initial concentration of NH3 to be

ignored and yet large enough compared with the OH- ion concentration in water to ignore

the dissociation of water. It can therefore be used to calculate the pOH of the solution.

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pOH = - log (1.3 x 10-3) = 2.89

Which, in turn, can be used to calculate the pH of the solution.

pH = 14 - pOH = 11.11

Equilibrium problems involving bases are relatively easy to solve if the value of Kb for the

base is known. The first step in many base equilibrium calculations involves determining

the value of Kb for the reaction from the value of Ka for the conjugate acid.

As an example, let's calculate the pH of a 0.030 M solution of sodium benzoate

(C6H5CO2Na) in water from the value of Ka for benzoic acid (C6H5CO2H): Ka = 6.3 x 10-5.

Benzoic acid and sodium benzoate are members of a family of food additives whose

ability to retard the rate at which food spoils has helped produce a 10-fold decrease in

the incidence of stomach cancer. To save time and space, we'll abbreviate benzoic acid as

HOBz and sodium benzoate as NaOBz. Benzoic acid, as its name implies, is an acid.

Sodium benzoate is a salt of the conjugate base, the OBz- or benzoate ion.

Whenever sodium benzoate dissolves in water, it dissociates into its ions.

H2O NaOBz(aq) Na+(aq) + OBz-(aq)

The benzoate ion then acts as a base toward water, picking up a proton to form the

conjugate acid and a hydroxide ion.

OBz-(aq) + H2O(l) HOBz(aq) + OH-(aq)

The base-ionization equilibrium constant expression for this reaction is therefore written

as follows.

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The next step in solving the problem involves calculating the value of Kb for the OBz- ion

from the value of Ka for HOBz.

The Ka and Kb expressions for benzoic acid and its conjugate base both contain the ratio

of the equilibrium concentrations of the acid and its conjugate base. Ka is proportional to

[OBz-] divided by [HOBz], and Kb is proportional to [HOBz] divided by [OBz-].

Two changes have to made to derive the Kb expression from the Ka expression: We need

to remove the [H3O+] term and introduce an [OH-] term. We can do this by multiplying

the top and bottom of the Ka expression by the OH- ion concentration.

Rearranging this equation gives the following result.

The two terms on the right side of this equation should look familiar. The first is the

inverse of the Kb expression, the second is the expression for Kw.

Rearranging this equation gives the following result.

Ka x Kb = Kw

According to this equation, the value of Kb for the reaction between the benzoate ion and

water can be calculated from Ka for benzoic acid.

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Now that we know Kb for the benzoate ion, we can calculate the pH of an 0.030 M NaOBz

solution with the techniques used to handle weak-acid equilibria. We start, once again, by

building a representation for the problem.

OBz-(aq) + H2O(l) HOBz(aq) + OH-

(aq) Kb = 1.6 x 10-10

Initial: 0.030 M 0 0

Equilibrium: 0.030 - C C C

We then substitute this information into the Kb expression.

Because Kb is relatively small, we assume that C is small compared with 0.030.

We then solve the approximate equation for the value of C.

C 2.2 x 10-6

The assumption that C is small is obviously valid. We can therefore use C to calculate

the pOH of the solution.

[OH-] = C 2.2 x 10-6

pOH = - log [2.2 x 10-6] = 5.66

The problem asked for the pH of the solution, however, so we use the relationship

between pH and pOH to calculate the pH.

pH = 14 - pOH = 8.34

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Two assumptions were made in this calculation.

• We assumed that C was small enough compared with the initial concentration of the base that it could be ignored in the [0.030 - C] term.

• We assumed that all of the OH- ion at equilibrium came from the reaction between the benzoate ion and water. (In other words, we ignored the contribution to the OH- ion concentration from the dissociation of water.)

We have already confirmed the validity of the first assumption. What about the second?

The OH- ion concentration obtained from this calculation is 2.1 x 10-6 M, which is 21 times

the OH- ion concentration in pure water. According to LeChatelier's principle, however,

the addition of a base suppresses the dissociation of water. This means that the

dissociation of water makes a contribution of significantly less than 5% to the total OH-

ion concentration in this solution. It can therefore be legitimately ignored.

Two factors affect the OH- ion concentration in aqueous solutions of bases: Kb and Cb. We

can ignore the dissociation of water when KbCb for a weak base is larger than

1.0 x 10-13. When KbCb is smaller than 1.0 x 10-13, we have to include the

dissociation of water in our calculations.

Practice Problem 5:

Calculate the HOAc, OAc-, and OH- concentrations at equilibrium in an 0.10 M NaOAc solution. (HOAc: Ka = 1.8 x 10-5)

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Polyprotic Acids An acid that provide more than one proton. It dissociates in a stepwise manner. One proton at a time. For a typical weak polyprotic acid Ka1>> Ka2 > Ka3

Acid Ka1 Ka2 Ka3 sulfuric acid (H2SO4) 1.0 x 103 1.2 x 10-2 chromic acid (H2CrO4) 9.6 3.2 x 10-7 oxalic acid (H2C2O4) 5.4 x 10-2 5.4 x 10-5 sulfurous acid (H2SO3) 1.7 x 10-2 6.4 x 10-8 phosphoric acid (H3PO4) 7.1 x 10-3 6.3 x 10-8 4.2 x 10-13 glycine (C2H6NO2) 4.5 x 10-3 2.5 x 10-10 citric acid (C6H8O7) 7.5 x 10-4 1.7 x 10-5 4.0 x 10-7 carbonic acid (H2CO3) 4.5 x 10-7 4.7 x 10-11 hydrogen sulfide (H2S) 1.0 x 10-7 1.3 x 10-13

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In most polyprotic acids the only the first step contributes to the [H+] concentration. Calculate the pH of a 5.0 M H3PO4 solution. Weak Acid Mixtures Only the acid with the largest Ka value will contribute an appreciable [H+]. Determine the pH based on this acid and ignore the rest Percent Dissociation The percent ionization. 100 % for strong acids. Weak acid example Calculate the percent ionization for a 0.1 M phenol (Ka = 1.6 x 10-10) Soluble aluminum compounds form the following acid Al+3 + H2O à Al(H2O)6

+3

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Which then undergoes the following acid base equilibrium Al(H2O)6

+3 + H2O à Al(H2O)5OH2+ + H3O+

Structure and Acid-Base Properties Acidity depends on the strength of the attraction for electrons. Acidity increases with stronger attractions for electrons. 1. Ionic charge. More positive ions are stronger acids and more negative ions are stronger bases. Higher charged cations (Al+3, Fe+3 e.c.t) are Lewis acids.

2. Oxidation Number. The greater the oxidation number of the central atom in similar formulas the stronger the acid

HNO2 < HNO3, H3PO3 < H2PO4

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3. Electronegativity. The greater the

electronegativity of the central atom in acids with similar formulas the stronger the acid.

H3BO3 < H2CO3 < H2SO3 < HNO3

Acid-Base Properties of Oxides When a covalent oxide dissolves in water , an acidic solution forms. SO2(g) + H2O(l) à H2SO3(aq) CO2(g) + H2O(l) à H3CO3 2NO2(g) + H2O(l) à HNO2(aq) + HNO2(aq) When an ionic oxide dissolves in water, a basic solution is formed. CaO(s) + H2O(l) à Ca(OH)2 K2O(s) + H2O(l) à 2KOH(aq) Other strong bases include hydrides, nitrides and carbides. Hydride ion, H- : NaH(s) + H2O à H2(g) + Na+(aq) + OH-(aq)

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Nitride ion N3-: Mg3N2(s) + 6H2O à 2NH3(g) + 3Mg2+(aq) + 6OH-

(aq) Carbide ion C-: Ca2C2(s) + 2H2O à C2H2(g) + Ca2+(aq) + 2OH-(aq)

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HOCl OCl- + H+ Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The acid dissociation constant, Ka, for the reaction represented above is 3.2 x 10-8.

a. Calculate the [H+] of a 0.14-molar solution of HOCl. b. Write the correctly balanced net ionic equation for the reaction that occurs when NaOCl is dissolved in water and calculate the numerical value of the equilibrium constant for the reaction. c. Calculate the pH of a solution made by combining 40.0 ml of 0.14-molar HOCl

and 10.0 ml of a 0.56 molar NaOH. d. How many millimoles of solid naOH must be added to 50.0 ml of 0.20-molar

HOCl to obtain a buffer solution that has a pH of 7.49? Assume that the addition of the solid NaOH results in a negligible change in volume.

e. Calculate the PH of such a solution if the concentration of HOCl in the solution is 0.065 molar.

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The overall dissociation of oxalic acid, H2C2O4, is represented below. The overall dissociation constant is also indicated. H2C2O4 ------ 2H+ + C2O42- K = 3.78 x 10-6

a. What volume of 0.400 molar NaOH is required to neutralize completely a 5.00 x 10-3 mole sample of pure oxalic acid?

b. Give the equation representing the first and second dissociations of oxalic acid. Calculate the value of the first dissociation constant, K1, for oxalic acid if the value of the second dissociation constant, K2 is 4.40 x 10-5.

c. To a 0.015 molar solution of oxalic acid, a strong acid is added until the pH is 0.5. Calculate the [C2O4

2-] in the resulting solution. (Assume the change in volume is negligible.)

d. Calculate the equilibrium constant, Kb, for the reaction that occurs when solid Na2C2O4 is dissolved in water.

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NH3(aq) + H2O à NH4+(aq) + OH- (aq)

In an aqueous solution, ammonia reacts as represented above. In 0.0180 M NH3 (aq) at 25 oC, the hydroxide ion concentration, [OH-] is 5.60 x 10-4 M. In answering the following, assume that temperature is constant at 25 oC and that volumes are additive. a. Write the equilibrium-constant expression for the reatction represented above. b. Determine the percent ionization of NH3 in 0.0180 M NH2(aq) c. Determine the value of the base ionization constant, Kb, for NH3(aq) d. Determine the percent ionization of NH3 in 0.0180 M NH3(aq) e. In an experiment, a 20.0 ml sample of 0.0180 M NH3 (aq) was placed in a flask

and titrated to the equivalence point and beyond using 0.0120 m HCl (aq) i. Determine the volume of 0.0120 M HCl (aq) that was added to reach

the equivalence point. ii. Determine the pH of the solution in the flask after a total of 15.0 ml of

0.0120 M HCl (aq) was added. iii. Determine the pH of the solution in the flask after a total of 40.0 ml of

0.0120 M HCl (aq) was added.

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Common Ion Effect Adding a common ion to a solution will decrease the ionization of a weak electrolyte shifting the equilibrium. Example HC2H3O2(aq) + H2O(l) == C2H3O2(aq) + H3O+ Adding sodium acetate to a solution of acidic acid will increase the acetate ion and shift the equilibrium to the left. The common ion effect is important in polyprotic acids. The first step produces protons which according to LeChaterlier’s principle will decrease the likelihood of the second ionization.

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Buffers A buffered solution is one that resists a change in its pH when either hydroxide ions or protons are added. A buffer may contain a weak acid and its salt or a weak base and its salt. buffer solutions are weak acid or weak base solutions therefore calculations use the same procedures What is the pH of an aqueous mixture containing 0.20 M acetic acid and 0.10 M sodium acetate

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Calculations involving buffers can be solved using the Henderson-Hasselbalch equation

This is the well-known Henderson-Hasselbalch equation that is often used to perform the calculations required in preparation of buffers for use in the laboratory, or other applications. Notice several interesting facts about this equation.

First, if the pH = pKa, the log of the ratio of dissociate acid and associated acid will be zero, so the concentrations of the two species will be the same. In other words, when the pH equals the pKa, the acid will be half dissociated.

Second, notice that as the pH increases or decreases by one unit relative to the pKa, the ratio of the dissociate form to the associated form of the acid changes by factors of 10. That is, if the pH of a solution is 6 and the pKa is 7, the ratio of [ A-]/[ HA] will be 0.1, will if the pH were 5, the ratio would be 0.01 and if the pH were 7, the ratio would be 1.

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Buffered Solutions 1. Contains a weak acid and the corresponding

weak base. 2. When hydrogen ions are added to the solution

they react with the weak base. 3. When hydroxide ions are added to the

solution they react with the weak acid. 4. The pH of the buffer depends on the ratio of

the concentrations of the weak acid and the weak base. If it remains constant the pH remains constant.

Buffer Capacity

A buffer can only absorb so much strong acid or base before failing to resist the change in pH. The buffer capacity is

the #of moles of weak acid present in the solution if base is being added

the #of moles of weak base present in the solution if acid is being added

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Chemical Type Formula Molecular weight

Ka pKa Kb pKb

Acetic Acid solution 1.0M

weak acid HC2H3O2 60.05 1.8 x 10-5

4.75 ---- ---

Sodium acetate anhydrous

conjugate base of above acid

NaC2H3O2 82.03 ------ --- ---- ---

Sodium phosphate monobasic, monohydrate

weak acid NaH2PO4*H2O 138.00 6.2 x 10-8

7.21 ---- ---

Sodium phosphate, dibasic heptahydrate

conjugate base of above acid

Na2HPO4.7H2O 268.1 4.7 x

10-13 12.32 ---- ----

Citric acid monohydrate

weak acid C6H8O7.H20 210.14 7.5 x

10-4 3.128 ---- ---

Sodium citrate dihydrate

conjugate base of above acid

Na3C6H5O7.2H2O 294.11 ------ --- ---- ---

Ammonia solution 2.0M

weak base

NH4OH 17.3 ----- --- 1.7 x

10-5

4.76

Ammonium Chloride

conjugate acid of above base

NH4Cl 53.5 ------ --- ---- ---

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Calculate the quantity of HA/A- needed to prepare a buffer with a pH of 9.0. Exercise 1 Calculate the pH of 0.500 L of a buffer solution composed of 0.50 M formic acid and 0.70 M sodium formate before and after adding 10.0 mL of 1.00 M HCl Exercise 2 Using an acetic acid\sodium acetate buffer solution, what ratio of acid to conjugate base will you need to maintain the pH at 5.00? Explain how you would make up such a solution. Exercise 3 A buffered solution contains 0.25 M NH3 ( Kb = 1.8 × 10–5) and 0.40 M NH4Cl. a) Calculate the pH of this solution. b) Calculate the pH when 0.10 mol of gaseous HCl is added to 1.0 L of the buffered solution.

Exercise  4    A chemist needs a solution buffered at pH 4.30 and can choose from the following list of acids and their soluble salt: a. chloroacetic acid Ka = 1.35 × 10–3

b. propanoic acid Ka = 1.3 × 10–5

c. benzoic acid Ka = 6.4 × 10–5

d. hypochlorus acid Ka = 3.5 × 10–8 Calculate the ratio of A/B required for each system to yield a pH of 4.30. Which system works best?

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Acid/base Titrations A method to determine an unknown concentration of an acid or base. An acid-base indicator changes color at the end point of the titration.

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Measuring pH during a titration gives a titration curve. Titration Curves A titration curve is drawn by plotting data attained during a titration, titrant volume on the x-axis and pH on the y-axis. The titration curve serves to profile the unknown solution. In the shape of the curve lies much chemistry and an interesting summary of what we have learned so far about acids and bases. The titration of a strong acid with a strong base produces the following titration curve:

The pH before the titration begins is that of a strong acid. As the strong base is added, the pH rises slightly as the strong acid concentration decreases. Near the equivalence point the pH changes rapidly. The pH at the equivalence point is always 7 for a strong acid-strong base titration. After the equivalence point the pH is that of a strong base. It rises slowly because the strong base concentration is increasing.

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Titration of a weak acid and a strong base

The pH before the titration begins is that of the weak acid. The next region is the “buffered named because the pH changes slightly with addition of the strong base. The point where the amount of titrant is equal to half the amount required to reach the equivalence point is the PKa of the weak acid because according to the Henderson-Hasselbalch equation:

pH = pka + log(1) ; pH = pka The equivalence point the pH is greater than 7 Beyond the equivalence point only the concentration of the strong base needs to be considered to calculate the pH of the solution

Calculations involving titrations 30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acid

1. Write the balanced chemical equation for the reaction NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l)

2. Extract the relevant information from the question: NaOH V = 30mL , M = 0.10M HCl V = 25.0mL, M = ?

3. Check the data for consistency NaOH V = 30 x 10-3L , M = 0.10M HCl V = 25.0 x 10-3L, M = ?

4. Calculate moles NaOH n(NaOH) = M x V = 0.10 x 30 x 10-3 = 3 x 10-3 moles

5. From the balanced chemical equation find the mole ratio NaOH:HCl 1:1

6. Find moles HCl NaOH: HCl is 1:1 So n(NaOH) = n(HCl) = 3 x 10-3 moles at the equivalence point

7. Calculate concentration of HCl: M = n ÷ V n = 3 x 10-3 mol, V = 25.0 x 10-3L M(HCl) = 3 x 10-3 ÷ 25.0 x 10-3 = 0.12M or 0.12 mol L-1

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Problem 2

50mL of 0.2mol L-1 NaOH neutralised 20mL of sulfuric acid. Determine the concentration of the acid

Problem 3

25.0mL of 0.05M Ba(OH)2 neutralised 40.0mL of nitric acid. Determine the concentration of the acid.

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• Confidence Building Questions o What are the pH of solutions of 10, 1.0, 0.10, 0.010 and 0.0010 M HCl?

o What are [H+] and the pH at the half equivalent point when a solution of 1.0 M HCl is titrated by a 1.0 NaOH solution?

o What are [Na+] at the half equivalent point when a solution of 1.0 M HCl is titrated by a 1.0 NaOH solution?

o What are [H+] and the pH at the equivalent point when a solution of 1.0 M HCl is titrated by a 1.0 NaOH solution?

o What are [Na+] and [Cl-] at the equivalent point when a solution of 1.0 M HCl is titrated by a 1.0 NaOH solution?

o What are [H+] and the pH of 1.0 M acetic acid solution? Ka = 1.8e-5

o What is the pH of the above solution when half of the acid is neutralized by NaOH in the titration?

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o What is the pH of the end point or equivalent point when 1.0 M acetic acid is titrated by 1.0 M NaOH solution?

What is the pH of a 0.5 M sodium acetate solution?

o A 0.10 M acetic acid solution is titrated with a 0.10 M NaOH solution to the equivalent point. What is the concentration of the acetate ion?

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Weak Acid strong base titration

No base is added, pH = 2.8

10 ml NaOH is added, pH = 4.0

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25 ml of 0.10 M NaOH has been added. pH = 4.67

40 ml of NaOH is added

50 ml of NaOH is added, pH = 8.73

60 ml of NaOH added, pH = 12.0

75 ml NaOH added, pH = 12.3

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Examples: You titrate 100 ml of 1.00 M sodium chlorate, NaClO with 1.00M HCl. What is the pH after you add 0 ml, 50 ml, 100 ml and 150 ml of acid? Kb for ClO- is 3.6*10-7.

pH = 10.78, 7.55, 3.4, 0.70 Exercise 5 For the titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH calculate the pH of the solution at the following selected points of the titration:

a) NO NaOH has been added:

b) 10.0 mL of NaOH has been added:

c) 20.0 mL (total as opposed to additional) of NaOH has been added:

d) 50.0 mL (total) of NaOH has been added:

e) 100.0 mL (total) of NaOH has been added: f) 200.0 mL (total) of NaOH has been added: Example 2 Write the hydrolysis reactions for the following salts: KNO2

NH4NO3

CH3NH3Cl NaClO2

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Exercise 6 Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka = 6.2 × 10–10) when dissolved in water. If a 50.0 mL sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution

a) after 8.00 mL of 0.100 M NaOH has been added.

b) at the halfway point of the titration. c) at the equivalence point of the titration.