~ 1 ~

Acids and bases are considered to be strong when

they dissociate or ionize fully (100%) in solution.

This allows them to act as strong electrolytes in

water and have a signficant impact on pH.

Weak acids and bases dissociate, but to a much

lesser extent. This limits their impact on pH and

allows them to act only as weak electrolytes.

Strong Acids: HCl, HBr, HI, HNO3, HClO3, HClO4

H2SO4 (first proton ONLY, 2nd

is weak)

Strong Bases: All Group 1 metal hydroxides (MOH)

Ca(OH)2, Sr(OH)2, and Ba(OH)2

Ex.1) How would the pH of a 0.100 M HCl sol'n compare

with that of a 0.100 M acetic acid (HAc) sol'n?

23.1.1 Acid/Base Strength and Major Species

~ 2 ~

Knowing this, we can predict major species (MS)

present in a solution containing acids or bases. For

example:

HCl is a strong acid and would fully dissociate:

Major Species: H+ Cl

- H2O

Being comparatively weak, HAc partially dissociates.

Not enough H+ or Ac

- is produced to be significant:

Major Species: HAc H2O

List the major species present in each of the

following solutions:

Ex.2) 0.010 M NaOH MS:

Ex.3) 0.90 M HF MS:

~ 3 ~

Ex.4) 1.40 M H2SO4 MS:

Ex.5) 0.175 M Ba(OH)2 MS:

Ex.6) 1.0e-4 M Al(OH)3 MS:

Ex.7) 2.5 M NH4Cl MS:

Ex.8) 1.00 M HClO4 MS:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Give the MS present in a nitric acid sol'n:

SC.2) Dissolved NaOH is reacted with an excess of of HCl.

Predict the surviving MS:

Answers: SC.1 / H= NO3

- H2O SC.2 / H

= Cl

- H2O Na

+

~ 4 ~

In water, acids dissociate or ionize in water to form

a proton and conjugate base. Weak acids exist in

equilibrium with their conjugates ( ↔ ), while strong

acids dissociate fully ( → ):

HA(aq) ↔ H+(aq) + A

-(aq)

Show the following acids dissociating water:

Ex.1) Acetic acid, HAc

Ex.2) Hydrochloric acid, HCl

Ex.3) Ammonium ion, NH4+

23.1.2 Acid Dissociation Equations in Water

~ 5 ~

Organic acids often show the protons being donated

at the end of the formula. The proton being

donated is bold in the examples below:

Ex.4) Acetic acid, CH3COOH

Ex.5) Pyridinium ion, C5H5NH+

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Anilinium ion, C6H5NH3+

SC.2) Hydrocyanic acid, HCN

Answers: SC.1 / C6H5NH3+(aq) ↔ H

+(aq) + C6H5NH2(aq)

SC.2 / HCN(aq) ↔ CN-(aq) + H

+(aq)

~ 6 ~

Strong bases are typically group 1 and 2 metal

hydroxides which dissociate completely in water to

give hydroxide ions:

MOH(aq) → OH-(aq) + M

+(aq)

Show the following bases dissociating in water:

Ex.1) NaOH

Ex.2) Ba(OH)2

Ex.3) KOH

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Show Ca(OH)2 dissociating in water:

23.1.3 Strong Base Equations in Water

Answers: SC.1 / Ca(OH)2(aq) → Ca2+

(aq) + 2 OH-(aq)

~ 7 ~

Weak bases typically react with water to produce a

hydroxide ion and conjugate weak acid:

A-(aq) + H2O(l) ↔ OH

-(aq) + HA(aq)

Show the following acting as a base in water:

Ex.1) Fluoride ion, F-

Ex.2) Ammonia, NH3

Ex.3) Hypochlorite ion, OCl-

23.1.4 Weak Base Equations in Water

~ 8 ~

Organic bases often contain a highly electronegative

element which accepts the proton in reactions. I've

made these bold in the examples below:

Ex.4) Methylamine, CH3NH2

Ex.5) Acetate ion, CH3COO-

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Aniline, C6H5NH2

SC.2) Phosphate ion, PO43-

Answers: SC.1 / C6H5NH2(aq) + H2O(l) ↔ OH-(aq) + C6H5NH3

+(aq)

SC.2 / PO43-

(aq) + H2O(l) ↔ HPO42-

(aq) + OH-(aq)

~ 9 ~

Within an acid-base equation, acids and bases exist

in pairs known as "conjugates". By donating a

proton, what remains of the acid, typically an anion,

is now known as a conjugate base:

HA(aq) ↔ H+(aq) + A

-(aq)

By accepting a proton, bases become conjugate

acids. This also changes the charge:

A-(aq) + H

+(aq) ↔ HA(aq)

The equations below contain two conjugate pairs.

Identify the pairs and label the CA and CB in each.

Ex.1) HAc(aq) + HCO3-(aq) ↔ H2CO3(aq) + Ac

-(aq)

Ex.2) H2O(l) + H2O(l) ↔ H3O+(aq) + OH

-(aq)

23.1.5 Identifying Conjugate Acid/Base Pairs

~ 10 ~

Ex.3) NH3(aq) + H2O(l) ↔ NH4+(aq) + OH

-(aq)

In a conjugate pair, HA/A- , the acid will always have

exactly one more proton than its conjugate base.

Amphiprotic substances will form 2 pairs, depending

on whether they initially act as an acid or as a base.

Give the conjugate(s) for each of the following:

Ex.4) NH3 ↔

Ex.5) H2CO3 ↔

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Write and label the conjugates of the hydrogen

carbonate ion, HCO3- :

SC.2) Given that HCl is a very strong acid, what can we

infer about the nature / strength of its conjugate?

Answers: SC.1 / H2CO3 is the CA, CO32-

is the CB SC.2 / chloride is an incredibly weak base.

~ 11 ~

Ionic compounds, or salts, often have acidic/basic

properties according to the following conditions:

Neutral: cation is from a SB (ex. group 1), and anion

is a conjugate of a SA.

Acidic: cation is CA of a WB, anion is neutral.

Basic: cation is neutral, anion is CB of a WA.

Classify each of the following salts as a WA, WB, or

neutral. Indicate which part of the salt, if any, has

acidic or basic tendencies:

Ex.1) NaF Ex.2) NH4Cl

Ex.3) KCl Ex.4) NaAc

23.1.6 Acidity and Basicity of Salts

~ 12 ~

Occasionally, a salt may contain both acidic and

basic components. In these instances, compare K

values to determine whether the sol'n will be acidic,

basic, or neutral (table 21 in your data booklet).

Ex.5) NH4Ac NH4+ (Ka = 5.6e-10) / Ac

- (Kb = 5.6e-10)

Ex.6) NH4NO2 NO2- (Kb = 1.4e-11)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) NH4NO3 SC.2) KNO3

SC.3) NH4F (HF has Ka = 6.6e-4)

Answers: SC.1 / acidic SC.2 / neutral SC.3 / acidic

~ 13 ~

Covalent oxides consist of a non-metal paired with

oxygen. In a reaction with water, many of these

non-metals oxides (NMOs) produce an oxoacid.

Oxoacids are polys and typically contain 3+ oxygens.

NMO(g) + H2O(l) → Oxoacid(aq)

CO2(g) + H2O(l) → H2CO3(aq)

When an NMO combines with water, "tack on" an

extra oxygen and only keep the hydrogen atoms

necessary to satisfy the charge. For example:

NO2 → NO3-

→ HNO3

Ex.1) SO2 → →

Ex.2) SO3 → →

Ex.3) ClO2 → →

23.1.7 Behavior of Covalent Oxides in Water

~ 14 ~

For larger, heavily-oxidized NMOs, assume the acid

produced will have the same oxidation numbers:

+5 -2 +1 -2 +1 +5 -2 +1 +3 -2

N2O5 + H2O → HNO3 not HNO2

Ex.4) P4O10 + H2O →

Ex.5) Cl2O7 + H2O →

Ex.6) N2O3 + H2O →

NMOs in lower oxidation states tend to have little

or no acidic nature. If you cannot "strip" a poly acid

of hydrogens and one oxygen to produce an NMO,

it is unlikely to have acidic behavior:

Ex.7) Which carbon oxides are acidic, which are not?

~ 15 ~

Ex.8) Which sulfur oxides are acidic, which are not?

Ex.9) What is the lowest oxidation number sulfur can

have in a sulfur oxide and still be acidic?

Ex.10) Would N2O be acidic in water? Why or why not?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Fossil fuels contain nitrogen and sulfur compounds.

Why would the combustion of these substances be

an environmental concern?

SC.2) Which sulfur oxide, SO2 or SO3, is most acidic?

Answers: SC.1 / the nitrogen oxides and sulfur oxides which are produced as a by-product of

combustion can both combine with atmospheric water to form acid rain. SC.2 / SO3 is most

highly oxidized and will produce the stronger sulfuric acid, as opposed to sulfurous.

~ 16 ~

Ionic oxides are compounds consisting of a metal

and oxygen and are often also known as metal

oxides. In a reaction with water, metal oxides form

metal hydroxides:

MO(s) + H2O(l) → MOH(aq)

Show the following ionic oxides reacting with water:

Ex.1) MgO:

Ex.2) Fe2O:

Ex.3) CaO:

Ex.4) Based on the equation above, how does the

addition of a metal oxide influence the pH of pure

water:

23.1.8 Behavior of Ionic Oxides in Water

~ 17 ~

In additional being called ionic oxides, most metal

oxides are also known as basic oxides due to their

common behavior in water.

Which basic oxide will produce the following

hydroxides in water?

Ex.5) Sr(OH)2 Ex.6) KOH

Ex.7) LiOH Ex.8) Ba(OH)2

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Show Na2O reacting with water:

SC.2) Which ionic oxide produces cesium hydroxide in

water?

Answers: SC.1 / Na2O(s) + H2O(l) → 2 NaOH(aq) SC.2 / Cs2O(s)

~ 18 ~

"p" is a mathematical function commonly found

when working with very large or small numbers,

such as concentrations (pH, pOH) and equilibrium

constants (Ka, Kb):

pX = - log10X 10 -pX

= X

Convert each of the following into their respective

p-values: Ka → pKa , [H+] → pH, etc. Only numbers

past the decimal are significant in "log" answers:

Ex.1) Ka = 1.3 x 106 Ex.2) [OH

-] = 1.3e-6

Ex.3) [H+] = 4.5e-2 Ex.4) Kb = 1.8 x 10

-5

Reverse the following p-calcs. Remember sigfigs!

Ex.5) pH = 10.23 Ex.6) pKa = -8.552

23.1.9 p - Calculations

~ 19 ~

Ex.7) pOH = 0.54 Ex.8) pKb = 13.29

Ex.9) What is the relationship between the size of the

original number and the corresponding p-value?

Larger numbers produce....

Smaller numbers produce...

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Estimate the pKa of an acid with a Ka of 1.2e-3:

SC.2) Calculate the actual pKa value from the SC.1:

SC.3) Give the [OH-] in a solution with pOH = 4.57

Answers: SC.1 / a little less than 3 SC.2 / 2.92 SC.3 / 2.69e-5

~ 20 ~

Due to the fact that they only partially dissociate or

ionize in water, weak acids and bases exist in a state

of equilibrium with their products. As a result, we

can use RICE to predict the extent to which

dissociation will occur as well as the amount of H+

or OH- which will be produced. Specialized versions

of K, the equilibrium constant, are used:

Ka or Kb = acid or base dissociation constants

Equation → RICE → K expression → solve for x

Ex.1) Calculate the [F-] in a 0.200 M HF solution once it

reaches equilibrium (Ka for HF = 6.6e-4) :

23.2.1 Weak Acid/Base Equilibria

~ 21 ~

Ex.2) What is the hydroxide concentration present in a

0.900 M solution of NH3 (Kb = 1.8e-5) :

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Calculate the [H+] present in a 0.100 M solution of

benzoic acid, C6H5COOH (Ka = 6.3e-5) :

Answers: SC.1 / [H+] = 2.51e-3 M

~ 22 ~

The percent dissociation for an acid or base in

solution is calculated by comparing the initial

molarity of the substance against the amount which

is consumed as the reaction reaches equilibrium. In

most cases, the value of "x" in the RICE expression is

the amount consumed. For a weak acid:

% dissociation = x • 100%

[HA]0

A similar equation for a weak base would simply

substitute "A" for "HA". In general, strong acids and

bases dissociate 100%, while weak acids and bases

dissociate partially, typically less than 5%.

Ex.1) A 0.0800 M solution of HAc contains [Ac-] = 1.2e-3

M at equilibrium. Determine the % dissociation:

23.2.2 Calculating Percent Dissociation

~ 23 ~

Ex.2) Calculate the % dissociation of a 0.500 M HAc

solution (Ka = 1.8e-5):

Ex.3) % dissociation of 1.0 M HCl (Ka = 1.3e6):

Ex.4) How is the value of "K" related to % dissociation?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Determine the % dissociation in a 0.800 M

methylamine solution (CH3NH2 , Kb = 4.5e-4):

Answers: SC.1 / 2.37% dissociation

~ 24 ~

Unlike strong acids, weak acids dissociate partially

and establish an equilibrium with their conjugate

bases. To determine pH, we often have to create a

RICE table and calculate the amount of H+ produced

via dissociation. For example, if you wanted to find

the pH of 0.100 M HAc (Ka = 1.8e-5):

Major Species: HAc H+ Ac

- H2O

R HAc(aq) ↔ H+(aq) + Ac

-(aq)

I 0.100 0 0

C -x +x +x

E 0.100 x x

1.8e-5 = x2

x = 1.3e-3 M

0.1

The [H+] produced by HAc is much higher than that

produced by water (1e-7), thus:

pH = -log(1.3e-3) = 2.89

23.2.3 pH of Weak Acid Solutions

~ 25 ~

Ex.1) Determine the pH of 2.500 M HAc:

Ex.2) Calculate the pH of 1.0e-5 M HCN (Ka = 6.1e-10)

~ 26 ~

SC.1) Lemons contain citric acid (HC6H7O7 , Ka = 7.4e-4)

at a concentration of ~0.30 M. Estimate the pH of

lemon juice from this data:

Answers: SC.1 / pH = 1.82

~ 27 ~

Within a conjugate pair, the acid will have a Ka value

and the base will have a Kb value. You can convert

between values using the equation shown below:

Ka • Kb = 1 x 10-14

Given each of the following substances and their K

values, determine the following: the conjugate, its

K value, and the matching equation.

Ex.1) Ammonia, NH3, Kb = 1.8e-5 :

Ex.2) Pyridine, C5H5N, has Kb = 5.9e-6 :

23.2.4 Converting Between Ka and Kb

~ 28 ~

Ex.3) Carbonic acid, H2CO3, has Ka = 4.4e-7

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Nitrous acid, HNO2, has Ka = 4.5e-4

SC.2) Phosphate ion, PO43-

, has Kb = 0.024

Answers: SC.1 / CB = NO2- Kb = 2.2e-11 / NO2

-(aq) + H2O(l) ↔ HNO2(aq) + OH

-(aq)

SC.2 / CA = HPO42-

Ka = 4.2e-13 / HPO42-

(aq) ↔ PO43-

(aq) + H+(aq)

~ 29 ~

Unlike strong bases, weak bases dissociate partially

and establish an equilibrium with their conjugate

acids. To determine pH, we often have to create a

RICE table and calculate the [OH-] produced via

dissociation. For example, if you wanted to find

the pH of 3.0 M NH3 (Kb = 1.8e-5):

Major Species: HAc H+ Ac

- H2O

R NH3(aq) + H2O(l) ↔ NH4+(aq) + OH

-(aq)

I 3.0 0 0

C -x +x +x

E 3.0 x x

1.8e-5 = x2

x = 7.3e-3 M

3.0

The [OH-] produced is much higher than that

produced naturally by water (1e-7), thus:

pOH = -log(3.0e-3) = 2.14 pH = 14 - 2.14 = 11.86

23.2.5 pH of Weak Base Solutions

~ 30 ~

Ex.1) Household ammonia has a [NH3] of 0.500 M. What

is the pH of this ammonia solution?

Ex.2) Methylamine, CH3NH2, has a Kb value of 4.4 x 10-4

.

Find the pH of a 0.65 M solution of methylamine:

~ 31 ~

SC.1) Pyridine, C5H5N has a Ka value of 5.9e-6. Determine

the pH of a 0.0500 M solution of pyridine:

Answers: SC.1 / pH = 10.732 (depending on how you rounded, your answer may be different.)

~ 32 ~

An acid is considered strong if it dissociates fully

in water. This is responsible for the incredibly high

Ka values for strong acids and allows us to make

several assumptions. For example, if you wanted to

find the pH of a 0.015 M solution of HCl:

Major Species: H+ Cl

- H2O

HCl fully dissociates, none remains as HCl(aq)

The H+

produced here is far greater than is normally

present in water (0.015 M >>> 1e-7 M) and chloride

is a neutral ion. We determine pH directly from the

[H+] produced by the dissociation of HCl:

pH = - log (0.015) = 1.82

Ex.1) Determine the pH of a 4.1e-5 mol dm-3

HNO3 sol'n:

Major Species: pH =

23.2.6 pH of Strong Acid Solutions

~ 33 ~

Ex.2) Calculate the pH of 9x10-2

M HClO4:

Major Species: pH =

Be wary - if the concentation of the acid is too low,

water becomes the greatest contributor to pH:

Ex.3) Determine the pH of a 9.05e-9 M HBr:

Major Species: pH =

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) pH of 0.10 M HNO3?

SC.2) pH of 6.90e-10 mol dm-3

HCl ?

SC.3) Estimate the [ ] level at which a strong acid no

longer contributes significantly to pH:

Answers: SC.1 / pH = 1.000 SC.2 / 7.0 SC.3 / any [ ] less than e-9, typically.

~ 34 ~

Like strong acids, bases must dissociate or ionize

fully to be considered strong. Assuming a high

enough concentration, we can typically determine

the pH indirecly from the [OH-] which is produced.

In a 2.8e-3 mol dm-3

NaOH solution:

Major Species: Na+ OH

- H2O

NaOH fully dissociates, none remains as NaOH(aq)

The OH- produced is far greater than is normally

present in water (2.8e-3 M >>> 1e-7 M) and sodium

is a neutral ion. We determine pH indirectly using

the [OH-]. (Find pOH, subtract from 14)

pH = 14 - (- log (2.8e-3)) = 11.45

Ex.1) Determine the pH of a 0.80 M KOH solution:

Major Species: pH =

23.2.7 pH of Strong Base Solutions

~ 35 ~

Ex.2) Calculate the pH of 3.34e-5 M Ba(OH)2:

Major Species: pH =

Be wary - if the concentation of the base is too low,

water becomes the greatest contributor to pH:

Ex.3) Determine the pH of 8.8x10-10

M LiOH:

Major Species: pH =

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) pH of 2.3e-11 mol dm-3

Ca(OH)2?

SC.2) pH of 0.78 M NaOH:

SC.3) At what concentration does an SB no longer

contribute significantly to pH?

Answers: SC.1 / pH = 7.00 SC.2 / 13.89 SC.3 / any [ ] less than e-9, typically.

~ 36 ~

A buffer is a solution which contains a weak acid

and its conjugate base. A buffered solution is

extremely resistant to changes in pH and can be

found in a variety of biological (homeostasis!) and

industrial systems. Buffers work by converting SA

and SB into WA and WB which have a substantially

smaller impact of pH:

A buffer made from acetic acid and sodium acetate

would produce an equlibrium between the

following two substances:

HAc(aq) ↔ Ac-(aq)

Ex.1) If added to this buffer, a strong base such as NaOH

would react with __________ and increase the

concentration of __________.

Ex.2) If added to this buffer, a strong acid such as HCl

would react with __________ and increase the

concentration of __________.

23.3.1 Buffers and How They Work

~ 37 ~

A mixture of ammonia and ammonium chloride are

dissolved in water. Describe how the buffer will

respond to each of the following:

Ex.3) The addition of a strong acid will convert _________

into _________ and slightly _________ the pH of

the solution.

Ex.4) The addition of a strong base will convert ________

into _________ and slightly _________ the pH of

the solution.

Buffer capacity is a rough measure of how long a

buffer can continue to neutralize SA and SB before

running out of either the WA or its conjugate base.

When too much of a strong acid or base is added, a

buffer is said to have "collapsed" and no longer

resists changes in pH.

Ex.5) In terms of molarity, how could you make a high-

capacity buffer?

~ 38 ~

Explain why a buffer will eventually collapse if you

continue to add a strong acid or base:

Ex.6) The strong acid will eventually convert all of the ___

into its conjugate ______. Any further _____ added

will not react and will instead be in excess. As a

result the pH will ________ rapidly.

Ex.7) The strong base will eventually convert all the ____

into its conjugate ______. Any further _____ added

will not react and will instead be in excess. As a

result the pH will ________ rapidly.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Could a buffer be made from nitric acid and sodium

nitrate?

SC.2) Name the sodium salt which would form a buffer

with carbonic acid:

Answers: SC.1 / no, nitric acid is a SA and sodium nitrate is neutral. SC.2 / NaHCO3, sodium

hydrogen carbonate or sodium bicarbonate.

~ 39 ~

Buffers are typically prepared in one of three ways.

Options 2 and 3 are most common as it is rare to

have a weak acid and its conjugate both in stock:

1.) Dissolve a weak acid and is conjugate base (usually

a salt) in water.

2.) Start with a weak acid and add a smaller amount of

strong base such as NaOH, producing a mixture of

the surviving weak acid and its conjugate base.

3.) Start with a weak base (typically a salt), and add a

small amount of a strong acid such as HCl. This

produces a mixture of conjugate weak acid and the

surviving weak base.

Ex.1) What could you add to a solution of sodium cyanide

to produce a buffer?

Ex.2) Acetic acid solution and a small amount of _______

could be mixed to produce a buffer.

23.3.2 Basics of Buffers Preparation

~ 40 ~

Determine the products of each of the following

buffer preparation mixtures. Does a buffer result?

Ex.3) 1.40 mol NaAc →

0.80 mol HCl

Ex.4) 0.900 mol NaOH →

1.000 mol H3PO4

Ex.5) 2.40 mol HAc →

2.80 mol KOH

Ex.6) 0.100 mol NaF →

0.075 mol HCl

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) 0.200 mol HCl + 0.240 mol NH3

SC.2) Name two ways you could make a buffer from 0.100

mol citric acid:

Answers: SC.1 / 0.200 mol NH4+ or NH4Cl and 0.040 mol NH3 SC.2 / Add a salt of citrate such

as sodium citrate OR add less than 0.100 mol of a strong base such as NaOH.

~ 41 ~

The easiest way of measuring the pH of a buffer

solution is to use the Henderson-Hasselbalch or "H-

H" equation shown below:

pH = pKa + log (A-/HA)

The value of A- and HA within this equation

represent the moles (or molarity) of the acid (HA)

and its conjugate base (A-) within the buffer.

Determine the pH of each of the following buffer

solutions when prepared as described:

Ex.1) 0.45 mol NaF is added to a solution containing 0.12

mol HF (Ka = 6.6e-4) in a total volume of 250 mL:

23.3.3 Measuring the pH of Buffer Sol'ns

~ 42 ~

Ex.2) 70.0 mL of 0.200 M HAc (Ka = 1.8e-5) is added to

100.0 mL of 0.100 M KAc:

Ex.3) 0.100 mol HAc + 0.040 mol NaOH

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) What is the pH of a sol'n containing 0.500 mol NaAc

and 0.250 mol HCl?

SC.2) A solution with a pH of 5.61 contains 0.200 mol of

an unknown acid and 0.350 mol of its conjugate

base. What is the Ka for this unknown acid?

Answers: SC.1 / pH = 4.74 SC.2 / Ka = 4.3e-6

~ 43 ~

A buffer is said to be at the halfway point when

[HA] = [A-]. You can think of this as the point at

which the buffer is "centered" and is therefore most

effective at resisting large pH changes due to the

addition of strong acids or bases. When this occurs,

the log (A/HA) term in the H-H equation becomes

equal to zero (log 1 = 0). As a result,

pH = pKa (At the halfway point ONLY)

(1-3) Given a buffer made of HAc/Ac- (Ka = 1.8e-5):

Ex.1) What is the pH at the halfway point for this buffer?

By comparing the current pH to its center, we can

infer which species is in the majority:

Ex.2) At a pH of 5.000, [HAc] ___ [Ac-]

Ex.3) At a pH of 4.200, [HAc] ___ [Ac-]

23.3.4 The Halfway Point

~ 44 ~

Ex.4) 0.15 mol NaOH is added to a solution containing

0.50 mol HF (pKa = 3.14). What can be inferred

about the resulting solution?

[HF] ___ [F-] pH ___ pKa

A buffer's capacity or ability to resist changes in pH

is highest near its halfway point, as represented by

its pKa value. HAc, for example, is most effective at

pH = 4.74. For this reason, we tend to choose

buffers whose pKa values are very close to the pH

value we wish to use or create.

Ex.5) HAc buffers at pH = 6.0 are possible. How could this

be achieved?

Ex.6) The buffer in Ex.5 would have a substantially lower

capacity that one at a pH of 4.74. Why?

~ 45 ~

Ex.7) You wish to produce a buffer at pH = 9.0 for a

biochemical study. Which of the following

substances would be most appropriate?

HF (Ka = 7.2e-4)

HCN (Ka = 6.2e-10)

HSO4- (Ka = 1.2e-2)

HC2H3O2 (Ka = 1.8e-5)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

SC.1) Given that Ka = 6.2-10 for HCN, calculate the half-

way point for a HCN / CN- buffer:

SC.2) Which of the following would be true for a HCN

buffer at a pH of 9.00?

pH ___ pKa [HCN] ___ [CN-]

SC.3) What about a HCN buffer at pH = 10.5?

pH ___ pKa [HCN] ___ [CN-]

Answers: SC.1 / pKa = 9.21 SC.2 / pH < pKa [HCN] > [CN-] SC.3 / pH > pKa [HCN] < [CN

-]

~ 46 ~

Buffers are commonly made to have a highly

specific or "targeted" pH for a variety of uses. This

requires a system of equations and allows us to

solve for the exact amounts of HA and A- which

must be present in the solution:

Ex.1) You wish to prepare a HAc (Ka = 1.8e-5) buffer at a

pH of 5.25 with a total molarity of 1.00 M. What are

the [HAc] and [Ac-] necessary to do so?

Solve the H-H equation for "A/HA"

Create an equation for the total capacity of the

buffer, such that "HA + A = total molarity or moles"

23.3.5 Targeted Preparation of Buffers

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Substitute the first equation into the second and

solve for either A or HA.

Solve for the remaining value, either A or HA

Ex.2) What mass of NaOH, in g, would have to be added

to 500.0 mL of 0.200 M HAc in order to produce a

buffer with pH = 4.10? (Assume no change in V)

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Ex.3) Given a 1.00 M NH3 (Kb = 1.8e-5) solution and a 1.00

M HCl solution, how many mL of each will have to

be combined to produce a buffer with pH = 10.00

and a total volume of 2.00 L?

Ex.4) What masses of malic acid (HMal, Mr = 134.09 g

mol-1

, Ka = 3.5e-4) and potassium malate (Mr =

172.18 g mol-1

) would have to be added to 1.00 L of

water to produce a 0.500 M buffer at pH = 3.00?

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SC.1) What mass of NaOH, in g, would have to be added

to 1.50 L of 1.00 M HAc to produce a buffer at pH =

5.00?

SC.2) What volumes of 1.00 M HF (Ka = 7.2e-4) and 1.00

M NaF would be combined to produce 250.0 mL of

a 1.00 M buffer with pH = 3.00?

Answers: SC.1 / ~38 g NaOH SC.2 / ~150 mL HF sol'n and 100 mL NaF sol'n