3.6: ACIDS AND BASES… Workbook pgs 145- 148… Buffered Solutions…

BUFFER SOLUTIONS Buffer

Causes solutions to be resistant to a change in pH when a strong acid or base is added

Mixture of a weak acid and salts of conjugate bases If you add 0.010 mol of NaOH to 1.0 L of pure water the

pH drops from 7 to 2 If you add 0.010 mol of NaOH to 1.0 L of blood (pH of 7.4)

the pH will drop to 7.3 Two requirements for a buffer:

An acid capable of reacting with added OH- ions and a base that can consume added H3O+ ions.

The acid and base must not react with each other (acetic acid and acetate ion or ammonia and

ammunium ions)

BUFFERS If a small amount of hydroxide is added to an

equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

BUFFERS: Similarly, if acid is added, the F− reacts with

it to form HF and water.

MATH TIME: What is the pH of a 0.700M solution of acetic acid?

Ka = 1.8 x 10-5

pH = 2.45 What is the pH of a 0.600M solution of sodium

acetate? Kb = 5.6 x 10-10

pH = 9.26 What is the pH of an acetic acid/sodium acetate

buffer with [CH3CO2H] = 0.700M and [CH3CO2-] =

0.600M? Ka = 1.8 x 10-5

pH = 4.68

BUFFER CALCULATIONS: Consider the equilibrium constant expression

for the dissociation of a generic acid, HA:

HA + H2O H3O+ + A−

[H3O+] [A−][HA]

Ka =

© 2009, Prentice-Hall, Inc.

BUFFER CALCULATIONS

Rearranging slightly, this becomes

[A−][HA]Ka = [H3O+]

Taking the negative log of both side, we get

[A−][HA]−log Ka = −log [H3O+] + −log

pKa

pHacid

base

© 2009, Prentice-Hall, Inc.

BUFFER CALCULATIONS So

pKa = pH − log [base][acid]

• Rearranging, this becomes

pH = pKa + log [base][acid]

• This is the Henderson–Hasselbalch equation.

HENDERSON-HASSELBALCH EQUATION: For when the concentrations of the acid and

conjugate base are quite large… … or when you want a short cut to a buffer

solution

PRACTICE: What is the pH of a buffer that is 0.12 M in

lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 10−4.

pH = pKa + log [base][acid]

pH = −log (1.4 10−4) + log(0.10)(0.12)

pH = 3.85 + (−0.08) = 3.77

PRACTICE: Benzoic acid (C6H5CO2H, 2.00g) and sodium

benzoate (NaC6H5CO2, 2.00g) are dissolved in enough water to make 1.00L of solution. Calculate the pH of the solution using the Henderson-Hasselbalch equation. Ka = 6.3 x 10-5

2.00g benzoic acid = 0.0164 mol 2.00g sodium benzoate = 0.0139 mol pH = 4.20 + log(0.0139/0.0164) = 4.13

PREPARING BUFFER SOLUTIONS Buffers work on a pH range:

The pH range is the range of pH values over which a buffer system works effectively.

It is best to choose an acid with a pKa close to the desired pH

PREPARING BUFFER SOLUTIONS: To be useful, a buffer solution must have 2

characteristics: pH control:

pH = pKa + log[conjugatebbase]/[acid] Acid is chosen whose pKa is near the intended value of

the pH Exact value of pH is then achieved by adjusting the

acid-to-conjugate base ratio Buffer Capacity:

Buffer should have the ability to keep the pH approximately constant after the addition of reasonable amounts of acid and base.

PREPARING BUFFER SOLUTIONS: You wish to prepare 1.0L of a buffer solution

with a pH of 4.30. A list of possible acids (and their conjugate bases) follows:

Which combination should be selected, and what should the ratio of acid to conjugate base be?

Acid Conjugate Base

Ka pKa

HSO4- SO4

- 1.2 x 10-2 1.92CH3CO2H CH3CO2

- 1.8 x 10-5 4.75HCO3 CO3

- 4.8 x 10-11 10.32

PREPARING BUFFER SOLUTIONS:

[CH3CO2−]

[CH3CO2H]Ka = [H3O+]

[CH3CO2−]

[CH3CO2H]Ka =

[H3O+]

[CH3CO2−]

[CH3CO2H]5.0 x 10-5 =

1.8 x 10-5

PREPARING BUFFER SOLUTIONS:

Ratio of conjugate base to acid = 0.36

pH = pKa + log [base][acid]

PREPARING BUFFER SOLUTIONS: The relative number of moles of acid and

conjugate base is important in determining the pH of a buffer solution, not the concentration. Why?

Volume cancels out! That means that diluting a buffer solution will not

change its pH