Buffer Solutions Buffers Convert a Strong Acid Into a Weak Acid And

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Buffer solutions

Buffers convert a strong acid into a weak acid and convert a strong base into a weak base.

So buffers can resist changes in pH. Fig. 16.1, p. 575

Buffers work because they consist of a weak acid and its conjugate base (or a weak base and its conjugate acid).

Example: A buffer solution contains both acetic acid and its conjugate base (acetate).

Addition of strong acid yields more weak acid. + - H3O + CH3COO  H2O(l) + CH3COOH

Addition of strong base yields more weak base. - - OH + CH3COOH  H2O(l) + CH3COO

Chemistry 103 Spring 2011

Note these are not equilibrium reactions. The strong component reacts completely to consume part of the buffer and form products. Fig. 16.2, p. 582

The best buffers have about equal amounts of - each conjugate: [CH3COOH] ≈ [CH3COO ], to resist strong acid and strong base equally.

Example: 1.0 L of a buffer solution contains

0.10 moles of CH3COOH and 0.080 moles of NaCH3COO. What is the pH of the buffer solution? + - HA + H2O(l) H3O + A I

2 Chemistry 103 Spring 2011

Practice: Calculate the pH of a buffer solution that has 0.15 mol lactic acid CH3CHOHCOOH -4 (HA, Ka = 1.0 x 10 ) and 0.20 mol sodium lactate NaCH3CHOHCOO (NaA) and enough water to make 1.00 L of solution.

3 Chemistry 103 Spring 2011

Alternatively, + - HA + H2O(l) H3O + A

+ - + Ka = [H3O ] [A ] = [H3O ] [conj. base] [HA] [conj. acid]

Take -log of both sides of equation

+ pKa = -log[H3O ] + (-log([conj. base]/[conj. acid]))

pKa = pH – log([conj. base]/[conj. acid])

pH = pKa + log([conj. base]/[conj. acid]) Henderson-Hasselbalch equation (H-H equation)

- Example: CH3COOH – CH3COO buffer (*see earlier Example) pH = -log(1.8 x 10-5) + log (0.08 M / 0.10 M)

4 Chemistry 103 Spring 2011

Limitations of H-H equation: 1. The ratio [conj. base]/[conj. acid] should be between 0.10 and 10. In other words, one component does not exceed the other by more than a factor of 10. (so the solution is truly a buffer, and not mainly an acid or base solution)

[conj. base]/[conj. acid] ≤ 10 AND [conj. base]/[conj. acid] ≥ 0.1

- Example: CH3COOH – CH3COO buffer (*) pH = -log(1.8 x 10-5) + log (0.08 M / 0.10 M)

[conj. base]/[conj. acid] = 0.08

2. The concentrations of [conj. base] and

[conj. acid] should exceed the Ka by a factor of at least 100. (because we assume +x and –x are small compared to the concentrations)

Example: [conj. base] = 0.08 M + x ≈ 0.08 M [conj. acid] = 0.10 M – x ≈ 0.10 M

5 Chemistry 103 Spring 2011

Changes to buffer solutions

Addition of strong acid yields more weak acid. + - H3O + CH3COO  H2O(l) + CH3COOH

Addition of strong base yields more weak base. - - OH + CH3COOH  H2O(l) + CH3COO

Example: 1.0 L of a buffer solution contains

0.10 moles of CH3COOH and 0.080 moles of NaCH3COO. What is the pH of the solution after 0.030 moles of KOH are added?

- - OH + CH3COOH  H2O(l) + CH3COO I 0.030 mol 0.10 - 0.080 C End

+ - CH3COOH + H2O(l) H3O + CH3COO

6 Chemistry 103 Spring 2011

Ka =

Solve for x, convert to pH.

OR use H-H equation, pH = pKa + log([conj. base]/(conj. acid]) pH =

7 Chemistry 103 Spring 2011

Practice: Using the lactic acid-acetate buffer solution from the previous Practice problem, calculate the pH after addition of 0.050 mol HCl to the buffer.

8 Chemistry 103 Spring 2011

Choosing buffers based on pKa

In many experiments, we need a buffer that maintains the solution at a specific pH.

We may need an acidic or a basic buffer, depending on the experiment.

The Henderson-Hasselbalch equation can help us choose a buffer that has the pH we want.

pH = pKa + log([conj. base]/[conj. acid])

With equal amounts of conjugate acid and base (preferred so buffers can resist base and acid equally), then …

pH = pKa + log(1) = pKa + 0 = pKa

So choose conjugates with a pKa closest to our target pH.

9 Chemistry 103 Spring 2011

Example: You need a buffer with pH of 7.80. Which conjugate acid-base pair should you use, and what is the molar ratio of its components?

10 Chemistry 103 Spring 2011

Practice: Choose the best conjugate acid-base pair for preparing a buffer with pH 5.00. What is the molar ratio of the buffer components?

11 Chemistry 103 Spring 2011 buffer capacity: the amount of strong acid or strong base that can be added to a buffer without changing its pH by more than 1 unit.

Example: What is the capacity of the buffer prepared in Practice, p. 6 (lecture notes)?