AP Chemistry · Lesson 2

Buffers and the Henderson-Hasselbalch equation

Know Know it

A buffer contains a weak acid and its conjugate base, both in substantial amounts (or a weak base and its conjugate acid). The acid reacts with added hydroxide, HA+OH−→A−+H2O\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}, and the base reacts with added hydronium, A−+H3O+→HA+H2O\text{A}^- + \text{H}_3\text{O}^+ \rightarrow \text{HA} + \text{H}_2\text{O}. Either way, the strong acid or base is used up and only the ratio of A−^- to HA changes.

The pH of a buffer comes from the Henderson-Hasselbalch equation: pH=pKa+log⁡[A−][HA]\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}. When the two are equal, pH=pKa\text{pH} = \text{p}K_a. Because both share the same volume, you may use a ratio of moles in place of a ratio of molarities.

A buffer works best within about one pH unit of its pKa\text{p}K_a (a base-to-acid ratio between 0.1 and 10). To make a buffer at a target pH, pick a weak acid whose pKa\text{p}K_a is close to it. Buffer capacity grows with the amounts of acid and base present.

Worked example Easy

A buffer is 0.30 M in the conjugate base A−\text{A}^- and 0.15 M in the weak acid HA, which has pKa=4.20\text{p}K_a = 4.20. What is the pH of the buffer?

  1. 3.90
  2. 4.50Answer
  3. 4.20
  4. 6.20

How to solve it

  1. Ratio of base to acid: 0.30/0.15=2.00.30/0.15 = 2.0.
  2. log⁡2.0=0.30\log 2.0 = 0.30.
  3. pH=4.20+0.30=4.50\text{pH} = 4.20 + 0.30 = 4.50; more base than acid puts the pH above the pKa\text{p}K_a.

Why each choice is right or wrong

  • A. Incorrect. A pH of 3.90 puts the acid on top of the ratio, log⁡(0.15/0.30)\log(0.15/0.30), but the base goes in the numerator.
  • B. Correct. pH=4.20+log⁡(0.30/0.15)=4.20+0.30=4.50\text{pH} = 4.20 + \log(0.30/0.15) = 4.20 + 0.30 = 4.50.
  • C. Incorrect. A pH of 4.20 is the pKa\text{p}K_a, which is the pH only when acid and base are equal; here the base is twice the acid.
  • D. Incorrect. A pH of 6.20 adds the ratio 2 itself to the pKa\text{p}K_a instead of its log, 0.30.

Trap React first, then use the buffer equation

When strong acid or base is added to a buffer, do the neutralization in moles first: the added hydroxide turns that many moles of HA into A−^-. Only then put the new amounts into the Henderson-Hasselbalch equation. Plugging the old amounts in, or adding the hydroxide to the wrong side, gives the wrong pH.

Explain Explain it

Task verbs: calculate, explain

Work a buffer problem in two steps. Stoichiometry: in moles, subtract the added strong acid or base from the side it reacts with and add the same amount to the other side. Equilibrium: put the new mole ratio into the Henderson-Hasselbalch equation.

This also shows why a buffer resists change. Adding 0.010 mol of hydroxide to pure water raises the pH by many units, but in a buffer with 0.10 mol each of HA and A−^- the ratio only goes from 1.0 to 0.11/0.09, about 1.2, a change of under 0.1 pH unit.

Worked example Medium

A 1.00 L buffer contains 0.100 mol of a weak acid HA (pKa=6.80\text{p}K_a = 6.80) and 0.100 mol of its sodium salt NaA. A student adds 0.020 mol of HCl; the volume change is negligible.

What is the pH after the HCl is added?

  1. 6.98
  2. 6.80
  3. 1.70
  4. 6.62Answer

How to solve it

  1. Stoichiometry: A−+H3O+→HA+H2O\text{A}^- + \text{H}_3\text{O}^+ \rightarrow \text{HA} + \text{H}_2\text{O}. A−\text{A}^-: 0.100 - 0.020 = 0.080 mol. HA: 0.100 + 0.020 = 0.120 mol.
  2. Equilibrium: pH=6.80+log⁡(0.080/0.120)=6.80−0.18=6.62\text{pH} = 6.80 + \log(0.080/0.120) = 6.80 - 0.18 = 6.62.

Why each choice is right or wrong

  • A. Incorrect. A pH of 6.98 adds the 0.020 mol to the base instead of the acid; added HCl turns A−\text{A}^- into HA.
  • B. Incorrect. A pH of 6.80 assumes a buffer holds the pH exactly, but the ratio of base to acid does change a little.
  • C. Incorrect. A pH of 1.70 is the pH of 0.020 M HCl in water, which ignores the buffer that uses up the acid.
  • D. Correct. HCl turns 0.020 mol of A−\text{A}^- into HA, giving 0.080 mol of A−\text{A}^- and 0.120 mol of HA, so the pH is 6.80+log⁡(0.080/0.120)=6.626.80 + \log(0.080/0.120) = 6.62.

Explain Your turn

A buffer is made from the weak acid HA and its sodium salt NaA. Explain how this buffer keeps its pH nearly the same when a small amount of NaOH is added. Include a chemical equation for the reaction that happens.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Writes the reaction of the added hydroxide with the weak acid
  • Says that the added hydroxide is used up by the weak acid, so it does not build up in the solution

Model answer

The weak acid in the buffer reacts with the added hydroxide: HA+OH−→A−+H2O\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}. The strong base is used up by HA and replaced by A−^-, a much weaker base, so hydroxide does not build up in the solution. The ratio of A−^- to HA rises only slightly, so the pH rises only a little.

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