Henderson-Hasselbalch EquationHenderson-Hasselbalch 方程

Big Idea · Structure and Properties (SAP)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 8~45 min

Learning goals

  • Derive the Henderson–Hasselbalch equation from the Ka expression, and use it to calculate the pH of a buffer
  • Do the neutralization first, then use the ratio of the amounts of the two components after reaction to find the pH
  • Choose a suitable conjugate acid–base pair for a target pH, and calculate the ratio and amounts needed

Before you start

  • pH and pKa (U8.7)
  • Properties of buffers (U8.8)

The Henderson–Hasselbalch equation

Taking the negative logarithm of both sides of [H₃O⁺] = Ka × [HA] ÷ [A⁻] gives the Henderson–Hasselbalch equationHenderson–Hasselbalch 方程:

pH=pKa+log⁡[AX−][HA]\text{pH} = \text{p}K_a + \log\frac{[\ce{A-}]}{[\ce{HA}]}

Both components are in the same solution, so the ratio of concentrations equals the ratio of amounts, and you can substitute moles directly. The equation shows that:

  • when [A⁻] = [HA], pH = pKa;
  • the more of the base component, the higher the pH; each factor of 10 in the ratio changes the pH by 1;
  • when the ratio is between 0.1 and 10, the pH is within pKa ± 1, which is the useful range of a buffer.

The equation assumes that the equilibrium concentrations of the two components equal the amounts added, that is, that the ionization of the weak acid and the reaction of the conjugate base with water are both negligible. This holds when the concentrations aren’t too small and the ratio isn’t too extreme.

For a weak-base buffer (such as NH₃ / NH₄⁺), treat NH₄⁺ as HA and NH₃ as A⁻, and use the pKa of NH₄⁺ (14.00 − pKb = 9.26).

Drag the amounts of the two components and compare the equation’s result with the exact calculation; then try a target pH and see which pair is best:

Try it: the Henderson–Hasselbalch equation

Example 1: pH of an ammonia buffer

A solution has [NH₃] = 0.20 M and [NH₄Cl] = 0.30 M. For NH₃, Kb = 1.8 × 10⁻⁵. Find the pH.

Step-by-step solution

  1. Identify the acid and base components. The Henderson–Hasselbalch equation needs the pKa of the conjugate acid. Here the weak acid HA is NH₄⁺ (0.30 M) and the conjugate base A⁻ is NH₃ (0.20 M):

    pH=pKa+log⁡[NHX3][NHX4X+]\text{pH} = \text{p}K_a + \log\frac{[\ce{NH3}]}{[\ce{NH4+}]}

  2. Get the pKa of NH₄⁺ from Kb. For a conjugate acid–base pair, pKa + pKb = 14.00:

    pKb=−log⁡(1.8×10−5)=4.74pKa=14.00−4.74=9.26\begin{aligned} \text{p}K_b &= -\log(1.8 \times 10^{-5}) = 4.74 \\ \text{p}K_a &= 14.00 - 4.74 = 9.26 \end{aligned}

  3. Substitute.

    pH=9.26+log⁡0.200.30=9.26+(−0.18)=9.08\begin{aligned} \text{pH} &= 9.26 + \log\frac{0.20}{0.30} \\ &= 9.26 + (-0.18) = \boxed{9.08} \end{aligned}

  4. Check. There is more of the acid component (NH₄⁺), so the pH should be below pKa: 9.08 < 9.26, which makes sense.

Example 2: making a buffer at pH 4.00

A 1.00 L buffer at pH 4.00 is to be made from HCOOH (Ka = 1.8 × 10⁻⁴, pKa = 3.74) and HCOONa, and it contains 0.100 mol HCOOH. What mass of HCOONa (M = 68.01 g/mol) is needed?

Step-by-step solution

  1. Write the Henderson–Hasselbalch equation with the known pH and pKa.

    4.00=3.74+log⁡[HCOOX−][HCOOH]4.00 = 3.74 + \log\frac{[\ce{HCOO-}]}{[\ce{HCOOH}]}

  2. Find the ratio. Rearrange, then take the antilog:

    log⁡[HCOOX−][HCOOH]=4.00−3.74=0.26[HCOOX−][HCOOH]=10 0.26=1.82\begin{aligned} \log\frac{[\ce{HCOO-}]}{[\ce{HCOOH}]} &= 4.00 - 3.74 \\ &= 0.26 \\ \frac{[\ce{HCOO-}]}{[\ce{HCOOH}]} &= 10^{\,0.26} = 1.82 \end{aligned}

  3. Moles of HCOO⁻. Both share one volume, so the mole ratio is also 1.82:

    n(HCOOX−)=1.82×0.100=0.182 mol\begin{aligned} n(\ce{HCOO-}) &= 1.82 \times 0.100 \\ &= 0.182\ \text{mol} \end{aligned}

  4. Convert to mass. HCOONa dissociates completely, so 1 mol of HCOONa gives 1 mol of HCOO⁻:

    m=0.182×68.01=12.4 gm = 0.182 \times 68.01 = \boxed{12.4\ \text{g}}

  5. Check. The target pH is above pKa, so there should be more of the base component: 1.82 > 1, which makes sense. The target is within pKa ± 1 (2.74–4.74), so this acid–base pair is a suitable choice.

Practice: after mixing (or at the given ratio), how does the pH of the buffer compare with pKa?
  1. [A⁻] = 2[HA]
  2. n(HA) = n(A⁻)
  3. 0.10 mol HA + 0.05 mol A⁻
  4. 0.20 mol HA + 0.10 mol NaOH
  5. 0.20 mol A⁻ + 0.05 mol HCl
  6. 0.10 mol HA + 0.08 mol NaOH
  7. 0.30 mol A⁻ + 0.20 mol HCl
  8. A 1 : 1 buffer diluted tenfold with water

Common mistakes

Mistake 1: “Substitute the amounts added directly, without working out the reaction first.” If a strong acid or base was added, first find how much HA and A⁻ remain after the reaction, then substitute.

Mistake 2: “The ratio is [HA] ÷ [A⁻].” The base component A⁻ goes on top. To check: with more of the base component, the pH should be above pKa.

Mistake 3: “The equation works for any weak acid solution.” It applies only to buffers in which both components are present in large amounts. With only the weak acid (no A⁻), use Ka and an ICE table (U8.3).

Quick check

What is the pH of a buffer with [HF] = 0.15 M and [NaF] = 0.30 M? (pKa of HF = 3.19)

A pH 10.00 buffer is to be made from NH₄Cl and NH₃ (pKa of NH₄⁺ = 9.26). What should [NH₃] ÷ [NH₄⁺] be?

Which pair is most suitable for a buffer at pH 10.0?

A 1.00 L buffer contains 0.050 mol HA and 0.050 mol A⁻. After 0.010 mol NaOH is added, the pH equals:

In which case is the pH from the Henderson–Hasselbalch equation least reliable?

A buffer has pH = pKa − 1. What is [A⁻] ÷ [HA]?

AP-style free response

AP-style free response

A student needs to prepare a buffer at pH 5.00. The weak acids available are HCOOH (pKa 3.74) and CH₃COOH (pKa 4.74). The student has 50.0 mL of 0.300 M weak acid solution and a 0.200 M NaOH solution.

(a) Which weak acid should the student choose? Explain.

(b) Calculate the required [A⁻] ÷ [HA].

(c) Calculate the volume of 0.200 M NaOH that must be added.

(d) 1.00 mmol HCl is then added to the finished buffer. Calculate the pH.

(e) If the student mistakenly added 80.0 mL of NaOH, would the solution still be a buffer? Explain.

Model answer and scoring

(a) CH₃COOH. Its pKa (4.74) is 0.26 from the target pH, within ± 1; the pKa of HCOOH is 1.26 away, outside the useful range.

(b) log of the ratio = 5.00 − 4.74 = 0.26, so the ratio = 100.26 = 1.82.

(c) There are 15.0 mmol of weak acid in all. If x mmol NaOH is added, x ÷ (15.0 − x) = 1.82, so x = 9.68 mmol and the volume = 9.68 ÷ 0.200 = 48.4 mL.

(d) After the reaction there are 8.68 mmol CH₃COO⁻ and 6.32 mmol CH₃COOH, so pH = 4.74 + log(8.68 ÷ 6.32) = 4.88.

(e) No. 80.0 mL of NaOH contains 16.0 mmol OH⁻, more than the 15.0 mmol of CH₃COOH, so all the weak acid is neutralized and the solution contains only CH₃COO⁻ and excess OH⁻.

Scoring: (a) pKa close to the target pH; (b) ratio; (c) setup and volume; (d) amounts after reaction and pH; (e) the weak acid is used up.

Summary

  • pH = pKa + log([A⁻] ÷ [HA]); within one solution, the ratio of amounts can replace the ratio of concentrations.
  • When strong acid or base has been added, work out the reaction first, then substitute the amounts after reaction.
  • Choosing a buffer pair: make pKa as close as possible to the target pH (within ± 1), then use the ratio 10pH − pKa to find the amounts.

AP tip: When using the Henderson–Hasselbalch equation, state clearly which species is A⁻ and which is HA, and check the direction of the result: more of the base component means a pH above pKa. For questions about choosing a buffer pair, give the reason “pKa is close to the target pH.”

Practice this chapter: Unit 8 · Acids and Bases · 12 questionsPractice this topic: U8.9 · Henderson-Hasselbalch Equation · 5 questions

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