Buffer Capacity缓冲容量

Big Idea · Structure and Properties (SAP)AP Science Practice 6 · ArgumentationAP link · Unit 8~40 min

Learning goals

  • Explain what buffer capacity means, the amount of strong acid or base a buffer can neutralize before its pH changes significantly
  • Explain that buffer capacity depends on the amounts of the two components, so a more concentrated or larger buffer has a larger capacity
  • Explain how the ratio of the two components affects the capacity for acid and for base, and calculate it

Before you start

  • Properties of buffers (U8.8)
  • Henderson-Hasselbalch equation (U8.9)

Buffer capacity

A buffer can resist only a limited amount of acid or base. Buffer capacity缓冲容量 is the amount of strong acid or strong base a buffer can neutralize before its pH changes significantly (often taken as a change of 1).

  • Capacity for acid depends on the amount of conjugate base A⁻, which uses up added H₃O⁺.
  • Capacity for base depends on the amount of weak acid HA, which uses up added OH⁻.

So the larger the amounts of the two components (a higher concentration or a larger volume), the larger the buffer capacity. Two buffers with the same pH can have very different capacities: the pH depends on the ratio, while the capacity depends on the amounts.

From the Henderson–Hasselbalch equation, a change of 1 in pH means the ratio [A⁻] ÷ [HA] becomes 10 times or a tenth of what it was. When the two components are close to 1 : 1, the capacities for acid and base are equal; when one component is in excess, the capacity in that direction is larger.

Compare acetate buffers with different concentrations and ratios, and add HCl or NaOH:

Try it: buffer capacity

Each is 100 mL of acetate buffer, ignoring volume changes; pH is calculated exactly from the charge balance.

Example 1: same pH, different capacity

There are two 100 mL acetate buffers (pKa 4.74): A has 0.50 M each of CH₃COOH and CH₃COO⁻, and B has 0.050 M of each. 20 mmol HCl is added to each (ignore the volume change). Find the pH of each.

Step-by-step solution

  1. Starting point. Both have a 1 : 1 ratio and log 1 = 0, so both have pH = pKa = 4.74. Amount of each component (mL × mol/L = mmol):

    A: 100×0.50=50 mmolB: 100×0.050=5.0 mmol\begin{aligned} \text{A:} &\ 100 \times 0.50 = 50\ \text{mmol} \\ \text{B:} &\ 100 \times 0.050 = 5.0\ \text{mmol} \end{aligned}

  2. HCl added to A. The added H₃O⁺ is used up by CH₃COO⁻: CH₃COO⁻ + H₃O⁺ → CH₃COOH + H₂O (in mmol):

    CH₃COO⁻H₃O⁺CH₃COOH
    Before502050
    Change−20−20+20
    After30070
    pH=4.74+log⁡3070=4.74−0.37=4.37\begin{aligned} \text{pH} &= 4.74 + \log\frac{30}{70} \\ &= 4.74 - 0.37 = \boxed{4.37} \end{aligned}

    A drop of only 0.37.

  3. HCl added to B. There is only 5.0 mmol CH₃COO⁻, not enough to use up 20 mmol H₃O⁺ (in mmol):

    CH₃COO⁻H₃O⁺CH₃COOH
    Before5.0205.0
    Change−5.0−5.0+5.0
    After01510

    What’s left is strong acid, which sets the pH (H₃O⁺ from the acetic acid is negligible):

    [HX3OX+]=15 mmol100 mL=0.15 MpH=−log⁡0.15=0.82\begin{aligned} [\ce{H3O+}] &= \frac{15\ \text{mmol}}{100\ \text{mL}} = 0.15\ \text{M} \\ \text{pH} &= -\log 0.15 = \boxed{0.82} \end{aligned}

  4. Conclusion. The two solutions start at the same pH, but B’s capacity is too small, and 20 mmol HCl overwhelms it: the pH plunges from 4.74 to 0.82.

Example 2: different ratio, different capacity in each direction

A 1.00 L buffer contains 0.30 mol CH₃COOH and 0.10 mol CH₃COO⁻ (pKa 4.74). How much NaOH, and how much HCl, can it neutralize before its pH changes by 1?

Step-by-step solution

  1. Starting pH.

    pH=4.74+log⁡0.100.30=4.74−0.48=4.26\begin{aligned} \text{pH} &= 4.74 + \log\frac{0.10}{0.30} \\ &= 4.74 - 0.48 = 4.26 \end{aligned}

  2. A pH change of 1 means a 10-fold change in the ratio. By the Henderson–Hasselbalch equation, raising the pH by 1 raises log(ratio) by 1, multiplying the ratio by 10; lowering the pH by 1 divides it by 10:

    up by 1: 13×10=103down by 1: 13÷10=130\begin{aligned} \text{up by 1:} &\ \tfrac{1}{3} \times 10 = \tfrac{10}{3} \\ \text{down by 1:} &\ \tfrac{1}{3} \div 10 = \tfrac{1}{30} \end{aligned}

  3. Adding NaOH: let n mol be added. OH⁻ turns n mol of CH₃COOH into CH₃COO⁻:

    0.10+n0.30−n=1033(0.10+n)=10(0.30−n)0.30+3n=3.00−10n13n=2.70n=0.21 mol\begin{aligned} \frac{0.10 + n}{0.30 - n} &= \frac{10}{3} \\ 3(0.10 + n) &= 10(0.30 - n) \\ 0.30 + 3n &= 3.00 - 10n \\ 13n &= 2.70 \\ n &= \boxed{0.21\ \text{mol}} \end{aligned}

  4. Adding HCl: let n mol be added. H₃O⁺ turns n mol of CH₃COO⁻ into CH₃COOH:

    0.10−n0.30+n=13030(0.10−n)=0.30+n3.00−30n=0.30+n31n=2.70n=0.087 mol\begin{aligned} \frac{0.10 - n}{0.30 + n} &= \frac{1}{30} \\ 30(0.10 - n) &= 0.30 + n \\ 3.00 - 30n &= 0.30 + n \\ 31n &= 2.70 \\ n &= \boxed{0.087\ \text{mol}} \end{aligned}

  5. Conclusion. With more weak acid (CH₃COOH), the buffer can neutralize more base: its capacity for base (0.21 mol) is about 2.4 times its capacity for acid (0.087 mol). For equal capacity in both directions, the two components should be present in equal amounts (a 1 : 1 ratio).

Practice: which buffer has the larger capacity?
  1. For acid: 1.0 L with 1.0 mol each of HA and A⁻, or 1.0 L with 0.10 mol each
  2. For acid: 100 mL at 0.10 M each, or 1.0 L at 0.10 M each (the whole bottle)
  3. For acid: 0.10 mol HA + 0.30 mol A⁻, or 0.30 mol HA + 0.10 mol A⁻
  4. For base: 0.10 mol HA + 0.30 mol A⁻, or 0.30 mol HA + 0.10 mol A⁻
  5. For acid: 0.20 mol each of NH₃ and NH₄⁺, or 0.20 mol each of CH₃COOH and CH₃COO⁻ (same volume)
  6. For base: 100 mL at 0.20 M each, or 100 mL at 0.020 M each
  7. For base: 0.10 mol HA + 0.10 mol A⁻, or 0.15 mol HA + 0.05 mol A⁻
  8. For acid: acetate buffers with the same pH, at 0.50 M or 0.050 M (1.0 L each)

Common mistakes

Mistake 1: “Buffers with the same pH have the same capacity.” The pH depends only on the ratio of the two components; the capacity depends on their amounts. 1 : 1 acetate buffers at 0.50 M and 0.050 M have the same pH but capacities that differ by a factor of 10.

Mistake 2: “Buffer capacity depends only on the amount of weak acid.” Capacity for base depends on the weak acid HA; capacity for acid depends on the conjugate base A⁻.

Mistake 3: “A buffer can neutralize any amount of acid.” Once the added H₃O⁺ exceeds the amount of A⁻, the A⁻ is used up, buffering stops, and the pH drops sharply.

Quick check

Which of these 1.0 L acetate buffers has the largest capacity?

A buffer contains 0.40 mol HA and 0.10 mol A⁻. It better resists:

A buffer contains 0.20 mol NH₃ and 0.30 mol NH₄⁺. Beyond how much added HCl is the NH₃ used up?

Buffers X and Y have the same pH, and the concentrations of both components in X are 10 times those in Y. The same small amount of HCl is added to equal volumes of X and Y. Then:

For a buffer to have equal capacity for acid and for base, [HA] ÷ [A⁻] should be:

A 1.00 L buffer contains 0.100 mol CH₃COOH and 0.100 mol CH₃COO⁻ (pKa 4.74). After 0.050 mol NaOH is added, the pH is about:

AP-style free response

AP-style free response

There are two 100 mL ammonia buffers: X has 0.10 M each of NH₃ and NH₄Cl, and Y has 1.0 M of each. The pKa of NH₄⁺ is 9.26.

(a) Calculate the pH of X and of Y.

(b) 5.0 mmol HCl is added to each of X and Y (ignore the volume change). Calculate each pH.

(c) Which buffer has the larger capacity? Use your results from (b) to explain.

(d) What is the most HCl, in mmol, that can be added to X before its pH falls by 1?

(e) If 15.0 mmol HCl is added to X, calculate the pH and explain what has happened.

Model answer and scoring

(a) Both have a 1 : 1 ratio, so pH = pKa = 9.26.

(b) X: NH₃ 10 → 5.0 mmol and NH₄⁺ 10 → 15 mmol, so pH = 9.26 + log(5.0 ÷ 15) = 8.78. Y: NH₃ 100 → 95 mmol and NH₄⁺ 100 → 105 mmol, so pH = 9.26 + log(95 ÷ 105) = 9.22.

(c) Y. It contains 10 times as much NH₃ and NH₄⁺ as X, so the same HCl changes its ratio only slightly and its pH falls by only 0.04, while the pH of X falls by 0.48.

(d) When the pH has fallen by 1, the ratio is 0.1: (10 − n) ÷ (10 + n) = 0.1, so n = 8.2 mmol.

(e) X has only 10 mmol NH₃, which is all converted to NH₄⁺, leaving 5.0 mmol H₃O⁺. [H₃O⁺] = 5.0 ÷ 100 = 0.050 M, so pH = 1.30 (the ionization of NH₄⁺ is negligible). The capacity of X has been exceeded, and buffering has stopped.

Scoring: (a) pH = pKa; (b) amounts after reaction and both pH values; (c) explanation in terms of the amounts of the components; (d) setup and n; (e) excess strong acid sets the pH.

Summary

  • Buffer capacity is the amount of strong acid or base a buffer can neutralize before its pH changes significantly. Capacity for acid depends on A⁻; capacity for base depends on HA.
  • The more of both components (higher concentration, larger volume), the larger the capacity; buffers with the same pH can have different capacities.
  • At 1 : 1 the capacities for acid and base are equal; whichever component is in excess gives a larger capacity in that direction.

AP tip: When comparing buffer capacities, talk about “the amounts of the two components,” not “the pH.” In explanations, write that added H₃O⁺ is used up by A⁻ (or OH⁻ by HA), and that the more of each component there is, the less the ratio and the pH change.

Practice this chapter: Unit 8 · Acids and Bases · 12 questionsPractice this topic: U8.10 · Buffer Capacity · 6 questions

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