What’s in a buffer
A buffer contains comparable amounts of a conjugate acid–base pair: a weak acid and its conjugate base (such as CH₃COOH and CH₃COO⁻), or a weak base and its conjugate acid (such as NH₃ and NH₄⁺). Both components must be plentiful so that the solution can deal with both added acid and added base.
There are two ways to make one:
- dissolve a weak acid together with one of its salts (such as CH₃COOH and CH₃COONa);
- neutralize part of a weak acid with a strong base (or part of a weak base with a strong acid), so that both components remain (U8.4).
Why the pH changes so little
When a little strong acid is added, the conjugate base turns the H₃O⁺ into the weak acid; when a little strong base is added, the weak acid turns the OH⁻ into the conjugate base:
Both reactions have very large K and go to completion. The added strong acid or base is exchanged for a weak acid or base, and the pH of the buffer depends only on the ratio of the two components:
As long as the amount added is much smaller than the amounts of the two components, the ratio changes only a little, and so does the pH. When the buffer is diluted with water, the amounts of both components stay the same, so the ratio and the pH stay essentially the same.
Add the same amount of HCl or NaOH to three beakers, and compare pure water, an unbuffered solution with the same starting pH, and the buffer:
The volume change from the added acid or base is ignored; pH is calculated exactly from the charge balance.
Example 1: adding NaOH to a formate buffer
1.00 L of solution contains 0.200 mol HCOOH and 0.150 mol HCOONa (Ka = 1.8 × 10⁻⁴). 0.020 mol of solid NaOH is added (ignore the volume change). How much does the pH change? What would the pH be if the same NaOH were added to 1.00 L of pure water?
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[H₃O⁺] in a buffer. Rearrange the Ka expression for HCOOH ⇌ H⁺ + HCOO⁻. The acid and its conjugate base share one volume, so their concentration ratio equals their mole ratio:
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Starting pH.
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The added OH⁻ is used up by the weak acid. HCOOH + OH⁻ → HCOO⁻ + H₂O goes to completion (in mol):
HCOOH OH⁻ HCOO⁻ Before 0.200 0.020 0.150 Change −0.020 −0.020 +0.020 After 0.180 0 0.170 -
New pH.
The pH rises by only 3.72 − 3.62 = 0.10.
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Comparison: the same NaOH in pure water. NaOH dissociates completely, so [OH⁻] = 0.020 M:
A rise of 5.30 from 7.00, more than 50 times the change in the buffer.
Example 2: diluting a buffer
100 mL of the buffer from Example 1 (before the NaOH was added) is diluted with water to 1.00 L. How does the pH change?
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Amounts in the 100 mL taken. The original 1.00 L holds 0.200 mol HCOOH and 0.150 mol HCOO⁻; 100 mL is a tenth of that:
Adding water doesn’t change these amounts.
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The ratio is unchanged. Both are in the same volume:
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The pH doesn’t depend on volume.
The same as before dilution: diluting a buffer doesn’t change its pH.
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The cost. The diluted 1.00 L holds only a tenth as much of each component as 1.00 L of the original buffer ([HCOOH] = 0.0200 M, [HCOO⁻] = 0.0150 M), so it can handle much less added acid or base. Adding 0.020 mol of NaOH now, for example, would use up all the HCOOH and the buffer would fail (buffer capacity, U8.10).
- HCl added to CH₃COOH / CH₃COO⁻
CH₃COO⁻ + H₃O⁺ → CH₃COOH + H₂O.
- KOH added to CH₃COOH / CH₃COO⁻
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.
- HNO₃ added to NH₄⁺ / NH₃
NH₃ + H₃O⁺ → NH₄⁺ + H₂O.
- NaOH added to NH₄⁺ / NH₃
NH₄⁺ + OH⁻ → NH₃ + H₂O.
- HCl added to H₂PO₄⁻ / HPO₄²⁻
HPO₄²⁻ + H₃O⁺ → H₂PO₄⁻ + H₂O.
- Water added to HF / F⁻
The amounts don't change; only the concentrations decrease.
- NaOH added to HCO₃⁻ / CO₃²⁻
HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O.
- NaCl added to HCOOH / HCOO⁻
Neither Na⁺ nor Cl⁻ reacts with the buffer components.
Common mistakes
Mistake 1: “The pH of a buffer doesn’t change at all.” The pH changes slightly. If the added strong acid (or base) exceeds the amount of conjugate base (or weak acid), the buffer fails and the pH changes sharply.
Mistake 2: “Diluting a buffer changes its pH noticeably.” The ratio of the amounts of the two components stays the same, so the pH stays essentially the same; what decreases is the buffer capacity per unit volume.
Mistake 3: “A mixture of HCl and NaCl is a buffer.” Cl⁻ is the conjugate base of a strong acid and can’t react with added H₃O⁺, so there is no buffering.
Quick check
A small amount of NaOH is added to an NH₄⁺ / NH₃ buffer. The reaction that occurs is:
The added OH⁻ is used up by the weak acid in the buffer, NH₄⁺.
Which pair of solutions gives a buffer when mixed?
HCN is a weak acid and CN⁻ is its conjugate base.
Which substance, added to a solution containing 0.10 mol CH₃COOH, gives a buffer?
Neutralizing half leaves 0.05 mol each of CH₃COOH and CH₃COO⁻. A neutralizes it exactly, leaving only CH₃COO⁻.
The pH of a buffer changes very little when a small amount of strong acid is added mainly because:
[H₃O⁺] = Ka × n(HA) ÷ n(A⁻); a small change in the ratio means a small change in pH.
A buffer contains 0.10 mol HA and 0.10 mol A⁻. After 0.15 mol HCl is added:
0.10 mol A⁻ can use up only 0.10 mol H₃O⁺; the remaining 0.05 mol of strong acid makes the pH drop sharply.
A buffer is diluted with an equal volume of water. Its pH:
The ratio of the amounts of the two components doesn't change, so [H₃O⁺] = Ka × n(HA) ÷ n(A⁻) doesn't either.
AP-style free response
A lab prepares 1.00 L of phosphate buffer containing 0.050 mol NaH₂PO₄ and 0.050 mol Na₂HPO₄. For H₂PO₄⁻, Ka = 6.2 × 10⁻⁸ (Ka₂ of H₃PO₄, OpenStax).
(a) Calculate the pH of the buffer.
(b) Write a net ionic equation showing how the buffer uses up a small amount of added HCl.
(c) 0.010 mol HCl is added (ignore the volume change). Calculate the pH.
(d) What is the pH if the same amount of HCl is added to 1.00 L of pure water? Compare (c) and (d).
(e) About how many moles of HCl can this buffer neutralize before it stops working? Explain.
Model answer and scoring
(a) n(H₂PO₄⁻) = n(HPO₄²⁻), so [H₃O⁺] = Ka = 6.2 × 10⁻⁸ M and pH = 7.21.
(b) HPO₄²⁻(aq) + H₃O⁺(aq) → H₂PO₄⁻(aq) + H₂O(l).
(c) After the reaction there are 0.060 mol H₂PO₄⁻ and 0.040 mol HPO₄²⁻. [H₃O⁺] = 6.2 × 10⁻⁸ × 0.060 ÷ 0.040 = 9.3 × 10⁻⁸ M, so pH = 7.03.
(d) [H₃O⁺] = 0.010 M, so pH = 2.00. The pH of the water falls by 5.00, while the buffer’s falls by only 0.18.
(e) About 0.050 mol. Added H₃O⁺ is used up by HPO₄²⁻; once the 0.050 mol of HPO₄²⁻ is gone, any more HCl stays in solution and the pH drops sharply (in practice, buffering is already weak when the HPO₄²⁻ is nearly used up).
Scoring: (a) pH = pKa; (b) equation; (c) amounts after reaction and pH; (d) comparison calculation; (e) reasoning from the amount of conjugate base.
Summary
- A buffer contains comparable amounts of a conjugate acid–base pair. Added H₃O⁺ is used up by the conjugate base, and added OH⁻ by the weak acid.
- [H₃O⁺] = Ka × n(HA) ÷ n(A⁻): adding a little strong acid or base changes the ratio only slightly, and dilution doesn’t change it.
- When the added strong acid (or base) exceeds the conjugate base (or weak acid), the buffer fails.
AP tip: To explain buffering, write the specific reaction, such as “the added H₃O⁺ reacts with CH₃COO⁻ to form CH₃COOH,” then say that the ratio of the two components changes only slightly, so the pH changes only slightly. In calculations, find the amounts after the reaction first, then substitute into Ka.