E. Chemical Systems and Equilibrium · Acid-Base and Solubility EquilibriaBuffer Solutions: How They Resist pH Change
E3.8 E2.1 E2.4About 75 minutes
Learning goals- I can describe the chemical characteristics of a buffer: a weak acid and its conjugate base (or a weak base and its conjugate acid) in similar, large amounts.
- I can explain with equations how a buffer resists changes in pH when a small amount of acid or base is added.
- I can calculate the pH of a buffer using Ka (or the Henderson-Hasselbalch equation), before and after adding a strong acid or base.
- I can explain buffer capacity and choose a weak acid to make a buffer of a given pH.
Words to know
buffer
A solution that resists changes in pH when small amounts of acid or base are added, or when it is diluted.
buffer capacity
The amount of acid or base a buffer can absorb before its pH changes significantly. It increases with the concentrations of the buffer components.
Henderson-Hasselbalch equation
pH = pKa + log([A⁻] ÷ [HA]), a rearranged form of the Ka expression used for buffers.
buffer range
The pH range in which a buffer works well, about pKa ± 1.
A solution that refuses to change
Add 0.010 mol of HCl to a litre of pure water and the pH crashes from 7.00 to 2.00. Add the same amount to a litre of a solution containing acetic acid and sodium acetate, and the pH barely moves: from 4.74 to 4.66. Your blood, the ocean, swimming pools, shampoos, contact lens solutions and many medicines rely on this kind of solution, called a buffer. In this lesson you will find out what makes a buffer work and how to calculate its pH.
What is in a buffer?
A buffer contains both members of a conjugate pair, each in fairly large amounts:
- a weak acid and its conjugate base: CH₃COOH with CH₃COONa; H₂CO₃ with NaHCO₃; H₂PO₄⁻ with HPO₄²⁻
- or a weak base and its conjugate acid: NH₃ with NH₄Cl
A strong acid with its conjugate base (HCl with NaCl) is not a buffer: Cl⁻ is too weak a base to remove added H₃O⁺.
How a buffer works
The two partners each have a job:
- When acid is added, the conjugate base removes it: A⁻(aq) + H₃O⁺(aq) → HA(aq) + H₂O(l)
- When base is added, the weak acid removes it: HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l)
These reactions go essentially to completion (each one is a strong acid or strong base reacting with a weak partner). The added H₃O⁺ or OH⁻ is converted into more of the buffer's own weak acid or weak base. The ratio of A⁻ to HA changes a little, so the pH changes only a little.
Calculating the pH of a buffer
Start from Ka and solve for [H₃O⁺]:
Ka = [H₃O⁺][A⁻] ÷ [HA], so [H₃O⁺] = Ka × [HA] ÷ [A⁻]
Taking −log of both sides gives the Henderson-Hasselbalch equation:
pH = pKa + log([A⁻] ÷ [HA])
In a buffer, so much A⁻ is already present that it pushes the ionization of HA far to the left (the common ion effect, which you will meet again with solubility). So you can use the initial amounts of HA and A⁻ directly, without an ICE table, as long as both are much larger than [H₃O⁺]. Because both are in the same volume, you can even use moles instead of concentrations in the ratio.
Two things to notice:
- When [A⁻] = [HA], log 1 = 0, so pH = pKa.
- Diluting a buffer changes both concentrations by the same factor, so the ratio and the pH stay (almost) the same.
Worked exampleExample 1: pH of a buffer
Find the pH of a buffer that is 0.20 mol/L CH₃COOH and 0.10 mol/L CH₃COONa. Ka = 1.8 × 10⁻⁵ (pKa = 4.74).
- Weak acid HA = CH₃COOH (0.20 mol/L); conjugate base A⁻ = CH₃COO⁻ (0.10 mol/L; Na⁺ is a spectator).
- With Ka: [H₃O⁺] = 1.8 × 10⁻⁵ × 0.20 ÷ 0.10 = 3.6 × 10⁻⁵ mol/L, pH = 4.44.
- Or with Henderson-Hasselbalch: pH = 4.74 + log(0.10 ÷ 0.20) = 4.74 − 0.30 = 4.44.
pH = 4.44. More acid than base puts the pH below pKa.
Worked exampleExample 2: Adding a strong acid to a buffer
1.00 L of buffer contains 0.10 mol CH₃COOH and 0.10 mol CH₃COO⁻ (pH 4.74). Find the pH after adding 0.010 mol of HCl (assume no volume change). Compare with adding it to 1.00 L of pure water.
- Stoichiometry first. The added H₃O⁺ reacts completely with acetate: CH₃COO⁻ + H₃O⁺ → CH₃COOH + H₂O.
- After reaction: CH₃COO⁻ = 0.10 − 0.010 = 0.09 mol; CH₃COOH = 0.10 + 0.010 = 0.11 mol.
- Then equilibrium. pH = 4.74 + log(0.09 ÷ 0.11) = 4.74 − 0.09 = 4.66 (4.6576 before rounding).
- In pure water: [H₃O⁺] = 0.010 mol/L, so pH = 2.00.
The buffer's pH drops only 0.09 unit, to 4.66. The same acid drops pure water's pH by 5 units, to 2.00.
pH as HCl is added to 1.00 L of pure water (dashed) and to 1.00 L of an acetic acid/acetate buffer (solid). The dots mark 0.010 mol of HCl: pH 2.00 in water, pH 4.66 in the buffer.Worked exampleExample 3: Adding a strong base
To the same 1.00 L buffer (0.10 mol CH₃COOH, 0.10 mol CH₃COO⁻), add 0.010 mol of NaOH instead. Find the pH.
- Stoichiometry: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. CH₃COOH = 0.10 − 0.010 = 0.09 mol; CH₃COO⁻ = 0.11 mol.
- pH = 4.74 + log(0.11 ÷ 0.09) = 4.74 + 0.09 = 4.83.
pH = 4.83, up only 0.09 unit. (Pure water would jump to pH 12.00.)
Worked exampleExample 4: Designing a buffer
You need a buffer with pH 5.00 made from acetic acid and sodium acetate. What ratio [CH₃COO⁻] ÷ [CH₃COOH] is needed? What mass of sodium acetate (M = 82.03 g/mol) should be added to 1.00 L of 0.10 mol/L acetic acid?
- 5.00 = 4.74 + log(ratio), so log(ratio) = 0.26 (more exactly 0.255) and ratio = 10⁰·²⁵⁵ = 1.8.
- [CH₃COO⁻] = 1.8 × 0.10 = 0.18 mol/L, so 0.18 mol in 1.00 L.
- Mass = 0.18 mol × 82.03 g/mol = 15 g.
Ratio 1.8 : 1; add about 15 g of sodium acetate. (Using exact pKa 4.745 gives the same answer.)
Buffer capacity and buffer range
A buffer is not a bottomless pit. If you add more acid than there is A⁻ to react with, the buffer is used up and the pH falls quickly. Buffer capacity is the amount of acid or base a buffer can absorb before its pH changes a lot. It is greater when:
- the concentrations of HA and A⁻ are larger (a 1.0 mol/L buffer resists ten times more acid than a 0.10 mol/L buffer with the same pH), and
- the amounts of HA and A⁻ are similar (ratio near 1).
A buffer works best within about one pH unit of its pKa (ratios from 1 : 10 to 10 : 1). To make a buffer at a particular pH, choose a weak acid whose pKa is close to the pH you want. For pH 7.4 (blood or cell cultures), the H₂PO₄⁻/HPO₄²⁻ pair (pKa 7.20) is a good choice; for pH 9 to 10, NH₄⁺/NH₃ (pKa 9.25).
In a lab, a buffer can also be made by partly neutralizing a weak acid with a strong base (or a weak base with a strong acid). That is exactly what happens in the middle of a weak acid titration, which you will see in the lesson on weak acid titration curves.
Just for funA buffer walks into a party where everyone keeps dumping acid in the punch. The buffer just shrugs. Its pH barely moves, at least until its capacity runs out.
Hands-on activityWhich glass resists the vinegar?
You need: red cabbage indicator, 2 clear glasses, white vinegar, baking soda, a teaspoon, an eyedropper or a ¼ teaspoon measure, and a sheet of white paper.
Grown-up note: supervise the cabbage indicator preparation (knife and hot water). Don't drink the mixtures.
- Glass A: ½ cup of water + 2 tablespoons of indicator. Glass B: ½ cup of water + ¼ teaspoon of baking soda + 2 tablespoons of indicator. Stand both on white paper.
- Add vinegar to glass A one dropper-full (or ¼ teaspoon) at a time, stirring. Count how many additions it takes to turn clearly pink.
- Repeat with glass B. Watch for fizzing.
- Compare the counts.
What to look for: glass A turns pink after only one or two additions. Glass B stays blue or purple for many additions, because the hydrogen carbonate ion removes the added acid: HCO₃⁻ + CH₃COOH → H₂CO₃ + CH₃COO⁻, and the H₂CO₃ breaks down to CO₂ (the fizz) and water. Eventually the HCO₃⁻ runs out and glass B turns pink too: you have used up its buffer capacity. Your blood uses the same HCO₃⁻/H₂CO₃ pair, with your lungs breathing away the CO₂ (more in the last lesson of this unit).
Your turn
1Which of these mixtures is a buffer? (a) HNO₃ and NaNO₃ (b) HF and NaF (c) NH₃ and NH₄Cl (d) NaOH and NaCl
(b) and (c). Each contains a weak acid or base with its conjugate partner. HNO₃ and NaOH are strong, so (a) and (d) are not buffers.
2Write the equation for the reaction that occurs when HCl is added to an NH₃/NH₄Cl buffer.
NH₃(aq) + H₃O⁺(aq) → NH₄⁺(aq) + H₂O(l). The weak base partner removes the acid.
3Find the pH of a buffer that is 0.050 mol/L benzoic acid and 0.075 mol/L sodium benzoate. (Ka = 6.3 × 10⁻⁵)
pKa = 4.20. pH = 4.20 + log(0.075 ÷ 0.050) = 4.20 + 0.18 = 4.38.
4Find the pH of a buffer that is 0.20 mol/L NH₃ and 0.30 mol/L NH₄Cl. (Ka of NH₄⁺ = 5.6 × 10⁻¹⁰)
pKa = 9.25. pH = 9.25 + log(0.20 ÷ 0.30) = 9.25 − 0.18 = 9.08. (Here NH₃ is the base A⁻ and NH₄⁺ is the acid HA.)
5A buffer has equal concentrations of HClO and NaClO. What is its pH? (Ka of HClO = 2.9 × 10⁻⁸)
pH = pKa = −log(2.9 × 10⁻⁸) = 7.54.
60.500 L of a buffer contains 0.050 mol CH₃COOH and 0.050 mol CH₃COO⁻. Find the pH after adding 0.0050 mol NaOH.
CH₃COOH: 0.050 − 0.0050 = 0.045 mol; CH₃COO⁻: 0.055 mol. pH = 4.74 + log(0.055 ÷ 0.045) = 4.74 + 0.09 = 4.83.
More practice: warm-up
1What two kinds of substance does an acid buffer contain?
A weak acid and its conjugate base (as a salt).
2In a buffer, which partner reacts with added OH⁻?
3What is the pH of a buffer when [A⁻] = [HA]?
4Write the Henderson-Hasselbalch equation.
pH = pKa + log([A⁻] ÷ [HA]).
5Does diluting a buffer with water change its pH much?
No. Both concentrations fall by the same factor, so the ratio, and the pH, stay about the same.
More practice: on your own
1Which weak acid would you choose to make a buffer of pH 3.5: HF (pKa 3.18), CH₃COOH (pKa 4.74) or H₂PO₄⁻ (pKa 7.20)? Explain.
HF, because its pKa (3.18) is within one unit of 3.5. The buffer will work best when the target pH is close to pKa.
2Buffer X is 0.010 mol/L in both CH₃COOH and CH₃COO⁻. Buffer Y is 0.10 mol/L in both. Both have pH 4.74. Find the pH of each after 0.0050 mol HCl is added to 1.00 L.
X: CH₃COO⁻ 0.005, CH₃COOH 0.015 mol; pH = 4.74 + log(0.005 ÷ 0.015) = 4.27. Y: 0.095 and 0.105 mol; pH = 4.74 + log(0.095 ÷ 0.105) = 4.70. Y has the greater buffer capacity.
3What mass of sodium acetate (82.03 g/mol) must be added to 500.0 mL of 0.20 mol/L acetic acid to make a buffer with pH 4.90?
log(ratio) = 4.90 − 4.745 = 0.155, ratio = 1.43. n(CH₃COOH) = 0.10 mol, so n(CH₃COO⁻) = 0.143 mol. m = 0.143 × 82.03 ≈ 11.7 g.
450.0 mL of 0.20 mol/L NH₃ is mixed with 50.0 mL of 0.10 mol/L HCl. Is the result a buffer? Find its pH.
n(NH₃) = 0.010 mol; n(H₃O⁺) = 0.0050 mol. Reaction: NH₃ + H₃O⁺ → NH₄⁺ + H₂O leaves 0.0050 mol NH₃ and 0.0050 mol NH₄⁺. Yes, it is a buffer, with equal amounts, so pH = pKa = 9.25.
5Explain why a mixture of HCl and NaCl cannot act as a buffer.
To be a buffer, the solution needs a base that will remove added acid. Cl⁻ is the conjugate base of a strong acid, so it is far too weak a base to react with H₃O⁺. Any added acid stays as H₃O⁺ and the pH falls.
Challenge
1Show how the Henderson-Hasselbalch equation is derived from the Ka expression.
Ka = [H₃O⁺][A⁻] ÷ [HA]. Take log: log Ka = log[H₃O⁺] + log([A⁻] ÷ [HA]). Multiply by −1: pKa = pH − log([A⁻] ÷ [HA]). Rearranged: pH = pKa + log([A⁻] ÷ [HA]).
2A phosphate buffer for a cell culture must have pH 7.40. Using H₂PO₄⁻ (Ka = 6.3 × 10⁻⁸) and HPO₄²⁻, find the required ratio [HPO₄²⁻] ÷ [H₂PO₄⁻].
pKa = 7.20. log(ratio) = 7.40 − 7.20 = 0.20, so ratio = 10⁰·²⁰ = 1.6 (1.58 with unrounded pKa).
31.00 L of buffer has 0.10 mol CH₃COOH and 0.10 mol CH₃COO⁻. What is the greatest amount of HCl it can absorb while keeping the pH at or above 3.74 (one unit below pKa)?
At pH 3.74, [A⁻] ÷ [HA] = 10⁻¹ = 0.1. With x mol of HCl: (0.10 − x) ÷ (0.10 + x) = 0.1, so 0.10 − x = 0.010 + 0.1x, which gives 1.1x = 0.090 and x = 0.082 mol.
Quiz
1. Which mixture makes a buffer?
2. When a small amount of NaOH is added to an acetic acid/acetate buffer, what happens?
3. What is the pH of a buffer with [HA] = 0.10 mol/L and [A⁻] = 0.10 mol/L, if Ka = 1.0 × 10⁻⁶?
4. To make a buffer with a pH near 9, the best choice is
5. Two acetate buffers have the same pH. Buffer P is 1.0 mol/L in each component; buffer Q is 0.10 mol/L in each. Which statement is true?
6. In a buffer with [A⁻] ÷ [HA] = 10, the pH is
Big idea (E3.8): a buffer contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in large, similar amounts. Each partner neutralizes one kind of added substance, so the pH changes very little. pH = pKa + log([A⁻] ÷ [HA]).
How to help: Ask your teen to point at the buffer diagram and explain which partner handles added acid and which handles added base. For calculation questions, check that they do the stoichiometry first (the strong acid or base reacts completely) and only then use Henderson-Hasselbalch. The vinegar activity shows capacity being used up in a way they can see.
Common difficulties:
- Thinking a strong acid plus its salt (HCl + NaCl) is a buffer.
- Putting the ratio upside down: it is base over acid, [A⁻] ÷ [HA].
- For NH₃/NH₄⁺ buffers, using Kb of NH₃ in the equation instead of pKa of NH₄⁺ (or remembering to convert).
- Forgetting that adding acid decreases A⁻ and increases HA by the same amount.