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General Chemistry II · Acids and bases

Buffer pH calculation practice (Henderson-Hasselbalch, with answers)

A buffer is a weak acid plus its conjugate base (or a weak base plus its conjugate acid), both in decent amounts. It holds its pH steady because the weak acid soaks up added OH− and the conjugate base soaks up added H3O+. Most buffer problems are two steps: decide whether you have a buffer, then plug a ratio into one equation. This page walks through that, then gives you problems to check yourself.

Step 1: Is it a buffer?

Before you reach for any equation, figure out what is actually in the beaker. Walk down these questions in order.

  1. Is there a weak acid or a weak base?

    • No (only strong acids, strong bases and their salts, like HCl + NaCl): not a buffer. A strong acid is fully ionized, so nothing is left to soak up added base.
    • Yes: go on.

  2. Is its conjugate partner there too, usually added as a salt?

    • Yes (CH3COOH + CH3COONa, or NH3 + NH4Cl): it's a buffer. Go to step 2.
    • No: go on.

  3. Was a strong acid or strong base added to it?

    • No: it's just a weak acid or weak base in water. Use an ICE table instead.
    • Yes, and fewer moles of strong than weak: let them react completely first. The leftover weak acid plus the conjugate base you just made is a buffer.
    • Yes, and exactly equal moles: that's the equivalence point. Only the conjugate is left, so it's not a buffer.
    • Yes, and more strong than weak: the extra strong acid or base sets the pH.

Quick check: buffer or not?

Pick an answer for each mix. Each one is 1.00 L of solution.

A. 0.10 mol HCl and 0.10 mol NaCl

Show answer for mix A

Not a buffer. HCl is a strong acid. You need a weak acid and its conjugate base.

B. 0.20 mol NH3 and 0.15 mol NH4Cl

Show answer for mix B

Buffer. A weak base (NH3) and its conjugate acid (NH4+) are both present.

C. 0.10 mol CH3COOH and 0.10 mol NaOH

Show answer for mix C

Not a buffer. CH3COOH + OH− → CH3COO− + H2O uses up all 0.10 mol of acid. Only the conjugate base is left.

D. 0.10 mol HF and 0.050 mol NaOH

Show answer for mix D

Buffer. HF + OH− → F− + H2O leaves 0.050 mol HF and 0.050 mol F−. Equal amounts, so pH = pKa. That's the half-equivalence point.

Step 2: The Henderson-Hasselbalch equation

pH = pKa + log([A−] ÷ [HA])A− is the conjugate base (on top). HA is the weak acid (on the bottom).

It comes straight from rearranging the Ka expression and taking the log of both sides. It assumes x is small, which holds when both partners are present in decent amounts. Three things make it fast:

  • Moles work as well as molarity. Both partners are in the same volume, so the volume cancels in the ratio. When strong acid or base is added, work in moles.
  • For a weak-base buffer, you still need pKa. The acid is the conjugate acid (NH4+ for NH3). Use pKa = 14.00 − pKb at 25 °C, then put the base (NH3) on top.
  • Sanity check the sign. More base than acid means pH is above pKa. More acid than base means below. Equal amounts means pH = pKa. A ratio of 10 adds exactly 1 pH unit.

The K values on this page come from the OpenStax Chemistry 2e appendix tables. Your textbook may list slightly different values. For example, some tables give HF as 6.8 × 10−4. Use the values your course gives you.

pKa values used on this page (25 °C)
Buffer pair (HA / A−)Constant from OpenStaxpKa
Acetic acid / acetate (CH3COOH / CH3COO−)Ka = 1.8 × 10−54.74
Formic acid / formate (HCOOH / HCOO−)Ka = 1.8 × 10−43.74
Hydrofluoric acid / fluoride (HF / F−)Ka = 6.4 × 10−43.19
Ammonium / ammonia (NH4+ / NH3)Kb(NH3) = 1.8 × 10−5, so pKb = 4.7414.00 − 4.74 = 9.26

Three worked examples, easy to harder

Example 1: A ready-made acid buffer

Find the pH of a solution that is 0.20 M acetic acid and 0.30 M sodium acetate. Ka = 1.8 × 10−5.

  1. Is it a buffer? Weak acid plus its conjugate base (acetate, from the salt). Yes.
  2. pKa = −log(1.8 × 10−5) = 4.74.
  3. Base on top: pH = 4.74 + log(0.30 ÷ 0.20) = 4.74 + log(1.5) = 4.74 + 0.18.

pH = 4.92. Check: more base than acid, so the pH is a little above pKa.

Example 2: Adding strong acid to a buffer

1.00 L of buffer holds 0.100 mol CH3COOH and 0.100 mol CH3COO−. You add 0.010 mol HCl. What is the new pH?

  1. The added H3O+ reacts with the base: CH3COO− + H3O+ → CH3COOH + H2O. It goes to completion.
  2. Track moles before and after:
    Moles in example 2
    StageCH3COO−H3O+ addedCH3COOH
    Before0.1000.0100.100
    Change−0.010−0.010+0.010
    After0.09000.110
  3. pH = 4.74 + log(0.090 ÷ 0.110) = 4.74 + (−0.087) = 4.65.

pH = 4.65, down only 0.09 from 4.74. Adding the same 0.010 mol HCl to 1.00 L of pure water would give 0.010 M H3O+, pH 2.00. (If you keep the unrounded pKa, 4.7447, you get 4.66. Both answers are reasonable; say which pKa you used.)

Example 3: A weak-base buffer

Find the pH of a solution that is 0.50 M NH3 and 0.25 M NH4Cl. Kb(NH3) = 1.8 × 10−5.

  1. Weak base plus its conjugate acid (NH4+), so it's a buffer.
  2. Convert to the acid's constant: pKb = 4.74, so pKa(NH4+) = 14.00 − 4.74 = 9.26.
  3. The base is NH3, so it goes on top: pH = 9.26 + log(0.50 ÷ 0.25) = 9.26 + log(2) = 9.26 + 0.30.

pH = 9.56. Check: a basic buffer, with more base than acid, so the pH is above 9.26.

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Six practice problems

Type the pH as a number, such as 4.92. Answers within 0.02 pH units count, so rounding pKa to 4.74 won't cost you. Problem 5 asks for a ratio instead. If you're wrong, the feedback names the usual slip when it spots one, and you can try again. Every problem has a full solution.

Problem 1 Equal amounts

A solution is 0.10 M HF and 0.10 M NaF. Ka(HF) = 6.4 × 10−4. What is its pH?

Show solution for problem 1
  1. pKa = −log(6.4 × 10−4) = 3.19.
  2. [F−] ÷ [HF] = 0.10 ÷ 0.10 = 1, and log(1) = 0.
  3. pH = 3.19 + 0 = 3.19. (With a table that lists 6.8 × 10−4, you'd get 3.17.)

Problem 2 More base than acid

A solution is 0.15 M formic acid (HCOOH) and 0.45 M sodium formate (HCOONa). Ka = 1.8 × 10−4. What is its pH?

Show solution for problem 2
  1. pKa = −log(1.8 × 10−4) = 3.74.
  2. Base on top: [HCOO−] ÷ [HCOOH] = 0.45 ÷ 0.15 = 3.0, and log(3.0) = 0.48.
  3. pH = 3.74 + 0.48 = 4.22.

Problem 3 Add strong base

Start with the buffer from example 2: 1.00 L holding 0.100 mol CH3COOH and 0.100 mol CH3COO−. This time you add 0.020 mol NaOH. What is the new pH?

Show solution for problem 3
  1. OH− reacts with the weak acid: CH3COOH + OH− → CH3COO− + H2O.
  2. After: CH3COOH 0.100 − 0.020 = 0.080 mol. CH3COO− 0.100 + 0.020 = 0.120 mol.
  3. pH = 4.74 + log(0.120 ÷ 0.080) = 4.74 + log(1.5) = 4.74 + 0.18 = 4.92.

Problem 4 Make a buffer from scratch

You mix 0.200 mol CH3COOH and 0.050 mol NaOH in enough water to make 1.00 L. Ka = 1.8 × 10−5. What is the pH?

Show solution for problem 4
  1. There's no salt yet, but there is less strong base than weak acid. React it first: the 0.050 mol OH− turns 0.050 mol CH3COOH into CH3COO−.
  2. After: CH3COOH 0.200 − 0.050 = 0.150 mol. CH3COO− = 0.050 mol.
  3. pH = 4.74 + log(0.050 ÷ 0.150) = 4.74 + log(1/3) = 4.74 − 0.48 = 4.26. (With the unrounded pKa you get 4.27. Both count.)

Problem 5 Work backward

You want an acetic acid / acetate buffer at pH 5.00. What ratio [CH3COO−] ÷ [CH3COOH] do you need? (pKa = 4.74.)

Show solution for problem 5
  1. 5.00 = 4.74 + log(ratio), so log(ratio) = 0.26.
  2. Undo the log: ratio = 100.26 = 1.8.
  3. So you need about 1.8 mol acetate for every 1 mol acetic acid. (With the unrounded pKa, 100.2553 = 1.80 as well.)

Problem 6 Weak base plus strong acid

You mix 0.25 mol NH3 and 0.10 mol HCl in enough water to make 1.00 L. Kb(NH3) = 1.8 × 10−5. What is the pH?

Show solution for problem 6
  1. React first: NH3 + H3O+ → NH4+ + H2O. After: NH3 0.25 − 0.10 = 0.15 mol, NH4+ = 0.10 mol.
  2. pKa(NH4+) = 14.00 − 4.74 = 9.26.
  3. pH = 9.26 + log(0.15 ÷ 0.10) = 9.26 + 0.18 = 9.44, or 9.43 with the unrounded pKa. Both count.

Common mistakes

  • Flipping the ratio. Base on top, acid on the bottom. If the pH lands on the wrong side of pKa, this is usually why.
  • Using pKb as if it were pKa. For NH3/NH4+, convert first: pKa = 14.00 − pKb. This ammonia buffer has a pH near 9.26.
  • Plugging in before reacting. Added strong acid or base reacts completely first. Do the before/after moles, then use the equation.
  • Mixing up moles and molarity. Either works, as long as both partners use the same one. When you mix two solutions, moles are safer because the volumes change.
  • Using the equation when it isn't a buffer. A weak acid alone, the equivalence point, and excess strong acid or base all need a different method.

Frequently asked questions

When does the Henderson-Hasselbalch equation work?

When both the weak acid and its conjugate base are present in reasonable amounts. A buffer has generally stopped being useful once one partner is less than about 10% of the other, so keep the base-to-acid ratio between about 0.1 and 10. Use it in the buffer region only, not for a weak acid alone or at the equivalence point.

What is buffer capacity?

It is how much strong acid or base a buffer can take before its pH changes a lot, usually by one unit. A buffer with more moles of the weak acid and its conjugate base has more capacity, even if its pH is the same. So 1.0 M acetic acid with 1.0 M acetate has more capacity than 0.10 M of each.

Why is pH = pKa at the half-equivalence point?

Halfway to the equivalence point, half of the weak acid has been turned into its conjugate base, so the moles of acid and base are equal. The ratio is 1, log(1) = 0, and pH = pKa.

What if I add more strong acid than the buffer has base?

Then the buffer is used up. All of the conjugate base turns into weak acid, and the extra strong acid sets the pH. Find the leftover moles of H3O+, divide by the total volume, and take the negative log. Henderson-Hasselbalch no longer applies.

Sources

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