AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/8/8-9)
Unit 8 · Topic 8.9
8.9 Henderson-Hasselbalch Equation
The Henderson–Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), gives a buffer's pH from the acid's pKa and the ratio of conjugate base to acid. When the two are equal, pH = pKa, and because small additions barely change the ratio, the pH barely moves.
Key terms
- Henderson-Hasselbalch equation
- pKa
- [A⁻]/[HA] ratio
- buffer pH
The equation
For a buffer made from a weak acid HA and its conjugate base A⁻: pH = pKa + log([A⁻] / [HA]). It comes straight from rearranging the Ka expression and taking logs. You won't be asked to derive it, but you should know how to use it.
Because both species are in the same volume, you can use the mole ratio instead of the concentration ratio: [A⁻]/[HA] = (mol A⁻)/(mol HA).
Reading the equation
- If [A⁻] = [HA], the log term is log 1 = 0, so pH = pKa.
- If there is more A⁻ than HA, the log term is positive and pH > pKa.
- If there is more HA than A⁻, the log term is negative and pH < pKa.
- A tenfold ratio moves the pH by exactly 1 unit from the pKa.
Using moles, and buffers made by partial neutralization
Because HA and A⁻ share the same volume, the volume cancels in the ratio. If a buffer contains 0.030 mol A⁻ and 0.010 mol HA, the ratio is 3.0 no matter what the volume is, and pH = pKa + log 3.0 = pKa + 0.48.
When a buffer is made by adding strong base to a weak acid (topic 8.4), react the base with the acid first, then use the leftover moles of HA and the moles of A⁻ formed in the equation.
Why small additions barely change the pH
The pH depends on the log of a ratio. Adding a little acid converts a little A⁻ to HA; adding a little base converts a little HA to A⁻. If both are present in large amounts, the ratio barely changes, and its log changes even less. So the pH changes much less than it would without the buffer. (Calculating the exact pH change after an addition isn't tested.)
Choosing an acid for a target pH
The most effective buffers have [A⁻]/[HA] near 1, so pick a weak acid whose pKa is close to the pH you want, usually within about 1 unit. Then fine-tune the ratio with the equation.
For a buffer made from a weak base B and its conjugate acid HB⁺, use the pKa of HB⁺ (pKa = 14.00 − pKb at 25 °C), with [B] in place of [A⁻] and [HB⁺] in place of [HA].
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
pH of an ammonia buffer
A buffer contains 0.20 M NH₃ and 0.30 M NH₄Cl. Kb of NH₃ is 1.8 × 10⁻⁵. Find the pH at 25 °C.
Show the solutionHide the solution
- Step 1: Here the weak acid is NH₄⁺ and its conjugate base is NH₃. You need the pKa of NH₄⁺.
- Step 2: pKb = −log(1.8 × 10⁻⁵) = 4.745, so pKa = 14.00 − 4.745 = 9.255.
- Step 3: pH = pKa + log([NH₃]/[NH₄⁺]) = 9.255 + log(0.20/0.30) = 9.255 − 0.176 = 9.08.
- Step 4: There is more acid form than base form, so the pH is a bit below the pKa, as expected.
Answer: pH ≈ 9.08
- Example 2Calculator allowed
Designing a buffer
You want 1.00 L of buffer at pH 5.00 using 0.100 M acetic acid (Ka = 1.8 × 10⁻⁵). How many moles, and how many grams, of sodium acetate (NaCH₃COO, molar mass 82.03 g/mol) should you dissolve? Assume the volume doesn't change.
Show the solutionHide the solution
- Step 1: pKa = −log(1.8 × 10⁻⁵) = 4.745.
- Step 2: 5.00 = 4.745 + log([A⁻]/[HA]), so log([A⁻]/[HA]) = 0.255 and [A⁻]/[HA] = 10^0.255 = 1.8.
- Step 3: [A⁻] = 1.8 × 0.100 M = 0.18 M, so you need 0.18 mol of sodium acetate in 1.00 L.
- Step 4: Mass = 0.18 mol × 82.03 g/mol = 15 g (14.8 g before rounding).
- Step 5: Check: the target pH is above the pKa, so you need more A⁻ than HA. It matches.
Answer: About 0.18 mol, or about 15 g, of sodium acetate
- Example 3Calculator allowed
Trap: using pKb instead of pKa
A student calculates the pH of a buffer with equal concentrations of NH₃ and NH₄⁺ as pH = pKb = 4.74. What's wrong, and what is the correct pH? (Kb of NH₃ = 1.8 × 10⁻⁵.)
Show the solutionHide the solution
- Step 1: Henderson–Hasselbalch as written uses the pKa of the acid in the pair. For NH₃/NH₄⁺, that's the pKa of NH₄⁺, not the pKb of NH₃.
- Step 2: pKa = 14.00 − 4.745 = 9.255.
- Step 3: With equal concentrations, log(1) = 0, so pH = 9.26.
- Step 4: A sanity check catches the error: a buffer made from a base and its conjugate acid in equal amounts can't have an acidic pH of 4.74.
Answer: pH = 9.26, not 4.74
Common mistakes
- Flipping the ratio: it's base over acid, [A⁻]/[HA].
- Using pKb in Henderson–Hasselbalch for a base buffer without converting to the pKa of the conjugate acid.
- Using the equation for solutions that aren't buffers, such as a weak acid alone or a strong acid with its salt.
- Forgetting to react any added strong acid or base first before using the equation (topic 8.4).
On the exam
- You'll use this equation to find a buffer's pH, to find the ratio needed for a target pH, or to explain why pH = pKa at the half-equivalence point.
- Questions may ask which acid from a list is best for a buffer at a given pH: pick the one whose pKa is closest.
Connected topics
Videos
Check yourself
4 questions on 8.9 Henderson-Hasselbalch Equation. Pick an answer to see if you got it, and why.
A buffer is 0.20 M CH₃COOH and 0.10 M CH₃COONa. The pKa of acetic acid is 4.74. What is the pH of the buffer?
A student wants to make a buffer with a pH of 5.00 using acetic acid (pKa = 4.74) and sodium acetate. Which ratio of [CH₃COO⁻] to [CH₃COOH] is needed?
A buffer is made by dissolving 0.30 mol of NH₃ and 0.10 mol of NH₄Cl in water to make 1.0 L of solution at 25 °C. The pKa of NH₄⁺ is 9.25. What is the pH of the buffer?
A buffer made from a weak acid, HA, and its conjugate base has [A⁻]/[HA] = 2.0 and a pH of 4.50 at 25 °C. What is the pKa of HA?
0 of 4 answered