Every buffer calculation so far has used [H₃O⁺] = Ka × [HA]/[A⁻]. The Henderson-Hasselbalch equation is the same relationship written in pH units, which makes it faster to use and, more importantly, easy to run backwards: given a target pH, it tells you how to make the buffer. Buffer preparation was one of the lowest-scoring free-response questions in recent years, so this page spends most of its time there.
Deriving the equation
Start from Ka for HA + H₂O ⇌ H₃O⁺ + A⁻ and take −log of both sides:
Ka = [H₃O⁺] × [A⁻]/[HA]
−log Ka = −log[H₃O⁺] − log([A⁻]/[HA])
pKa = pH − log([A⁻]/[HA])
pH = pKa + log([A⁻]/[HA])
This is the Henderson-Hasselbalch equation, and it is on the equations sheet. Three rules for using it:
- Base-10 log. pH and pKa are base-10, so the log here is base-10 too. Using ln, a mistake readers flag every year, gives an answer off by a factor of 2.303 in the log term.
- Base on top. The conjugate base A⁻ is in the numerator. Sanity check: more base than acid means pH above pKa.
- Moles are fine. Both species are in the same volume, so the ratio of concentrations equals the ratio of moles.
![pH against log([A⁻]/[HA]) for pKa 4.74: a straight line of slope 1 through (0, 4.74), (1, 5.74) and (−1, 3.74). The shaded band is ratios from 0.1 to 10. A note: use the base-10 log; ln would give pKa + 2.30 for a 10:1 ratio.](../figures/hh-ratio-line.svg)
The figure shows the equation as a straight line: pH against log([A⁻]/[HA]) has slope 1 and crosses pKa at a ratio of 1.
The equation needs real amounts of both HA and A⁻, so it is for buffers. It does not give the pH of a weak acid alone (A⁻ would be zero) or of a solution at the equivalence point (HA would be zero).
Calculating a buffer's pH
Worked example 1. A buffer is 0.40 M HNO₂ and 0.25 M NaNO₂ (Ka = 4.0 × 10⁻⁴). Find its pH.
Step 1. pKa = −log(4.0 × 10⁻⁴) = 3.398 (keep the extra digit until the end).
Step 2. pH = 3.398 + log(0.25/0.40) = 3.398 − 0.204 = 3.19.
Check. More acid than base, so the pH should be below pKa. It is. (With ln, you would get 2.93: wrong.)
Worked example 2: a weak base buffer. A buffer is 0.15 M NH₃ and 0.35 M NH₄Cl (Kb of NH₃ = 1.8 × 10⁻⁵). Find its pH.
Step 1. The acid of this pair is NH₄⁺. pKa = 14.00 − pKb = 14.00 − 4.74 = 9.26. Using pKb (4.74) here is a common slip that gives an acidic pH for an ammonia buffer.
Step 2. The base is NH₃: pH = 9.26 + log(0.15/0.35) = 9.26 − 0.37 = 8.89.
Worked example 3: after an addition. 0.0050 mol NaOH is added to a buffer of 0.0300 mol CH₃COOH and 0.0200 mol CH₃COO⁻ (pKa 4.74). Find the new pH.
Step 1, react first. OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O: acid 0.0250 mol, base 0.0250 mol.
Step 2. pH = 4.74 + log(0.0250/0.0250) = 4.74. (Before the addition: 4.74 + log(0.0200/0.0300) = 4.56.)
Designing a buffer
Run the equation backwards. Three steps:
- Choose the acid. Pick the conjugate pair whose pKa is closest to the target pH, within about 1 unit. That keeps the ratio between 0.1 and 10, so both components are present in useful amounts.
- Find the ratio. [A⁻]/[HA] = 10^(pH − pKa).
- Turn the ratio into amounts, either by weighing the salt or by partly neutralizing the acid with strong base.
Worked example 4: choose and weigh. Make 0.250 L of a buffer at pH 4.50 that contains 0.100 M acetic acid. How many grams of sodium acetate (82.03 g/mol) are needed? Options: acetic acid (pKa 4.74), formic acid (3.74), H₂PO₄⁻ (7.21).
Step 1. Acetic acid's pKa (4.74) is closest to 4.50.
Step 2. [A⁻]/[HA] = 10^(4.50 − 4.74) = 10^−0.24 = 0.575.
Step 3. mol HA = 0.250 L × 0.100 M = 0.0250 mol; mol A⁻ = 0.575 × 0.0250 = 0.0144 mol.
Step 4. Mass = 0.0144 mol × 82.03 g/mol = 1.18 g of sodium acetate.
Procedure. Weigh the salt on an analytical balance, dissolve it with the acetic acid in some distilled water in a 250.0 mL volumetric flask, dilute to the mark, stopper and invert to mix. Check the pH with a calibrated meter.
Worked example 5: partial neutralization. You have 50.0 mL of 0.200 M acetic acid and 0.100 M NaOH. How much NaOH makes a buffer at pH 4.94?
Step 1. Ratio: 10^(4.94 − 4.74) = 10^0.20 = 1.58.
Step 2. Start with 0.0100 mol HA. Adding x mol OH⁻ leaves (0.0100 − x) mol HA and makes x mol A⁻, so x / (0.0100 − x) = 1.58.
Step 3. x = 1.58 × 0.0100 / (1 + 1.58) = 0.00613 mol.
Step 4. Volume of NaOH = 0.00613 mol ÷ 0.100 mol/L = 0.0613 L = 61.3 mL, added from a buret rinsed with the NaOH solution.
The trap in Example 5 is to set x = ratio × 0.0100. The base you add makes A⁻ and removes HA, so both sides of the ratio change.
Where students lose points
- ln instead of log.
- The ratio upside down (acid on top).
- pKb in place of pKa for an NH₃/NH₄⁺ buffer.
- Forgetting to react an added strong acid or base before using the equation.
- Vague procedures: name the glassware (volumetric flask, buret, analytical balance) and say "dilute to the mark".
- Choosing an acid whose pKa is far from the target pH.