Henderson–Hasselbalch Calculator
Pick a weak acid and its conjugate base, then find the pH of a buffer, the ratio that gives a target pH, or the grams of each component to make a buffer of a set volume. The calculator also solves the buffer exactly, so it tells you when the Henderson–Hasselbalch approximation can be trusted. Runs entirely in your browser. Free, no sign-up.
At 25 °C. Write K as 1.8e-5 or 1.8 × 10^-5; pK as 4.74.
Results
4.74
Buffer pH
1.00
[A⁻] ÷ [HA]
4.74
pH without the approximation
The ratio is within 0.1 to 10, the range in which the buffer works well: neither component is less than about a tenth of the other.
The approximation agrees with the exact calculation to within 0.05 pH units for this buffer.
The working
pH = pKa + log₁₀([A⁻] ÷ [HA]) = 4.74 + log₁₀(0.100 ÷ 0.100) = 4.74
How to Use the Henderson–Hasselbalch Calculator
- 1. Choose the buffer pair. Pick a weak acid and its conjugate base from the list, or choose your own and type its Ka or pKa. The acid form is HA and the conjugate base is A⁻.
- 2. Choose the job. pH of a buffer uses the two concentrations. Ratio for a target pH gives [A⁻] ÷ [HA]. Recipe for a buffer gives grams of each component for a volume and total concentration.
- 3. Enter the numbers. Concentrations are in mol/L, the volume in mL, and the formulas of the two forms you weigh out can be edited, including hydrates.
- 4. Read the result and the checks. The pH comes with the ratio, whether it lies in the working range, and the exact pH from the full charge balance for comparison.
The Equation
For a weak acid HA, Ka = [H⁺][A⁻] ÷ [HA]. Take the negative logarithm of both sides and rearrange, and you have the Henderson–Hasselbalch equation:
Buffer pH
pH = pKa + log₁₀([A⁻] ÷ [HA])
Ratio
[A⁻] ÷ [HA] = 10^(pH − pKa)
Equal parts
[A⁻] = [HA] gives pH = pKa
Ten times as much base as acid raises the pH one unit above the pKa; ten times as much acid lowers it one unit below. Because only the ratio matters, diluting a buffer leaves its pH almost unchanged, although it lowers its capacity to absorb added acid or base.
| Buffer pair (acid form / base form) | pKa at 25 °C |
|---|---|
| Formic acid / formate | 3.74 |
| Acetic acid / acetate | 4.74 |
| Carbonic acid / hydrogen carbonate | 6.37 |
| Dihydrogen phosphate / hydrogen phosphate | 7.21 |
| Ammonium / ammonia | 9.25 |
| Hydrogen carbonate / carbonate | 10.33 |
These come from the Ka values in OpenStax Chemistry 2e, Appendix H; the ammonium value is Kw ÷ Kb for ammonia. Pick the pair whose pKa is closest to the pH you want.
Worked Examples
Buffer pH. The calculator opens on 0.10 M acetic acid and 0.10 M sodium acetate. The ratio is 1, log 1 = 0, so pH = pKa = 4.74.
Target pH. To make the acetate buffer pH 5.00: [A⁻] ÷ [HA] = 10^(5.00 − 4.74) = 1.80, so 64.3 % of the buffer is acetate and 35.7 % acetic acid. For 500 mL at a total of 0.100 M, that is 0.0500 mol altogether: 0.0179 mol of acetic acid (1.07 g of CH₃CO₂H) and 0.0321 mol of sodium acetate (2.64 g of CH₃CO₂Na). Dissolve both and make up to 500 mL.
When the Approximation Fails
The equation uses the concentrations you mixed as if they were the equilibrium concentrations. That is a good approximation when acid and base are both present in large amounts, but it breaks down at the extremes. OpenStax notes that a buffer stops being useful when one component falls below about a tenth of the other, which is a ratio outside 0.1 to 10, or a pH more than one unit from the pKa.
- Very dilute buffers. With 1 × 10⁻⁵ M each of acetic acid and acetate, the equation says pH 4.74, but the exact pH is 5.27, because the acid's own ionisation is no longer small beside its concentration.
- Extreme ratios. With 1 × 10⁻⁶ M acid and 1 M base, the equation says pH 10.74; the exact value is 9.36, because acetate reacts with water and sets a ceiling.
- Strong acid or base added. Work out what is left of HA and A⁻ after the reaction, then use those amounts.
The calculator solves the same mixture from the charge balance and reports the difference, so you can see how far the shortcut is off for your own numbers.
Frequently Asked Questions
- What is the Henderson–Hasselbalch equation?
- It is pH = pKa + log([A⁻] ÷ [HA]), the equilibrium expression of a weak acid rearranged to give the pH of a solution containing the acid and its conjugate base. It lets you calculate the pH of a buffer from the ratio of its two components.
- How do you make a buffer of a given pH?
- Choose an acid whose pKa is within one unit of the target, work out the ratio [A⁻] ÷ [HA] = 10^(pH − pKa), split your total concentration in that ratio, weigh out each form, dissolve and make up to volume. The recipe mode does all of this and also gives the exact resulting pH.
- What is the effective range of a buffer?
- About pKa ± 1. At pH = pKa the buffer is equally able to neutralise acid and base. Outside a ratio of 0.1 to 10 one component is too scarce to absorb much added acid or base, so choose a pair whose pKa is near your target.
- Can I use it for a weak base and its conjugate acid?
- Yes, using the pKa of the conjugate acid. For ammonia and ammonium, use the ammonium pair, pKa 9.25, with ammonium as the acid HA and ammonia as the base A. For a base's Kb, the conjugate acid has Ka = Kw ÷ Kb.
- Why does the buffer pH not change when I dilute it?
- The equation depends on the ratio of base to acid, which dilution does not change. Only very large dilutions matter, when the concentrations become too small for the approximation, which the exact check reveals. What dilution does change is the buffer's capacity.
- How accurate is the Henderson–Hasselbalch equation?
- For a 0.1 M buffer with a ratio near 1, the two agree to within 0.01 pH units. The gap grows at lower concentrations, at the ends of the 0.1 to 10 range and for stronger acids: for a 0.01 M formic acid and formate buffer at a ratio of 0.1 it reaches 0.33 pH units. Check the exact value in those cases, and calibrate a pH meter before relying on any calculated buffer.