Henderson–Hasselbalch buffer calculator
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Result
The Henderson–Hasselbalch equation assumes that [HA] and [A⁻] are much larger than [H⁺] and [OH⁻]. A buffer works best within pKa ± 1. Amounts in mol give the same ratio as concentrations.
Use the Henderson–Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), to work with buffers. Find the pH of a buffer from the concentrations of the weak acid and its conjugate base, find the base-to-acid ratio and the concentrations needed for a target pH, or find the pKa from a measured pH. You can give the strength of the acid as pKa or as Ka.
Step-by-step solution
Henderson–Hasselbalch equation
Substitute
Key formulas
- pH
- pH = −log[H⁺]
- Weak acid
- [H⁺] ≈ √(Ka·c)
- Buffer
- pH = pKa + log([A⁻]/[HA])
How to use the Henderson–Hasselbalch equation
- Identify the weak acid HA and its conjugate base A⁻, for example acetic acid and acetate.
- Find the pKa. If you are given Ka, use pKa = −log Ka.
- For the pH: pH = pKa + log([A⁻]/[HA]). You can use moles instead of concentrations, because both are in the same volume.
- For a target pH: [A⁻]/[HA] = 10^(pH − pKa). Split a total concentration c into [A⁻] = c·r/(1 + r) and [HA] = c/(1 + r).
- Check the result: when [A⁻] = [HA] the pH equals the pKa, and a buffer works best within pKa ± 1.
Questions and answers
When is the Henderson–Hasselbalch equation accurate?
When both [HA] and [A⁻] are much larger than [H⁺] and [OH⁻], which is usually true for buffers above about 0.01 mol/L with a pH between about 3 and 11. For very dilute solutions or fairly strong acids, solve the full equilibrium instead.
What happens when I add a strong acid or base to a buffer?
Work with moles. Added H⁺ turns A⁻ into HA, and added OH⁻ turns HA into A⁻. Update both amounts, then put the new ratio into the equation.
Can I use it for a base buffer such as ammonia?
Yes, with the pKa of the conjugate acid: for NH₄⁺/NH₃ the pKa is 9.25, so pH = 9.25 + log([NH₃]/[NH₄⁺]). You can also work with pOH = pKb + log([BH⁺]/[B]) and pH = 14 − pOH at 25 °C.
Why does a buffer only work near its pKa?
Buffering needs plenty of both HA and A⁻ to take up added acid or base. One pH unit from the pKa the ratio is already 10:1, and further away there is too little of one of the two forms left.
What is the pH of a buffer with equal amounts of acid and base?
It equals the pKa, because log 1 = 0. For acetic acid and sodium acetate at equal concentrations, the pH is about 4.76.
Practise it yourself
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