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A buffer is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resists changes in pH. The Henderson–Hasselbalch equation gives its pH from the acid's pKa and the ratio of the two concentrations.
Enter the pKa (or Ka) and the concentrations of the salt and the acid. For a basic buffer such as NH₃/NH₄Cl, use the pKa of the ammonium ion (9.25).
| I know the | pKa |
| pKa or Ka | 4.76 |
| [A⁻] conjugate base (salt) | 0.1 mol/L |
| [HA] weak acid | 0.1 mol/L |
| Buffer pH | 4.76 |
| pKa | 4.76 |
| Ka | 1.7378e-5 |
| [A⁻] ÷ [HA] | 1 |
With equal concentrations the pH equals the pKa: 4.76.
Try this example →| I know the | pKa |
| pKa or Ka | 4.76 |
| [A⁻] conjugate base (salt) | 0.2 mol/L |
| [HA] weak acid | 0.05 mol/L |
| Buffer pH | 5.36206 |
| pKa | 4.76 |
| Ka | 1.7378e-5 |
| [A⁻] ÷ [HA] | 4 |
A 4 : 1 ratio raises the pH to 5.36206.
Try this example →pH = pKa + log([A⁻]/[HA]), used to find the pH of a buffer.
log(1) = 0, so pH = pKa.
pKa = −log₁₀ Ka. Ka = 1.75 × 10⁻⁵ gives pKa = 4.76.
Yes. Both are in the same volume, so the ratio of moles equals the ratio of concentrations.