Buffer pH (Henderson–Hasselbalch)

Worked example

Find the pH of a solution made from 41 mmol of a weak acid (pKₐ = 7.83) and 75 mmol of its sodium salt.

1. 1. Henderson–Hasselbalch equation: solve for pH

   $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log_{10}\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}$

   = 8.092

Answer: pH = 8.09

Method

Both species share one volume, so the mole ratio equals the concentration ratio: pH = pKₐ + log₁₀(n(A⁻)/n(HA)).

Common mistakes

• The ratio is upside down: pH = pKₐ + log₁₀([A⁻]/[HA]).

Formulas used

• Henderson–Hasselbalch equation

  $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log_{10}\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}$

More equilibrium questions

• pH of a weak acid
• Kp from Kc
• Solubility from Ksp
• pH of a weak base
• Kc and its units for the Haber process
• Kc from equilibrium amounts and volume
• Kc from initial amounts and one equilibrium amount
• An equilibrium concentration from Kc
• Kp from equilibrium amounts and total pressure
• Kp and its units for the Haber process
• Ksp from a solubility in g dm⁻³
• Solubility with a common ion
• pH of benzoic acid from a weighed sample
• H₂ + I₂ ⇌ 2HI: equilibrium amounts