pH of weak acids and bases
Weak acids only give off a small share of their protons, so you need the pKa. For solutions that are not too dilute a simple approximation is enough.
Formulas and rules
- Weak acid: pH = ½ · (pKa − log c)
- Weak base: pOH = ½ · (pKb − log c), then pH = 14 − pOH
- pKa + pKb = 14 for a conjugate pair
- The smaller the pKa, the stronger the acid
Your turn
Three problems on this topic. Type numbers like 12.5 or 1,250 and elasticities with their sign.
Task
Work out the pH of a methylamine solution (CH₃NH₂, pKb = 3.36) with c = 0.2 mol/L at 25 °C.
pOH (2 decimal places)
pH (2 decimal places)
Numbers like 12.5 or 0.025, with as many decimals as asked. Masses and amounts count within 1%, pH values within ±0.02.
Worked example
Work out the pH of a solution of hydrofluoric acid (HF, pKa = 3.17) with c = 0.1 mol/L. Use the weak acid approximation.
Worked solution
- 1.Weak acid, barely dissociated: pH = ½ · (pKa − lg c)
- 2.pH = ½ × (3.17 − lg 0.1) = ½ × (3.17 − (−1)) = 2.09
Typical mistake: Using the strong acid formula (pH = −lg c). A weak acid only gives off a small share of its protons, so the pH is clearly higher.
Work out the pH of a methylamine solution (CH₃NH₂, pKb = 3.36) with c = 0.1 mol/L at 25 °C.
Worked solution
- 1.pOH = ½ × (pKb − lg c) = ½ × (3.36 − (−1)) = 2.18
- 2.pH = 14 − pOH = 14 − 2.18 = 11.82
Typical mistake: Getting the pH straight from pKb. The formula gives the pOH first; the pH is 14 − pOH.
Common questions
When does the approximation hold?
When the acid is weak and the solution is not very dilute, so only a small share dissociates and c(HA) ≈ c₀.