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Question
A buffer solution contains 0.3 mol dm–3 NH4OH (Kb = 1.8 × 10–5) and 0.4 mol dm–3 NH4Cl. Calculate pOH of the solution.
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Solution
Given: [Base] = 0.3 mol dm–3,
[Salt] = 0.4 mol dm–3,
Kb = 1.8 × 10–5 for the weak base
To find: pOH of the buffer solution
Formula: pOH = pKb + log10 `(["salt"])/(["base"])`
Calculation: pOH of basic buffer is given by Henderson-Hasselbalch equation:
pOH = pKb + log10 `(["salt"])/(["base"])`
pKb = – log10Kb
= – log10 (1.8 × 10–5) = 5 – log101.8
= 5 – 0.2553 = 4.7447
Substitution in the Henderson-Hasselbalch equation gives
pOH = 4.7447 + log10 `0.4/0.3` = 4.7447 + log10 1.333
= 4.7447 + 0.1248 = 4.8695
The pOH of the given buffer solution is 4.8695.
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