First slide

Basic Buffer

Question

A buffer solution contains 1 mole of ${({NH}_{4})}_{2} {SO}_{4}$ and 1 mole of ${NH}_{4} OH (K_{b} = 10^{- 5}) .$ The pH of solution will be

Moderate

A

5
B

9
C

5.3
D

8.7

Solution

$pOH = {pK}_{b} + \log \frac{[Salt]}{[Base]} = {pK}_{b} + \log \frac{[Cation]}{[Base]}$
${({NH}_{4})}_{2} {SO}_{4} ⇌ 2 {NH}_{4}^{+} + {SO}_{4}^{- 2} [{NH}_{4}^{+}] = 2 \times 1 = 2 P^{K_{b}} = - \log K_{b} = - \log 10^{- 5} = 5 ∴ pOH = 5 + \log 2 = 5.3 or pH = 14 - 5.3 = 8.7$

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