First slide

Ionic equilibrium

Question

$K_{a} for {CH}_{3} COOH is 1.8 \times 10^{- 5} and K_{b} for {NH}_{4} OH is 1.8 \times 10^{- 5}$ . The pH of 0.01 M ammonium acetate will be

Moderate

Solution

$W A - W B s a l t$
$P^{H} = 7 + \frac{1}{2} P^{K a} - \frac{1}{2} P^{K_{b}}$
$K_{a} = K_{b}$
$∴ P^{H} = 7$

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