Ionic EquilibriumMedium
Question
If the pH of 0.001 M potassium propionate solution be 8.0, then the dissociation constant of propionic acid will be
Options
A.10–3
B.10–2
C.10–2.5
D.10–5
Solution
Potassium propionate is the salt of strong base and weak acid and hence.
$P^{H} = 7 + \frac{1}{2}\left( P^{K_{a}} + \log C \right) $$$\text{Or, }8 = 7 + \frac{1}{2}\left( P^{K_{a}} + \log 0.001 \right) $$$\Rightarrow P^{K_{a}} = 5 \Rightarrow K_{a} = 10^{- 5}$
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