The dissociation constant of a substituted benzoic acid at $25^{\circ} \mathrm{C}$ is $1.0 \times 10^{-4}$. The pH of a 0.01 M solution of its sodium salt is

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Equilibrium

The dissociation constant of a substituted benzoic acid at

2 5^{\circ} C

1.0 \times 1 0^{- 4}

. The pH of a 0.01 M solution of its sodium salt is

Answer: 8

For the salt of a weak acid, hydrolysis occurs:

C_{6} H_{5} COO^{-} + H_{2} O ⇌ C_{6} H_{5} COOH + OH^{-}

Hydrolysis constant:

K_{h} = \frac{K _{w}}{K _{a}} = \frac{1 0 ^{- 14}}{1 0 ^{- 4}} = 1 0^{- 10}

For 0.01 M solution:

[OH^{-}] = K_{h} \times C = 1 0^{- 10} \times 1 0^{- 2} = 1 0^{- 6}

[H^{+}] = \frac{1 0 ^{- 14}}{1 0 ^{- 6}} = 1 0^{- 8}

M

pH = 8

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Integer answer