The dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10-4. The pH of 0.01 M solution of its sodium salt is

Question

The dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10-4. The pH of 0.01 M solution of its sodium salt is (A) 7 (B) 8 (C) 9 (D) 10

Tardigrade Admin · Accepted Answer

The hydrolysis reaction of conjugate base of acid is

A_{(a q)}^{-} + H_{2} O \to H O^{-} + H A

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

Since, degree of hydrolysis is negligible;

∴ [O H^{-}] = K_{h} C

= 1 0^{- 10} \times 1 0^{- 2} = 1 0^{- 12} = 1 0^{- 6}

∴ pO H = l o g 1 0^{- [O H^{-}]}

∴ pO H = 6

p H = 14 - 6 = 8

Q. The dissociation constant of a substituted benzoic acid at $2 5^{\circ} C$ is $1.0 \times 1 0^{- 4}$ . The $p H$ of $0.01 M$ solution of its sodium salt is

$7$

$8$

$9$

$10$

Q. The dissociation constant of a substituted benzoic acid at 25∘C is 1.0×10−4. The pH of 0.01M solution of its sodium salt is

7

8

9

10

The hydrolysis reaction of conjugate base of acid is A(aq)−​+H2​O→HO−+HA Kh​=Ka​Kw​​=10−410−14​=10−10 Since, degree of hydrolysis is negligible; ∴[OH−]=Kh​C​ =10−10×10−2​=10−12​=10−6 ∴pOH=log10−[OH−] ∴pOH=6 pH=14−6=8

Q. The dissociation constant of a substituted benzoic acid at $2 5^{\circ} C$ is $1.0 \times 1 0^{- 4}$ . The $p H$ of $0.01 M$ solution of its sodium salt is

$7$

$8$

$9$

$10$