The dissociation constant of a substituted benzoic acid at 250C is 1.0×10−4 Find the pH of a 0.01M solution of its sodium salt.

Ka=1×10−4 C6H5COOΘ+H2O⇌C6H5COOH+OH− 0.01(1−h) 0.01h 0.01h Kb=KwKa=0.01h21−h⇒h=10−4 [OH−]=0.01h=0.01×10−4=10−6 [H+]=10−8 PH=8

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The dissociation constant of a substituted benzoic acid at $25^{0} C$ is $1.0 \times 10^{- 4}$ Find the pH of a 0.01M solution of its sodium salt.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 8.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$K_{a} = 1 \times 10^{- 4} C_{6} H_{5} C O O^{Θ} + H_{2} O ⇌ C_{6} H_{5} C O O H + O H^{-} 0.01 (1 - h) 0.01 h 0.01 h K_{b} = \frac{K_{w}}{K_{a}} = \frac{0.01 h^{2}}{1 - h} \Rightarrow h = 10^{- 4} [O H^{-}] = 0.01 h = 0.01 \times 10^{- 4} = 10^{- 6} [H^{+}] = 10^{- 8} P^{H} = 8$