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.

Book Online Demo

Check Your IQ

Try Test

Courses

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

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 8.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

It is given to find out the pH of a 0.01M solution

pH = $\frac{1}{2}$ ( $p K_{w} + p K_{a}$ + log c)

Sodium salt of benzoic acid is a strong base-weak acid salt.

pH = $\frac{1}{2}$ ( $p K_{w} + p K_{a}$ + log c)

pH = $\frac{1}{2}$ [ $(- l o g K_{w}) + (- l o g K_{a})$ + log 0.01]

= $\frac{1}{2}$ [14 + 4 -2]

= $\frac{1}{2}$ [16] = 8.

Hence, the answer is 8.

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

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

Take Free Test