What will be the pH at the equivalence point during the titration of a 100 mL 0.2 M solution of CH3COONa with 0.2 M solution of HCl ?Ka=2×10−5

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

What will be the pH at the equivalence point during the titration of a $100 mL 0.2 M$ solution of ${CH}_{3} COONa$ with $0.2 M$ solution of $HCl$ ?

$[K_{a} = 2 \times 10^{- 5}]$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$3 + \log \sqrt{2}$

$3 - \log 2$

$3 - \log \sqrt{2}$

$3 + \log 2$

answer is A.

(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

The chemical formula for the sodium titration reaction is balanced.

$HCl$ and acetate are ${CH}_{3} COONa + HCl ⟶ {CH}_{3} COOH + NaCl$

To create 200 ml of 0.1 M sodium acetate, 100 ml of 0.2 M sodium acetate must be combined with 100 ml of 0.2 M $HCl$ .

$pH = \frac{1}{2} [{pK}_{a} - \log [{CH}_{3} COOH]]$

$pH = \frac{1}{2} [4.69 - \log 0.1] = 2.849 = 3 - \log \sqrt{2}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE