Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The ionization constant of nitrous acid is 4.5 × 10^–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

7.61     The ionization constant of nitrous acid is 4.5 × 10-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Answers (1)

best_answer

We have,
Ionization constant of nitrous acid = 4.5\times 10^{-5}
Concentration of sodium nitrite (NaNO_2) = 0.04 M
Degree of hydrolysis can be calculated as;
K_h = \frac{K_w}{K_a}=\frac{10^{-14}}{4.5\times 10^{-4}} = 0.22\times 10^{-10}
Sodium nitrite is a salt of sodium hydroxide (strong base) and the weak acid (HNO_2)
NO_2^-+H_{2}O\rightleftharpoons HNO_{2}+OH^-
Suppose x moles of salt undergoes hydrolysis, then the concentration of-
[NO_2^-]=0.04-x \approx 0.04
[HNO_2^-]=x, and 
[OH^-] = x

Therefore 
k_h = \frac{x^2}{0.04}=0.22\times 10^{-10}
from here we can calculate the value of x ;
\Rightarrow x = \sqrt{0.0088\times 10^{-10}} = 0.093 \times 10^{-5} = [OH^-]

\Rightarrow [H_3O^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{0.093\times 10^{-5}}
                                              =10.75 \times 10^{-9} M

Now p^H = -\log (10.75 \times 10^{-9} M) = 7.96

Therefore the degree of hydrolysis  

Posted by

manish

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question