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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 61
Total: 73
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.
Solution:  
Degree of hydrolysis, h = √
Kw
Ka.C
 = √
1×10−14
4.5×10−4×0.04
 = 2.36 × 10−5
pH =
1
2
pKw+
1
2
pKa+
1
2
‌log‌C
=
1
2
[−log‌10−14+(−log‌4.5×10−4)+log(4×10−2)]
= 7.975
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