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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 61 of 73
Marks: +1, -0
The ionization constant of nitrous acid is 4.5 × 10−410^{-4}. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Solution:  
Degree of hydrolysis, h = KwKa⋅C\sqrt{\frac{K_w}{K_a \cdot C}} = 1×10−144.5×10−4×0.04\sqrt{\frac{1 \times 10^{-14}}{4.5 \times 10^{-4} \times 0.04}} = 2.36 × 10−510^{-5}
pH = 12pKw+12pKa+12log⁡C\frac{1}{2} pK_w + \frac{1}{2} pK_a + \frac{1}{2} \log C
=
12[−log⁡10−14+(−log⁡4.5×10−4)+log⁡(4×10−2)]\frac{1}{2}[-\log 10^{-14}+(-\log 4.5 \times 10^{-4})+ \log (4 \times 10^{-2})]
= 7.975
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