Question

Question 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.

Class XI Equilibrium Page 229

Answer 1

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Answer 2

sodium nitrite is a salt of weak acid (HNO2) and strong base (NaOH).hence, salt sodium nitrite is basic .
so, we have to use formula for finding pH;
$pH=7+\frac{pK_a+logC}{2}$
Given, [NaNO2] = 0.04 M
Ka of HNO2 = 4.5 × 10^-4
PKa = -logKa = - log(4.5 × 10^-4)
= 4 - 0.6532 = 3.3468 ≈ 3.35

now, PH = 7 + {3.35 + log(0.04)}/2
= 7 + {3.35 + log(4 × 10^-2)}/2
= 7 + { 3.35 -2 + log4}/2
= 7 + {3.35 - 1.39 }/2 = 7.98

now, use formula for degree of Hydrolysis ,
$h = \sqrt{\frac{Kw}{K_aC}}$
h = √{10^-14/4.5 × 10^-4 × 0.04}
= √{10^-8/4.5 × 4}
= √{100/18} × 10^-5
= 2.36 × 10^-5