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# 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.

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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

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