Question

The ionization constant of nitrous acid is $4.5\times {10}^{-4}$. Calculate the $pH$ of $0.04M$ sodium nitrite solution and also its degree of hydrolysis.

Answer 1

Hint: Sodium nitrite is salt which is basic in nature because it is the salt of weak acid and strong base. So therefore, the sodium nitrite is higher than nature and on hydrolysis with water it produces hydroxide ions along with weak acid.

Complete answer:
Sodium nitrite is a basic salt as it is made up by nitrous acid and sodium hydroxide. Nitrous acid is a weak acid whereas sodium hydroxide is a strong base therefore overall sodium nitrite is basic salt.
The formula to calculate the $pH$ of the following reaction is:
 $pH=7+{\displaystyle \frac{1}{2}}pKa+{\displaystyle \frac{1}{2}}\log C$

No of moles of sodium nitrite is 0.04M
$$\left[NaN{O}_2\right]=0.04M$$

Ionization constant of nitrous acid, Ka=$4.5\times {10}^{-4}$
Strength of an acid is the negative log an acid constant
$$pKa=-logKa=-log(4.5\times {{10}^{-}}^4)=4-0.6532=3.3468\approx 3.35$$

Now,
$pH=7+3.35+{\displaystyle \frac{log(0.04)}{2}}=7+3.35+{\displaystyle \frac{log(4\times {{10}^{-}}^2)}{2}}=7+3.35-2+{\displaystyle \frac{log4}{2}}=7+3.35-{\displaystyle \frac{1.39}{2}}=7.98$
$pH$=7.98

Degree of hydrolysis is the amount of salt which will hydrolyze at equilibrium.

Now, use formula for degree of Hydrolysis,

$h=\sqrt{{\displaystyle \frac{kw}{k_a\times C}}}$


$$h=\sqrt{{\displaystyle \frac{{10}^{{}^{-}14}}{4.5\times {{10}^{-}}^4\times 0.04}}}=\sqrt{{\displaystyle \frac{{{10}^{-}}^8}{4.5\times 4}}}=\sqrt{{\displaystyle \frac{100}{18}}}\times {{10}^{-}}^5=2.36\times {{10}^{-}}^5$$

Therefore, the $pH$ of the solution is 7.98 and the degree of hydrolysis of sodium nitrite is $2.36\times {10}^{-5}$.

Note: Salts of weak acid and strong base have higher $pH$ than 7. The conjugate acid of basic salt is a weak acid that is its pH would be more than 7. That’s why sodium nitrate is basic salt as its conjugate acid is nitrous acid which is a weak acid whereas its base is a strong base which is sodium hydroxide.
The anion of this salt is reactive so on hydrolysis it reacts with water to form weak acid and hydroxide ions.
So, its pH is calculated by:

$pH=7+{\displaystyle \frac{1}{2}}pKa+{\displaystyle \frac{1}{2}}\log C$
Degree of hydrolysis is the amount of salt that can be hydrolyzed at equilibrium.
Degree of hydrolysis of salt is formulated by:

$h=\sqrt{{\displaystyle \frac{kw}{k_a\times C}}}$