Calculate the percentage of hydrolysis in 0.0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $NaOCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HOCN)$.For the salt of a strong base and a weak acid,the hydrolysis constant $K_h$ i

Vedclass · Accepted Answer

$0.01$

Vedclass · Answer

(B) $NaOCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HOCN)$.For the salt of a strong base and a weak acid,the hydrolysis constant $K_h$ is given by $K_h = \frac{K_w}{K_a}$.Given $K_b$ for $OCN^- = 10^{-10}$,we know $K_a 	imes K_b = K_w = 10^{-14}$.Thus,$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4}$.$K_h = \frac{10^{-14}}{10^{-4}} = 10^{-10}$.The degree of hydrolysis $h$ is given by $h = \sqrt{\frac{K_h}{C}}$,where $C = 0.01 \ M = 10^{-2} \ M$.$h = \sqrt{\frac{10^{-10}}{10^{-2}}} = \sqrt{10^{-8}} = 10^{-4}$.Percentage of hydrolysis $= h 	imes 100 = 10^{-4} 	imes 100 = 0.01 \%$.

Calculate the percentage of hydrolysis in $0.01 \ M$ aqueous solution of $NaOCN$ ($K_b$ for $OCN^{-} = 10^{-10}$)

Similar Questions

Find out $K_a$ for the acid when the degree of hydrolysis of $0.1 \ M \ CH_3COONa$ is $1 \ \%$.

The $pK_{a}$ of acetic acid and $pK_{b}$ of ammonium hydroxide are $4.76$ and $4.75$ respectively. Calculate the $pH$ of ammonium acetate solution.

Why is the $pH$ of an aqueous solution of copper sulphate less than $7$?

The aqueous solution of copper sulfate is acidic because...

The hydrolysis reaction $A^{-} + H_2O \rightleftharpoons HA + OH^{-}$ indicates that the salt is formed from:

Vedclass Test Series

Exam Paper Generator

Online Exam Module