The dissociation constants of CH_3COOH and HC | vedclass

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

Question

(A) The salts are formed from a weak acid and a strong base.The degree of hydrolysis ($h$ or $x$) is given by the formula: $x = \sqrt{\frac{K_h}{C}} =

Vedclass · Accepted Answer

$x_1 > x_2$

Vedclass · Answer

(A) The salts are formed from a weak acid and a strong base.The degree of hydrolysis ($h$ or $x$) is given by the formula: $x = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}{K_a 	imes C}}$.Assuming the concentration $(C)$ is the same for both salts,the relationship is $x \propto \sqrt{\frac{1}{K_a}}$.For $KCN$,the acid is $HCN$ $(K_a = 7.2 	imes 10^{-10})$.For $CH_3COOK$,the acid is $CH_3COOH$ $(K_a = 1.8 	imes 10^{-5})$.Since $K_a(CH_3COOH) > K_a(HCN)$,it follows that $x_1 > x_2$ because $x$ is inversely proportional to the square root of $K_a$.

The dissociation constants of $CH_3COOH$ and $HCN$ are $1.8 \times 10^{-5}$ and $7.2 \times 10^{-10}$ respectively. What is the relationship between the degree of hydrolysis of $KCN$ $(x_1)$ and $CH_3COOK$ $(x_2)$?

Similar Questions

Which is the correct expression for the hydrolysis constant $(K_h)$ of $NH_4CN$?

The approximate $pH$ of $0.01 \ M \ NaHA$ is calculated by: ($K_{a_1} = 10^{-6}$ and $K_{a_2} = 10^{-8}$ are ionization constants of $H_2A$)

$pH$ of $NH_4Cl$ solution is $5.28$. Calculate the degree of hydrolysis $(h)$ of its $0.02 \ M$ solution.

Calculate the percentage hydrolysis of $NaOCN$ in its $0.003 \ M$ aqueous solution ($K_a$ for $HOCN = 3.33 \times 10^{-4} \ M$) .....$\%$

What will be the $pH$ of an aqueous solution of $MgCl_2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module