Aqueous solutions of 1 M potassium acetate a | vedclass

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

Question

(B) Both $CH_3COOK$ and $HCOOK$ undergo hydrolysis in water to form their respective weak acids,$CH_3COOH$ and $HCOOH$.The degree of hydrolysis $(h)$

Vedclass · Accepted Answer

$[CH_3COOH] > [HCOOH]$

Vedclass · Answer

(B) Both $CH_3COOK$ and $HCOOK$ undergo hydrolysis in water to form their respective weak acids,$CH_3COOH$ and $HCOOH$.The degree of hydrolysis $(h)$ is given by $h = \sqrt{\frac{K_h}{C}}$.Since $K_h = \frac{K_w}{K_a}$,we have $h = \sqrt{\frac{K_w}{K_a 	imes C}}$.The concentration of the acid formed is $[Acid] = C 	imes h = \sqrt{\frac{K_w 	imes C}{K_a}}$.Since $K_a$ of $HCOOH$ $(1.8 	imes 10^{-4})$ is greater than $K_a$ of $CH_3COOH$ $(1.8 	imes 10^{-5})$,the concentration of $HCOOH$ formed will be less than the concentration of $CH_3COOH$ formed.Therefore,$[CH_3COOH] > [HCOOH]$.

Aqueous solutions of $1 \ M$ potassium acetate and $1 \ M$ potassium formate are taken in separate beakers. Which of the following relationships is correct?

Similar Questions

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

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

At $25 \ ^\circ C$,a weak acid $HX$ has the dissociation constant $K_a = 1 \times 10^{-5}$. It forms a salt $NaX$ on reaction with an alkali. The percentage of degree of hydrolysis of a $0.1 \ M$ solution of $NaX$ is $..... \%$.

$pH$ of $10^{-3} \ M$ $NaCl$ solution at $25 \ ^oC$ is

$A$ salt is formed from a weak acid and a weak base. If $pK_b > pK_a$,then the $pH$ of the aqueous solution is $.....$

Vedclass Test Series

Exam Paper Generator

Online Exam Module