If the K_b value in the hydrolysis reaction B | vedclass

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

Question

(C) The hydrolysis reaction for the cation $B^{+}$ is given by: $B^{+} + H_2O \rightleftharpoons BOH + H^{+}$.For the hydrolysis of a salt of a weak b

Vedclass · Accepted Answer

$1.0 	imes 10^{-8}$

Vedclass · Answer

(C) The hydrolysis reaction for the cation $B^{+}$ is given by: $B^{+} + H_2O ightleftharpoons BOH + H^{+}$.For the hydrolysis of a salt of a weak base and a strong acid,the hydrolysis constant $K_H$ is related to the dissociation constant of the base $K_b$ and the ionic product of water $K_w$ by the formula: $K_H = \frac{K_w}{K_b}$.Given $K_w = 1.0 	imes 10^{-14}$ and $K_b = 1.0 	imes 10^{-6}$.Substituting the values: $K_H = \frac{1.0 	imes 10^{-14}}{1.0 	imes 10^{-6}} = 1.0 	imes 10^{-8}$.

If the $K_b$ value in the hydrolysis reaction $B^{+} + H_2O \rightleftharpoons BOH + H^{+}$ is $1.0 \times 10^{-6}$,then the hydrolysis constant of the salt would be

Similar Questions

Calculate $K_h$ and $pH$ of $0.1 \ M$ $NH_4Cl$ solution. [ $K_w = 1 \times 10^{-14}$,$K_{NH_4OH} = 1.75 \times 10^{-5}$ ]

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 $..... \%$.

Which of the following is a basic salt?

What should be the degree of hydrolysis of $0.01 \ M$ Lithium formate solution at $25 \ ^oC$,if $K_h = 3.175 \times 10^{-5}$?

The ionization constant of ammonium hydroxide is $1.77 \times 10^{-5}$ at $298 \ K$. The hydrolysis constant of ammonium chloride is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module