At 25^{circ} C, the ionization constant for a | vedclass

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

Question

(A) The hydrolysis constant $(K_h)$ for the salt of a weak base and a strong acid is given by the formula: $K_h = \frac{K_w}{K_b}$ Given that the ioni

Vedclass · Accepted Answer

$2.00 	imes 10^{-5}$

Vedclass · Answer

(A) The hydrolysis constant $(K_h)$ for the salt of a weak base and a strong acid is given by the formula: $K_h = \frac{K_w}{K_b}$ Given that the ionization constant of the weak base $(K_b)$ is $5.00 	imes 10^{-10}$ and the ionic product of water $(K_w)$ at $25^{\circ} C$ is $1.00 	imes 10^{-14}$. Substituting the values: $K_h = \frac{1.00 	imes 10^{-14}}{5.00 	imes 10^{-10}} = 0.20 	imes 10^{-4} = 2.00 	imes 10^{-5}$ Thus, the hydrolysis constant of anilinium chloride is $2.00 	imes 10^{-5}$.

At $25^{\circ} C$, the ionization constant for anilinium hydroxide is $5.00 \times 10^{-10}$. The hydrolysis constant of anilinium chloride is

Similar Questions

Which of the following salts will not undergo hydrolysis?

Do you expect different products in solution when aluminium $(III)$ chloride and potassium chloride are treated separately with $(i)$ normal water,$(ii)$ acidified water,and $(iii)$ alkaline water? Write equations wherever necessary.

An aqueous solution of $CuSO_4$ will be:

If the dissociation constant of $HCN$ is $10^{-7}$,what are the values of $pH$,$h$ (degree of hydrolysis),and $[OH^-]$ for a $10^{-3} \ M$ solution of $KCN$,respectively?

An aqueous solution of sodium carbonate is alkaline because sodium carbonate is a salt of which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module