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

The aqueous solution of $FeCl_3$ is acidic in nature. This is due to:

An aqueous solution of $CH_3COONa$ will be

If the $pH$ of a solution is less than $7$,it will be ......

Which of the following relations is correct for a solution of $CH_3COONH_4$ salt? Concentration $0.001 \ M$,degree of hydrolysis $x_1$; Concentration $0.002 \ M$,degree of hydrolysis $x_2$.

Which of the following has the highest degree of hydrolysis?

Vedclass Test Series

Exam Paper Generator

Online Exam Module