The dissociation constant of NH_4OH is 1.8 ti | vedclass

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

Question

(B) The hydrolysis constant $(K_h)$ for a salt of a weak base and a strong acid $(NH_4Cl)$ is given by the formula: $K_h = \frac{K_w}{K_b}$.Given:$K_w

Vedclass · Accepted Answer

$5.55 	imes 10^{-10}$

Vedclass · Answer

(B) The hydrolysis constant $(K_h)$ for a salt of a weak base and a strong acid $(NH_4Cl)$ is given by the formula: $K_h = \frac{K_w}{K_b}$.Given:$K_w = 1.0 	imes 10^{-14}$ (ionic product of water at $25^{\circ}C$)$K_b = 1.8 	imes 10^{-5}$ (dissociation constant of $NH_4OH$)Substituting the values:$K_h = \frac{1.0 	imes 10^{-14}}{1.8 	imes 10^{-5}}$$K_h = 0.555 	imes 10^{-9} = 5.55 	imes 10^{-10}$.

The dissociation constant of $NH_4OH$ is $1.8 \times 10^{-5}$. Calculate the hydrolysis constant of $NH_4Cl$.

Similar Questions

Which of the following salts has the minimum $pH$ in an aqueous solution?

Which of the following $0.1 \ M$ aqueous solutions will be basic?

Which of the following is not an acidic salt?

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module