The formula K_h = K_w / K_a is applicable to | vedclass

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

Question

(C) The hydrolysis constant $K_h$ is related to the dissociation constants of the acid $(K_a)$ and base $(K_b)$ and the ionic product of water $(K_w)$

Vedclass · Accepted Answer

$CH_3COONa$

Vedclass · Answer

(C) The hydrolysis constant $K_h$ is related to the dissociation constants of the acid $(K_a)$ and base $(K_b)$ and the ionic product of water $(K_w)$.For a salt of a weak acid and a strong base (e.g.,$CH_3COONa$),the hydrolysis constant is given by $K_h = K_w / K_a$.For a salt of a strong acid and a weak base (e.g.,$NH_4Cl$ or $FeCl_3$),the formula is $K_h = K_w / K_b$.For a salt of a weak acid and a weak base (e.g.,$CH_3COONH_4$),the formula is $K_h = K_w / (K_a 	imes K_b)$.Thus,$K_h = K_w / K_a$ applies to $CH_3COONa$.

The formula $K_h = K_w / K_a$ is applicable to which of the following salts?

Similar Questions

The $pH$ of an aqueous solution of $CH_3COONa$ of concentration $C \ M$ is given by:

On adding solid potassium cyanide to water:

Which of the following salts on dissolution in water gives $pH$ less than $7$ at $298 \ K$?

For a weak acid with dissociation constant $K_a$ and its conjugate salt with hydrolysis constant $K_h$,which of the following represents their relationship?

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 $x_2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module