The degree of hydrolysis in hydrolytic equili | vedclass

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

Question

(A) For the hydrolysis of a salt of a weak acid and a strong base,the hydrolysis constant is given by $K_h = \frac{K_w}{K_a}$.Given $K_w = 1 	imes 10

Vedclass · Accepted Answer

$1 	imes 10^{-3}$

Vedclass · Answer

(A) For the hydrolysis of a salt of a weak acid and a strong base,the hydrolysis constant is given by $K_h = \frac{K_w}{K_a}$.Given $K_w = 1 	imes 10^{-14}$ and $K_a = 1 	imes 10^{-5}$,we have $K_h = \frac{10^{-14}}{1 	imes 10^{-5}} = 10^{-9}$.The degree of hydrolysis $\alpha$ is related to $K_h$ and concentration $C$ by the formula $\alpha = \sqrt{\frac{K_h}{C}}$ (assuming $\alpha \ll 1$).Given $C = 0.001 \ M = 10^{-3} \ M$,we have $\alpha = \sqrt{\frac{10^{-9}}{10^{-3}}} = \sqrt{10^{-6}} = 1 	imes 10^{-3}$.

The degree of hydrolysis in hydrolytic equilibrium $A^{-} + H_2O \rightleftharpoons HA + OH^{-}$ at salt concentration of $0.001 \ M$ is $(K_a = 1 \times 10^{-5})$

Similar Questions

$A$ $100 \ mL$, $0.1 \ M$ solution of ammonium acetate is diluted by adding $100 \ mL$ of water. The pH of the resulting solution will be ($pK_{a}$ of acetic acid is nearly equal to $pK_{b}$ of $NH_{4}OH$)

The aqueous solution of ammonium chloride is

The correct relationship between the hydrolysis constant $(K_h)$ and the dissociation constant $(K_b)$ for $MgCl_2$ is:

Which of the following aqueous solutions will be most basic?

If the dissociation constants of ammonium hydroxide $(NH_4OH)$ and acetic acid $(CH_3COOH)$ are both $1.8 \times 10^{-5}$ and $K_w = 10^{-14}$,what is the degree of hydrolysis of $CH_3COONH_4$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module