Calculate [H^{+}] of a 0.01 M A^{+}B^{-} sal | vedclass

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

Question

(D) The salt $A^{+}B^{-}$ is a salt of a strong base and a weak acid $HB$.Hydrolysis reaction: $B^{-} + H_2O \leftrightarrow HB + OH^{-}$.Hydrolysis c

Vedclass · Accepted Answer

$1.6 	imes 10^{-12} \ M$

Vedclass · Answer

(D) The salt $A^{+}B^{-}$ is a salt of a strong base and a weak acid $HB$.Hydrolysis reaction: $B^{-} + H_2O \leftrightarrow HB + OH^{-}$.Hydrolysis constant $K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{10^{-12}} = 10^{-2}$.For a salt of a weak acid and strong base,$K_h = \frac{C\alpha^2}{1-\alpha}$.Given $C = 0.01 \ M$,we have $\frac{0.01 \alpha^2}{1-\alpha} = 10^{-2}$,which simplifies to $\alpha^2 = 1 - \alpha$ or $\alpha^2 + \alpha - 1 = 0$.Solving for $\alpha$ using the quadratic formula: $\alpha = \frac{-1 + \sqrt{1 + 4}}{2} = \frac{\sqrt{5}-1}{2} \approx 0.618$.$[OH^{-}] = C\alpha = 0.01 	imes 0.618 = 6.18 	imes 10^{-3} \ M$.$[H^{+}] = \frac{K_w}{[OH^{-}]} = \frac{10^{-14}}{6.18 	imes 10^{-3}} \approx 1.6 	imes 10^{-12} \ M$.

Calculate $[H^{+}]$ of a $0.01 \ M$ $A^{+}B^{-}$ salt solution if the $K_a$ of $HB$ is $10^{-12}$.

Similar Questions

Which of the following salt solutions turns red litmus blue?

Which of the following is not an acidic salt?

The aqueous solution of which of the following salts has the lowest $pH$?

When solid potassium cyanide $(KCN)$ is added to water,then:

The aqueous solution of disodium hydrogen phosphate $(Na_2HPO_4)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module