If the dissociation constant of HCN is 1.3 ti | vedclass

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

Question

(A) For a salt of a weak acid and strong base,the degree of hydrolysis $h$ is given by $h = \sqrt{\frac{K_w}{K_a 	imes C}}$.Given: $K_w = 10^{-14}$,$

Vedclass · Accepted Answer

$10.4, 2.77\%$

Vedclass · Answer

(A) For a salt of a weak acid and strong base,the degree of hydrolysis $h$ is given by $h = \sqrt{\frac{K_w}{K_a 	imes C}}$.Given: $K_w = 10^{-14}$,$K_a = 1.3 	imes 10^{-9}$,$C = \frac{1}{100} = 0.01 \ M$.$h = \sqrt{\frac{10^{-14}}{1.3 	imes 10^{-9} 	imes 0.01}} = \sqrt{\frac{10^{-14}}{1.3 	imes 10^{-11}}} = \sqrt{0.769 	imes 10^{-3}} = \sqrt{7.69 	imes 10^{-4}} \approx 0.0277$.Percentage of hydrolysis $= h 	imes 100 = 0.0277 	imes 100 = 2.77\%$.The $pH$ of the salt solution is given by $pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C$.$pK_a = -\log(1.3 	imes 10^{-9}) = 9 - \log(1.3) = 9 - 0.1139 = 8.886$.$pH = 7 + \frac{1}{2}(8.886) + \frac{1}{2}\log(0.01) = 7 + 4.443 - 1 = 10.443 \approx 10.4$.

If the dissociation constant of $HCN$ is $1.3 \times 10^{-9}$ and $K_w = 1 \times 10^{-14}$,then the $pH$ and percentage of hydrolysis of $\frac{N}{100} \ KCN$ are respectively:

Similar Questions

Which of the following is a basic salt?

Which of the following aqueous solutions will be most basic?

An aqueous solution of $Al_2(SO_4)_3$ is .......

Assertion $(A)$: The aqueous solution of $CH_3COONa$ is alkaline in nature.
Reason $(R)$: Acetate ion undergoes anionic hydrolysis.
The correct answer is

The degree of hydrolysis of a $0.01 \ M$ ammonium acetate solution is = ........ [ $K_h = 3.175 \times 10^{-5}$ ]

Vedclass Test Series

Exam Paper Generator

Online Exam Module