A centimolar solution of a weak acid is 1.3 % | vedclass

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

Question

(D) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degr

Vedclass · Accepted Answer

$1.7 	imes 10^{-6}$

Vedclass · Answer

(D) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.Given: $C = 0.01 \ M$ (centimolar) and $\alpha = 1.3 \% = 0.013$.Substituting the values: $K_a = 0.01 	imes (0.013)^2$.$K_a = 0.01 	imes 0.000169 = 1.69 	imes 10^{-6}$.Rounding to the nearest option,the value is $1.7 	imes 10^{-6}$.

$A$ centimolar solution of a weak acid is $1.3 \%$ dissociated at $298 \ K$. What is its dissociation constant?

Similar Questions

$0.02 \, M$ monobasic acid dissociates $2 \%$,hence,$pH$ of the solution is

Calculate the $pH$ of a $0.02 \ M$ ammonia solution that is $5\%$ ionized.

$A$ weak base is $1.42 \%$ dissociated in its $0.05 \ M$ solution. Calculate its dissociation constant.

If a $0.1 \ N$ solution of $CH_3COOH$ is $1.3\%$ ionized,what will be the $pH$ of the solution? (Given: $\log 1.3 = 0.11$)

$A$ monoprotic acid in $1.00 \ M$ solution is $0.01\%$ ionized. The dissociation constant of this acid is

Vedclass Test Series

Exam Paper Generator

Online Exam Module