A monobasic acid is 5 % dissociated in its 0. | vedclass

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

Question

(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{	ext{Percent dissociation}}{100} = \frac{5}{100} = 0.05 = 5 	imes 10^{-2}$.Given

Vedclass · Accepted Answer

$5 	imes 10^{-5}$

Vedclass · Answer

(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{	ext{Percent dissociation}}{100} = \frac{5}{100} = 0.05 = 5 	imes 10^{-2}$.Given concentration $c = 0.02 \ M = 2 	imes 10^{-2} \ M$.For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 c$ (since $\alpha$ is very small,$1 - \alpha \approx 1$).Substituting the values: $K_a = (5 	imes 10^{-2})^2 	imes (2 	imes 10^{-2})$.$K_a = (25 	imes 10^{-4}) 	imes (2 	imes 10^{-2}) = 50 	imes 10^{-6} = 5 	imes 10^{-5}$.

$A$ monobasic acid is $5 \%$ dissociated in its $0.02 \ M$ solution. Calculate the dissociation constant of the acid.

Similar Questions

Dimethyl amine $(CH_3)_2NH$ is a weak base and its ionization constant is $5.4 \times 10^{-5}$. Calculate $[OH^{-}]$,$[H_3O^{+}]$,$pOH$ and $pH$ of its $0.2 \ M$ solution at equilibrium.

The dissociation constant of a weak monobasic acid is $2.7 \times 10^{-5}$. If the degree of dissociation of the acid is $3 \times 10^{-2}$,what is the concentration of the acid (in $M$)?

The $pH$ of $0.1 \ M$ acetic acid is $3$. The dissociation constant of the acid will be:

For a diprotic organic acid in aqueous solution,the ionization constants are $K_1 = 4.2 \times 10^{-7}$ and $K_2 = 4.8 \times 10^{-11}$. For a $0.034 \ M$ solution of this acid,select the correct statement.

The $pH$ of a $0.01 \, M$ solution of acetic acid having a degree of dissociation of $12.5 \%$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module