What is the molar concentration of a weak mon | vedclass

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

Question

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is $HA \rightleftharpoons H^+ + A^-$.Let $C$ be the initial molar concentration and $x

Vedclass · Accepted Answer

$0.002$

Vedclass · Answer

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is $HA ightleftharpoons H^+ + A^-$.Let $C$ be the initial molar concentration and $x$ be the degree of dissociation.$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(Cx)(Cx)}{C(1-x)} = \frac{Cx^2}{1-x}$.Since the acid is weak,$x \ll 1$,so $1-x \approx 1$,and $K_a \approx Cx^2$.Given $K_a = 5 	imes 10^{-8}$ and $x = 0.5 \% = 0.005 = 5 	imes 10^{-3}$.$C = \frac{K_a}{x^2} = \frac{5 	imes 10^{-8}}{(5 	imes 10^{-3})^2}$.$C = \frac{5 	imes 10^{-8}}{25 	imes 10^{-6}} = \frac{5}{25} 	imes 10^{-2} = 0.2 	imes 10^{-2} = 0.002 \ M$.

What is the molar concentration of a weak monobasic acid if its dissociation constant is $5 \times 10^{-8}$ and it undergoes $0.5 \%$ dissociation (in $M$)?

Similar Questions

The amount of acetic acid in $1 \ L$ of solution having $\alpha = 1\%$ and $K_a = 1.8 \times 10^{-5}$ is ............. $g$.

Calculate $[H_3O^{+}]$ in $0.02 \ M$ solution of monobasic acid if dissociation constant is $1.8 \times 10^{-5}$.

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.

$A$ weak monobasic acid is $20 \%$ dissociated in its $0.01 \ M$ solution. What is the dissociation constant of the weak acid?

The hydrogen ion concentration of a $0.006 \ M$ benzoic acid solution is $(K_a = 6 \times 10^{-5})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module