The degree of ionization of 0.10 M lactic ac | vedclass

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

Question

(B) Given: Concentration $C = 0.10 \ M$,degree of ionization $\alpha = 4.0 \% = 0.04$.For the dissociation of lactic acid $(CH_3CH(OH)COOH)$:$CH_3CH(O

Vedclass · Accepted Answer

$1.66 	imes 10^{-4}$

Vedclass · Answer

(B) Given: Concentration $C = 0.10 \ M$,degree of ionization $\alpha = 4.0 \% = 0.04$.For the dissociation of lactic acid $(CH_3CH(OH)COOH)$:$CH_3CH(OH)COOH ightleftharpoons CH_3CH(OH)COO^- + H^+$At equilibrium,the concentrations are:$[CH_3CH(OH)COOH] = C(1 - \alpha)$$[CH_3CH(OH)COO^-] = C\alpha$$[H^+] = C\alpha$The dissociation constant $K_a$ is given by:$K_a = \frac{[CH_3CH(OH)COO^-][H^+]}{[CH_3CH(OH)COOH]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha}$Substituting the values:$K_a = \frac{0.1 	imes (0.04)^2}{1 - 0.04} = \frac{0.1 	imes 0.0016}{0.96} = \frac{0.00016}{0.96} = \frac{1.6 	imes 10^{-4}}{0.96} \approx 1.66 	imes 10^{-4}$

The degree of ionization of $0.10 \ M$ lactic acid is $4.0 \%$. The value of $K_a$ is:

Similar Questions

For a $10 \ M \ CH_3COOH$ solution,if $K_a = 10^{-5}$,what will be the values of $[H^{+}]$ and $pH$ respectively?

$A$ monoprotic acid in a $0.1 \ M$ solution ionizes to $0.001 \%$. Its ionisation constant is

For a weak acid $HA$ with dissociation constant $10^{-9}$,the $pOH$ of its $0.1 \ M$ solution is:

The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its $0.05 \ M$ solution. Calculate the concentration of acetate ion in the solution and its $pH$.

What will be the $pH$ of a $0.1 \ M$ $NH_3$ solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module