In aqueous solution,the ionization constants | vedclass

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

Question

(C) For the first dissociation step: $H_2CO_3(aq) ightleftharpoons HCO_3^-(aq) + H^+(aq)$.Given $K_1 = 4.2 	imes 10^{-7}$ and initial concentration

Vedclass · Accepted Answer

The concentrations of $H^{+}$ and $HCO_3^-$ are approximately equal.

Vedclass · Answer

(C) For the first dissociation step: $H_2CO_3(aq) ightleftharpoons HCO_3^-(aq) + H^+(aq)$.Given $K_1 = 4.2 	imes 10^{-7}$ and initial concentration $C = 0.034 \ M$.Since $K_1$ is very small,$[H^+] \approx [HCO_3^-] = \sqrt{K_1 	imes C} = \sqrt{4.2 	imes 10^{-7} 	imes 0.034} \approx 1.195 	imes 10^{-4} \ M$.For the second dissociation step: $HCO_3^-(aq) ightleftharpoons CO_3^{2-}(aq) + H^+(aq)$.$K_2 = \frac{[CO_3^{2-}][H^+]}{[HCO_3^-]}$.Since $[H^+] \approx [HCO_3^-]$,we have $[CO_3^{2-}] = K_2 = 4.8 	imes 10^{-11} \ M$.Comparing the values,$[H^+] = [HCO_3^-] = 1.195 	imes 10^{-4} \ M$,which are approximately equal.

In aqueous solution,the ionization constants for carbonic acid are $K_1 = 4.2 \times 10^{-7}$ and $K_2 = 4.8 \times 10^{-11}$. Select the correct statement for a saturated $0.034 \ M$ solution of carbonic acid.

Similar Questions

The $pH$ of $0.1 \ M$ $HCN$ solution is $5.2$. Calculate the dissociation constant $K_a$ of this solution.

The $pH$ of $1\, N$ aqueous solutions of $HCl$,$CH_{3}COOH$ and $HCOOH$ follows the order

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

The $pK_a$ of acetylsalicylic acid (aspirin) is $3.5$. The $pH$ of gastric juice in the human stomach is about $2-3$ and the $pH$ in the small intestine is about $8$. Aspirin will be

Calculate $[H_3O^{+}]$ of a monobasic acid if it is $0.04 \%$ dissociated in $0.05 \ M$ solution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module