Question

What is the concentration of acetic acid which can be added to $0.5M$ $HCOOH$ solution so that dissociation of both is same $K_{{CH}_{3}COOH}=1.8\times {10}^{-5}$; $K_{HCOOH}=2\times {10}^{-4}$.

• A. $C_{{CH}_{3}COOH}=6.67M$
• B. $C_{{CH}_{3}COOH}=3.33M$
• C. $C_{{CH}_{3}COOH}=1.5M$
• D. $C_{{CH}_{3}COOH}=3M$

Answer 1

The correct option is A $C_{{CH}_{3}COOH}=6.67M$

${CH}_{3}COOH\rightleftharpoons {CH}_{3}CO{O}^{-}+H^{+};K_a=1.8\times {10}^{-5}$

$HCOOH\rightleftharpoons HCO{O}^{-}+H^{+};K_a=2.4\times {10}^{-4}$

Considering the dissociation of both acetic acid and HCOOH is same as given in the question.

$\left[H^{+}\right]$ by $HCOOH=\left[H^{+}\right]$ by ${CH}_{3}COOH$

$c_1{\alpha }_1=c_2{\alpha }_2$

or $\sqrt{\left(K_{{\alpha }_1}{c}_1\right)}=\sqrt{\left(K_{{\alpha }_2}{c}_2\right)}$

or $K_{{\alpha }_1}{c}_1=K_{{\alpha }_2}{c}_2$

or $2.4\times {10}^{-4}\times 0.5=1.8\times {10}^{-5}\times c$

$\therefore$ $c_{{CH}_{3}COOH}=6.67M$