Question

Both $HCOOH$ and $C{H}_{3}COOH$ solutions have equal pH. If $\frac{K_1}{K_2}$ (ratio of acid ionization constants) of these acids is $4$, their molar concentration ratio will be

• A. 2
• B. 0.5
• C. 4
• D. 0.25

Answer 1

Answer: (D)

As $pH$ of $HCOOH$ sol. $=pH$ of $C{H}_{3}COOH$ sol.
$\left[H^{+}\right]$ in $HCOOH$ sol. $=\left[H^{+}\right]$in $C{H}_{3}COOH$ sol.
$K_1=\frac{\left[HCO{O}^{-}\right]\left[H^{+}\right]}{[HCOOH]}$
$=\frac{{\left[H^{+}\right]}^2}{[HCOOH]}$
$K_2=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$
$=\frac{{\left[H^{+}\right]}^2}{\left[C{H}_{3}COOH\right]}$
$\Rightarrow K_1[HCOOH]=K_2\left[C{H}_{3}COOH\right]$
$\therefore \frac{[HCOOH]}{\left[C{H}_{3}COOH\right]}=\frac{K_2}{K_1}$
$=\frac{1}{4}=0.25$