Question

What is the conc. of acetic acid that must be added to 0.5M HCOOH solution so that the dissociation of both acids is the same? [An example of Isohydric solutions] $K_a$ of acetic acid $1.8\times {10}^{-5},K_a$ of formic acid = $2.4\times {10}^{-4}$

• A. 4.5 M
• B. 3.2 M
• C. 6.66 M
• D. 12.3 M

Answer 1

The correct option is C

6.66 M

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$

$HCOOH\rightleftharpoons HCO{O}^{-}+H^{+}$

Since the acids are isohydric, the [$H^{+}$] of acetic acid must be equal to $H^{+}$ of formic acid

[$H^{+}$] of $HCOOH$ = [$H^{+}$] of $C{H}_{3}COOH$

$C_1{\alpha }_1=C_2{\alpha }_2$

$\therefore$ $\sqrt{K_{a1}{C}_1}=\sqrt{Ka2C_2}$

$\sqrt{2.4\times {10}^{-4}\times 0.5}=\sqrt{1.8\times {10}^{-5}\times C_2}$
$\therefore 2.4\times {10}^{-4}\times 0.5=1.8\times {10}^{-5}\times C_2$

$C_2=\frac{2.4\times {10}^{-4}\times 0.5}{1.8\times {10}^{-5}}=\frac{2.4\times {10}^{-5}\times 5}{1.8\times {10}^{-5}}=\frac{12.0}{1.8}$ = 6.66M