Question

Ionization constant of ${CH}_{3}COOH$ is $1.7\times {10}^{-5}$ and concentration $H+$ ions is $3.4\times {10}^{-4}$. Then find out initial concentration of${CH}_{3}COOH$ molecules.

• A. $3.4\times {10}^{-4}$
• B. $3.4\times {10}^{-3}$
• C. $6.8\times {10}^{-4}$
• D. $6.8\times {10}^{-3}$

Answer 1

Answer: (D)

$6.8\times {10}^{-3}$

$\underset{\text{At equillibrium}}{\underset{\text{Initially}}{}}\underset{(1-x)}{\underset{1}{C{H}_{3}COOH}}⟶\underset{x}{\underset{0}{{C{H}_{3}COO}^{-}}}+\underset{x}{\underset{0}{H^{+}}}$

At equillibrium-

$\left[{C{H}_{3}COO}^{-}\right]=\left[H^{+}\right]=x$

Given that $\left[H^{+}\right]=3.4\times {10}^{-4}$

$\therefore \left[{C{H}_{3}COO}^{-}\right]=3.4\times {10}^{-4}$

$\left[C{H}_{3}COOH\right]=?$

Given that ionisation constant of $C{H}_{3}COOH$ is $1.7\times {10}^{-5}$, i.e.,

$K_a=1.7\times {10}^{-5}$

Fro the reaction, we have

$K_a=\frac{\left[H^{+}\right]\left[{C{H}_{3}COO}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

$\Rightarrow \left[C{H}_{3}COOH\right]=\frac{\left[H^{+}\right]\left[{C{H}_{3}COO}^{-}\right]}{K_a}=\frac{3.4\times {10}^{-4}\times 3.4\times {10}^{-4}}{1.7\times {10}^{-5}}=6.8\times {10}^{-3}$

Hence the initial concentration of $C{H}_{3}COOH$ molecules ould be $\left(D\right)6.8\times {10}^{-3}$