Question

The volume (in ml.) of 1.00 M $HCl$ that would have to be added to $100$ ml of $0.1$ M acetic acid solution to make the concentration of acetate ion equal to ${10}^{-5}M(K_a=2\times {10}^{-5})$ is:

• A. $10$ ml
• B. $20$ ml
• C. $30$ ml
• D. $40$ ml

Answer 1

Answer: (B)

$20$ ml

Let, $V$ ml of $HCl$ are to be added. Since $HCl$ is a strong acid, it will suppress the ionisation of acetic acid.

$\left[H^{+}\right]$ (considering $HCl$ as the only source) $=\frac{1\times V}{100+V}$

Also, $\left[C{H}_{3}COOH\right]=\frac{100\times 0.1}{(100+V)}$

$K_a=2\times {10}^{-5}=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$

$\Rightarrow 2\times {10}^{-5}=\frac{{10}^{-5}\times \left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}=\frac{{10}^{-5}\times V}{10}$

$\therefore V=20ml$

Hence, option $\left(B\right)$ is correct.