Question

Ka for ${CH}_{3}COOH$ is $1.8\times {10}^{-5}$. Find out the percentage dissociation of 0.2 M ${CH}_{3}COOH$ in 0.1 M HCl solution ________.

• A. 0.018
• B. 0.36
• C. 18
• D. 36

Answer 1

The correct option is A 0.018

The dissociation of acetic acid can be represented as:

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

HCl dissociated completely.

Now, let the concentration of acetate ion be $x$.

Therfore, hydrogen ion contribution from acetic acid$=x$

Since HCL dissociated completely, hydrogen ion contribution from HCl$=0.1M$

Total $\left[H^{+}\right]$ concetration$=x+0.1$

Now, $K_a$ for acetic acid$=1.8\times {10}^{-5}$

Hence,

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

$1.8\times {10}^{-5}=\frac{x(x+0.1)}{0.2}$

$0.36\times {10}^{-5}=x(x+0.1)$

$x^2+0.1x-0.36\times {10}^{-5}$

$x=0.35\times {10}^{-4}$

Percentage dissociation$=\frac{0.35\times {10}^{-4}\times 100}{0.2}=0.0175\approx 0.018$