Question

The dissociation constant of acetic acid at a given temperature is $1.69\times {10}^{-5}$. The degree of dissociation of $0.01M$ acetic acid in the presence of $0.01M$ $HCl$ is equal to:

• A. $0.41$
• B. $0.13$
• C. $1.69\times {10}^{-3}$
• D. $0.013$

Answer 1

Answer: (C)

$1.69\times {10}^{-3}$

$K=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$
Here, $\alpha$ is the degree of dissociation of $C{H}_{3}COOH$.
$\Rightarrow K=\frac{C\alpha \left[H^{+}\right]}{C(1-\alpha )}=\frac{\alpha \left[H^{+}\right]}{1-\alpha }$
$HCl$ dissociates completely.

$C{H}_{3}COOH$ itself is a weak acid and in the presence of $HCl$, its dissociation will be negligible.
Hence, $\left[H^{+}\right]$ can be approximated to $\left[HCl\right]$ ($H^{+}$ ions from acetic acid will be negligible).
As $\alpha$ is very less, the denominator can be approximated to 1.
$\Rightarrow 1.69\times {10}^{-5}=\alpha 0.01$
$\Rightarrow \alpha =\frac{1.69\times {10}^{-5}}{0.01}=1.69\times {10}^{-3}$