Question

The dissociation constant $\left(K_a\right)$ and the percentage dissociation $\left(\alpha \right)$ of a weak monobasic acid solution of $0.2M$ with a $pH=6$ are respectively:

• A. $5\times {10}^{-9},5\times {10}^{-12}$
• B. $5\times {10}^{-13},5\times {10}^{-4}$
• C. $5\times {10}^{-12},5\times {10}^{-6}$
• D. $5\times {10}^{-5},5\times {10}^{-2}$

Answer 1

The correct option is C $5\times {10}^{-12},5\times {10}^{-6}$

$\underset{C}{H}A\rightleftharpoons \underset{-}{H^{+}}+\underset{-}{A^{-}}$

$C-c\alpha C\alpha C\alpha$

$K_a=\frac{c\alpha \times C\alpha }{C(1-\alpha )}=\frac{C{\alpha }^2}{(1-\alpha )}$

For weak acid, $\alpha <<1$ thus, $1-\alpha \simeq 1$

$K_a=C{\alpha }^2$

Given, $C=0.2M$

as $\left[H^{+}\right]=C\alpha$

$pH=6⟹\left[H^{+}\right]={10}^{-6}M$

Thus, $C\alpha ={10}^{-6}\alpha =\frac{{10}^{-6}M}{0.2M}==⟹5\times {10}^{-6}$

Now, $K_a=C{\alpha }^2{K}_a=0.2\times {(5\times {10}^{-6})}^2=5\times {10}^{-12}$

Thus, $\alpha =5\times {10}^{-6}{K}_a=5\times {10}^{-12}M$