Question

The $pH$ of 0.1 M monobasic acid is 4.50. The acidity constant ($K_a$) of the monobasic acid is:

• A. $1.0\times {10}^{-7}$
• B. $1.0\times {10}^{-5}$
• C. $1.0\times {10}^{-4}$
• D. $1.0\times {10}^{-8}$

Answer 1

Answer: (D)

$1.0\times {10}^{-8}$

$pH=4.5\Rightarrow \left[H^{+}\right]={10}^{-4.5}=3.16\times {10}^{-5}$

Since the acid is monobasic, it will dissociate in the aqueous solution as

$HA\rightleftharpoons H^{+}+A^{-}$

Since, $0.1>>3.16\times {10}^{-5}$, $\left[HA\right]=0.1mol{L}^{-1}$

Also $\left[H^{+}\right]=\left[A^{-}\right]$

$\Rightarrow K_a=\frac{[H^{+}{]}^2}{\left[HA\right]}$

$\Rightarrow K_a=\frac{(3.16\times {10}^{-5}{)}^2}{0.1}=1\times {10}^{-8}$

Hence, option $D$ is the correct answer.