Question

The solution of weak monoprotic acid which is $0.10M$ has $pH=3$. Calculate $K_a$ of weak acid.

• A. $1.11\times {10}^{-4}$
• B. $1.11\times {10}^{-2}$
• C. $1.11\times {10}^{-3}$
• D. $1.11\times {10}^{-5}$

Answer 1

The correct option is D $1.11\times {10}^{-5}$

Let consider weak monoprotic acid $C{H}_{3}COOH$

$\begin{array}{cccccc}& C{H}_{3}COOH& \rightleftharpoons & C{H}_{3}CO{O}^{-}& +& H^{+}\\ initially& 0.1& & 0& & 0\\ ateqm& (0.1-x)& & x& & x\end{array}$

$H_a=\frac{x^2}{(0.1-x)}$

$pH=3$

So, $H^{+}={10}^{-3}=x$

$K_a=\frac{({10}^{-3}{)}^2}{(0.1-0.001)}=\frac{{10}^{-6}}{0.099}$

$K_a=10.1\times {10}^{-6}$

$K_a=1.01\times {10}^{-5}$

Option D is correct.