Question

What is $\left[H^{+}\right]$ in mol/L of a solution that is 0.20M in $C{H}_{3}COONa$ and 0.10 M in $C{H}_{3}COOH$?

$K_a$ for $C{H}_{3}COOH=1.8\times {10}^{-5}$

• A. $3.5\times {10}^{-4}$
• B. $1.1\times {10}^{-5}$
• C. $1.8\times {10}^{-5}$
• D. $9.0\times {10}^{-6}$

Answer 1

The correct option is D $9.0\times {10}^{-6}$

The ionisation of $C{H}_{3}COOHandC{H}_{3}COONa$ takes places as follows :-

$C{H}_{3}COONa\to C{H}_{3}CO{O}^{-}+N{a}^{+}$

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

Due to the presence of common $C{H}_{3}CO{O}^{-}$ ions, the equilibrium of the acid will shift towards left (by Le Chateliers's principle)

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

Here the $\left[C{H}_{3}CO{O}^{-}\right]=0.2M$ and $\left[C{H}_{3}COOH\right]=0.1M$ and $K_a=1.8\times {10}^{-5}$ and hence substituting the values in the above formula, we can get the value of concentration of $\left[H^{+}\right]$ ions.

$\therefore$ $1.8\times {10}^{-5}=\frac{\left[0.2\right]\left[H^{+}\right]}{\left[0.1\right]}$

$\therefore \left[H^{+}\right]=9.0\times {10}^{-6}mol/L$

Hence, the correct answer is option (D).