Question

What is the concentration of $C{H}_{3}COOH$ in a solution prepared by dissolving $0.01$ mol of $C{H}_{3}CO{O}^{-}N{H}_4^{+}$ in $1L$ of $H_{2}O$?

Use : $[K_{a\left(C{H}_{3}COOH\right)}=1.8\times {10}^{-5};K_{b\left(N{H}_{4}OH\right)}=1.8\times {10}^{-5}]$

• A. $5.55\times {10}^{-5}M$
• B. $1.0\times {10}^{-1}M$
• C. $6.4\times {10}^{-4}M$
• D. $5.41\times {10}^{-4}M$

Answer 1

The correct option is D $5.41\times {10}^{-4}M$

Given that a solution is prepared by dissolving $0.01$ mol of $C{H}_{3}CO{O}^{-}N{H}_4^{+}$ in $1L$ of $H_{2}O$.

The expression for the hydrolysis constant is $k_h=\frac{k_w}{k_a{k}_b}$,

where $K_w$ is the ionic product of water, $K_a$ is the dissociation constant of an acid and $K_b$ is the dissociation constant of a base.

Given $K_a=1.8\times {10}^{-5}$ and $K_b=1.8\times {10}^5$.

Substituting values in the above expression, we get

$k_h=\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}\times 1.8\times {10}^{-5}}=3.09\times {10}^{-5}$

The hydrolysis of the salt can be represented as follows:

$C{H}_{3}CO{O}^{-}N{H}_4^{+}+H_{2}O\to N{H}_{4}OH+C{H}_{3}COOH$

The inital concentrations of $C{H}_{3}CO{O}^{-}N{H}_4^{+},N{H}_{4}OHandC{H}_{3}COOH$ are $0.01M,0M$ and $0M$ respectively.

The equilibrium concentrations of $C{H}_{3}CO{O}^{-}N{H}_4^{+},N{H}_{4}OHandC{H}_{3}COOH$ are $0.01-xM,xM$ and $xM$ respectively.

The expression for the hydrolysis constant becomes

$k_h=\frac{\left[N{H}_{4}OH\right]\left[C{H}_{3}COOH\right]}{\left[C{H}_{3}CO{O}^{-}N{H}_4^{+}\right]}$

Substituting values in the above expression, we get

$3.09\times {10}^{-5}=\frac{x^2}{(0.01-x)}$

Hence, $x=5.41\times {10}^{-4}$

Hence, the concentration of $C{H}_{3}COOH$ is $5.41\times {10}^{-4}M$.