Question

What is the concentration of $N{H}_4^{+}$ ions in a solution that is $0.02MN{H}_3$ and $0.01MKOH$? $(K_b$ of $N{H}_3=1.8\times {10}^{-5}M)$

• A. $3.6\times {10}^{-5}M$
• B. $1.8\times {10}^{-5}M$
• C. $0.9\times {10}^{-5}M$
• D. $7.2\times {10}^{-5}M$

Answer 1

The correct option is A $3.6\times {10}^{-5}M$

A solution is $0.02MN{H}_3$ and $0.01MKOH$.

In the solution, $N{H}_3$ exits as $N{H}_{4}OH$.

The equilibrium reaction for the dissociation of ammonium hydroxide is given below:

$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

The expression for the equilibrium constant is as follows:

$K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$

Substituting values in the above expression, we get

$1.8\times {10}^{-5}=\frac{\left[N{H}_4^{+}\right]\left(0.01\right)}{\left(0.02\right)}$

Hence, $\left[N{H}_4^{+}\right]=3.6\times {10}^{-5}M$

Hence, the concentration of $N{H}_4^{+}$ is $3.6\times {10}^{-5}M$.