Question

Calculate the equilibrium concentration of $N{H}_3$ when the initial concentration 0.2 M $Z{n}^{2+}$ solution reduces to $1.0\times {10}^{-4}Z{n}^{2+}$.

Given that: $K_f$ of $Zn(N{H}_3{)}_4^{2+}=5\times {10}^8$

Note: $N{H}_3$ and $Zn(N{H}_3{)}_4^{2+}$ (assume no partial complexation)

• A. $14.5\times {10}^{-8}M$
• B. $3.2\times {10}^{-1}M$
• C. $15\times {10}^{-2}M$
• D. $4.5\times {10}^{-2}M$

Answer 1

The correct option is D $4.5\times {10}^{-2}M$

$\left[Z{n}^{2+}\right]$ final $=1\times {10}^{-4}M$

$\left[Zn\right(N{H}_3{)}_4^{2+}]eq=0.2-{10}^{-4}\approx 0.2$

$Z{n}^{2+}+4N{H}_3\rightleftharpoons Zn(N{H}_3{)}_4^{2+}$

$K_f=\frac{\left[Zn\right(N{H}_3{)}_4^{2+}]}{\left[Z{n}^{2+}\right][N{H}_3{]}^4}$

$5\times {10}^8=\frac{0.2}{\left({10}^{-4}\right)[N{H}_3{]}^4}$

$(N{H}_3{)}^4=4\times {10}^{-6}=400\times {10}^{-8}$

$\left(N{H}_3\right)=4.5\times {10}^{-2}M$