Question

How much $N{H}_3$ should be added to a solution of $0.0010M$ $Cu(N{O}_3{)}_2$ to reduce $\left[C{u}^{2+}\right]$ to ${10}^{-13}$? Neglect the amount of copper in complexes containing fewer than 4 ammonia molecules per copper atom.

• A. 0.15 mol/L
• B. 0.0055 mol/L
• C. 0.02 mol/L
• D. 0.32 mol/L

Answer 1

The correct option is D

0.32 mol/L

$Cu(N{H}_3{)}_4^{2+}$ $\rightleftharpoons$ $C{u}^{2+}+4N{H}_3$

$K_d$ = $\frac{\left[C{u}^{2+}\right][N{H}_3{]}^4}{Cu(N{H}_3{)}_4^{2+}}$ = $1.0\times {10}^{-12}$

Since the sum of the concentrations of coper in the complex and in the free ionic state must equal 0.0010 mol/L and since the amount of the free ion is very small, the concentration of the complex is taken to be 0.0010 mol/L.

Let $x$ = $\left[N{H}_3\right]$.

Then $\frac{\left({10}^{-13}\right)\left(x^4\right)}{0.0010}$ = $1.0\times {10}^{-12}$

or $x^4$ = $1.0\times {10}^{-2}$ or $x$ = $0.32$

The concentration of $N{H}_3$ at equilibrium is 0.32 mol/L. The amount of $N{H}_3$ used up in forming 0.0010 mol/L of complex is 0.0040 mol/L, an amount negligible compared with the amount remaining at equilibrium. Hence the amount of $N{H}_3$ to be added is $0.32mol/L$