Question

The complex ion Cu(NH₃)₄²⁺ is formed in a solution made of 0.0100 M Cu(NO₃)₂ and 0.300 M NH₃. What are the concentrations of Cu₂ , NH₃, and Cu(NH₃)₄²⁺ at equilibrium? The formation constant*, Kf, of Cu(NH₃)₄²⁺ is 1.70 × 10¹³

Answer 1

Given data Concentration of \(\left[\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\right]=0.010 \mathrm{M}\)

Concentration of \(\left[\mathrm{NH}_{3}\right]=0.30 \mathrm{M}\)

Equilibrium constant \(\left(k_{f}\right)=1.70 \times 10^{13}\)

Equilibrium reaction is written as \(\begin{array}{llcc} & \mathrm{Cu}^{2+} & + & 4 \mathrm{NH}_{3} & \rightleftarrows & {\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}} \\ \text { Initial } & 0.01 & 0.30 & 0 \\ \text { Change } & -0.01 & 0.30-4(0.01) & 0.01 \\ \text { Equilibrium } & \mathrm{x} & 0.26 & 0.01\end{array}\)

Equilibrium constant is written as

$$ \begin{aligned} K_{f} &=\frac{\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}}{\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{4}} \\ 1.70 \times 10^{13} &=\frac{[0.01]}{[\mathrm{x}][0.26]^{4}} \\ \mathrm{x} &=\frac{[0.01]}{\left[1.70 \times 10^{13}\right][0.26]^{4}} \\ \mathrm{x} &=1.3 \times 10^{-13} \end{aligned} $$

So final equilibrium concentration is

$$ \begin{aligned} \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} &=0.01 \mathrm{M} \\ \left[\mathrm{Cu}^{2+}\right] &=1.3 \times 10^{-13} \mathrm{M} \\ \left[\mathrm{NH}_{3}\right] &=0.30-4(0.01)=0.26 \mathrm{M} \end{aligned} $$