Question

Suppose 0.10 mol of Cu(NO₃)₂ and 1.50 mol of NH₃ are dissolved in water and diluted to a total volume of 1.00 L. Calculate the concentrations of Cu(NO₃)₄²⁺ and of Cu²⁺ at equilibrium.

Short answer

$\left[Cu{{\left(N{H}_3\right)}_4}^{2+}\right]=5.56\times {10}^{11}M$

Step-by-step solution

Complex-ion equilibria

When transition metal ions react with certain anions or neutral molecules, they form dative/coordinate covalent bonds and form coordinate complexes.

There exists an equilibrium between the dissociated ions and the undissociated complex compound.

The key equilibria in this reaction

Assume that entire Cu²⁺ will be consumed during the formation of [Cu(NH₃)₄]²⁺

$$\begin{gathered} Cu{\left(N{O}_3\right)}_2\left(aq\right)\square C{u}^{2+}\left(aq\right)+2N{O_3}^{-}\left(aq\right) \\ C{u}^{2+}\left(aq\right)+4N{H}_3\left(aq\right)\square {\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}\left(aq\right) \\ I0.10mol1.50mol0 \\ C-x-4x+x \\ E0.10-x1.50-4xx \\ {K}_f=1.1\times {10}^{12} \\ {K}_f=\frac{x}{\left(0.10-x\right){\left(1.50-4x\right)}^4} \\ 1.1\times {10}^{12}=\frac{x}{\left(0.10-x\right){\left(1.50-4x\right)}^4} \\ x=1.1\times {10}^{12}\times \times 0.10\times {\left(1.50\right)}^4 \\ x=5.56\times {10}^{11} \end{gathered}$$

Thus, entire will be consumed and the concentration of Cu(NH₃)₄²⁺ is

$\left[Cu{{\left(N{H}_3\right)}_4}^{2+}\right]=5.56\times {10}^{11}M$.