Chapter 15, Problem 102CP

Interpretation Introduction

Interpretation: The equilibrium concentrations of ${\text{NH}}_{\text{3}}$, ${\text{Cu}}^{\text{2+}}$, $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}^{\text{2+}}$, $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{\text{2+}}$, $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{\text{2+}}$ and $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{\text{2+}}$ in a solution prepared by mixing $500\text{mL}$ of $\text{3}\text{M}{\text{NH}}_{\text{3}}$ with $500\text{mL}$ of $2\times {10}^{-3}\text{M}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is to be calculated.

Concept introduction: The solubility product is the mathematical product of a substance’s dissolved ion concentration raised to its power of its stoichiometric coefficients. When sparingly soluble ionic compound releases ions in the solution, it gives relevant solubility product. The solvent is generally water.

Answer to Problem 102CP

The concentration of ${\text{NH}}_{\text{3}}$ is $\underset{\_}{1.50M}$.

The concentration of ${\text{Cu}}^{\text{2+}}$ is $\underset{\_}{1.77\times 10^{-17}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}^{\text{2+}}$ is $\underset{\_}{4.93\times 10^{-13}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{\text{2+}}$ is $\underset{\_}{2.866\times 10^{-9}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{\text{2+}}$ is $\underset{\_}{4.30\times 10^{-6}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{\text{2+}}$ is $\underset{\_}{1\times 10^{-3}M}$.

Explanation of Solution

Explanation

To determine: The equilibrium concentrations of ${\text{NH}}_{\text{3}}$, ${\text{Cu}}^{\text{2+}}$, $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}^{\text{2+}}$, $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{\text{2+}}$, $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{\text{2+}}$ and $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{\text{2+}}$ in a solution prepared by mixing $500\text{mL}$ of $\text{3}\text{M}{\text{NH}}_{\text{3}}$ with $500\text{mL}$ of $2\times {10}^{-3}\text{M}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

The initial concentration of ${\text{Cu}}^{\text{2+}}$ is $\underset{\_}{1\times 10^{-3}M}$ and the initial concentration of ${\text{NH}}_{\text{3}}$ is $\underset{\_}{1.50M}$.

The concentration of ${\text{Cu}}^{\text{2+}}$ is calculated by the formula,

${\left[{\text{Cu}}^{\text{2+}}\right]}_{\text{initial}}=\frac{\left(\text{Volume}\text{of}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{solution}\right)\times \left(\text{Concentration}\text{of}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{solution}\right)}{\left(\text{Volume}\text{of}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{solution}+\text{Volume}\text{of}{\text{NH}}_{\text{3}}\text{solution}\right)}$

Where,

• ${\left[{\text{Cu}}^{\text{2+}}\right]}_{\text{initial}}$ is the initial concentration of ${\text{Cu}}^{\text{2+}}$.

The volume of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution is $500\text{mL}$.

The volume of ${\text{NH}}_{\text{3}}$ solution is $500\text{mL}$.

The concentration of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution is $2\times {10}^{-3}\text{M}$.

Substitute the values of volume of solutions and concentration of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution in the above formula.

$\begin{array}{c}{\left[{\text{Cu}}^{\text{2+}}\right]}_{\text{initial}}=\frac{\left(500\text{mL}\right)\times \left(2\times {10}^{-3}\text{M}\right)}{\left(500\text{mL}+500\text{mL}\right)}\\ =\underset{\_}{1\times 10^{-3}M}\end{array}$

The concentration of ${\text{NH}}_{\text{3}}$ is calculated by the formula,

${\left[{\text{NH}}_{\text{3}}\right]}_{\text{initial}}=\frac{\left(\text{Volume}\text{of}{\text{NH}}_{\text{3}}\text{solution}\right)\times \left(\text{Concentration}\text{of}{\text{NH}}_{\text{3}}\text{solution}\right)}{\left(\text{Volume}\text{of}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{solution}+\text{Volume}\text{of}{\text{NH}}_{\text{3}}\text{solution}\right)}$

Where,

• ${\left[{\text{NH}}_{\text{3}}\right]}_{\text{initial}}$ is the equilibrium concentration of ${\text{NH}}_{\text{3}}$.

The volume of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution is $500\text{mL}$.

The volume of ${\text{NH}}_{\text{3}}$ solution is $500\text{mL}$.

The concentration of ${\text{NH}}_{\text{3}}$ solution is $3\text{M}$.

Substitute the values of volume of solutions and concentration of ${\text{NH}}_{\text{3}}$ solution in the above formula.

$\begin{array}{c}{\left[{\text{NH}}_{\text{3}}\right]}_{\text{initial}}=\frac{\left(500\text{mL}\right)\times \left(3\text{M}\right)}{\left(500\text{mL}+500\text{mL}\right)}\\ =\underset{\_}{1.50M}\end{array}$

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{\text{2+}}$ is $\underset{\_}{1\times 10^{-3}M}$.

Explanation

The given stepwise equilibria for the formation of complex ion is as follows,

${\text{Cu}}^{2+}\left(aq\right)+{\text{NH}}_{\text{3}}\left(aq\right)\rightleftharpoons {\text{CuNH}}_{\text{3}}{}^{2+}\left(aq\right)K_1=1.86\times {10}^4$ (1)

${\text{CuNH}}_{\text{3}}{}^{2+}\left(aq\right)+{\text{NH}}_{\text{3}}\left(aq\right)\rightleftharpoons \text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{2+}\left(aq\right)K_2=3.88\times {10}^3$ (2)

$\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{2+}\left(aq\right)+{\text{NH}}_{\text{3}}\left(aq\right)\rightleftharpoons \text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{2+}\left(aq\right)K_3=1.00\times {10}^3$ (3)

$\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{2+}\left(aq\right)+{\text{NH}}_{\text{3}}\left(aq\right)\rightleftharpoons \text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}\left(aq\right)K_4=1.55\times {10}^2$ (4)

Since the equilibrium constants for every reaction is high and there is an excess of ${\text{NH}}_{\text{3}}$, this means that the reaction goes to completion and the net reaction in solution is,

${\text{Cu}}^{2+}\left(aq\right)+4{\text{NH}}_{\text{3}}\left(aq\right)\rightleftharpoons \text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}\left(aq\right)$

The concentration of the species in the solution is calculated by the table given below.

$\begin{array}{cccccc}& {\text{Cu}}^{\text{2+}}(aq)& +& 4{\text{NH}}_{\text{3}}\left(aq\right)& \rightleftharpoons & \text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}\left(aq\right)\\ \text{Before}\text{reaction}& 1\times {10}^{-3}\text{M}& & 1.50\text{M}& & 0\\ \text{After}\text{reaction}& 0& & \begin{array}{l}1.50\text{M}-4\left(1\times {10}^{-3}\right)\text{M}\\ \approx 1.50\text{M}\end{array}& & 1\times {10}^{-3}\text{M}\end{array}$

The entire ${\text{Cu}}^{2+}$ will form $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}$ because reaction goes to completion. Hence the

concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{\text{2+}}$ is $\underset{\_}{1\times 10^{-3}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{\text{2+}}$ is $\underset{\_}{4.30\times 10^{-6}M}$.

The equilibrium constant for equation (4) is calculated by the formula,

$K_4=\frac{\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}\right]}{\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{2+}\right]\left[{\text{NH}}_{\text{3}}\right]}$

Where,

• $K_4$ is the equilibrium constant of equation (4).

Substitute the value of $K_4$ and the concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{2+}$ and ${\text{NH}}_{\text{3}}$ in the above formula.

$\begin{array}{c}1.55\times {10}^2=\frac{\left[1\times {10}^{-3}\right]}{\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{2+}\right]\left[1.50\right]}\\ \left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{2+}\right]=\underset{\_}{4.30\times 10^{-6}M}\end{array}$

Therefore, the concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{\text{2+}}$ is $\underset{\_}{4.30\times 10^{-6}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{\text{2+}}$ is $\underset{\_}{2.866\times 10^{-9}M}$.

The equilibrium constant for equation (3) is calculated by the formula,

$K_3=\frac{\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{2+}\right]}{\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{2+}\right]\left[{\text{NH}}_{\text{3}}\right]}$

Where,

• $K_3$ is the equilibrium constant of equation (3).

Substitute the value of $K_3$ and the concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{2+}$ and ${\text{NH}}_{\text{3}}$ in the above formula.

$\begin{array}{c}1\times {10}^3=\frac{\left[4.30\times {10}^{-6}\right]}{\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{2+}\right]\left[1.50\right]}\\ \left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{2+}\right]=\underset{\_}{2.866\times 10^{-9}M}\end{array}$

Therefore, the concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{\text{2+}}$ is $\underset{\_}{2.866\times 10^{-9}M}$.

The concentration of ${\text{CuNH}}_{\text{3}}{}^{\text{2+}}$ is $\underset{\_}{4.93\times 10^{-13}M}$.

The equilibrium constant for equation (2) is calculated by the formula,

$K_2=\frac{\left[\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{2+}\right]}{\left[{\text{CuNH}}_{\text{3}}{}^{2+}\right]\left[{\text{NH}}_{\text{3}}\right]}$

Where,

• $K_2$ is the equilibrium constant of equation (2).

Substitute the value of $K_2$ and the concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{2+}$ and ${\text{NH}}_{\text{3}}$ in the above formula.

$\begin{array}{c}3.88\times {10}^3=\frac{\left[2.87\times {10}^{-9}\right]}{\left[{\text{CuNH}}_{\text{3}}{}^{2+}\right]\left[\text{1}\text{.50}\right]}\\ \left[{\text{CuNH}}_{\text{3}}{}^{2+}\right]=\underset{\_}{4.93\times 10^{-13}M}\end{array}$

Therefore, the concentration of ${\text{CuNH}}_{\text{3}}{}^{2+}$ is $\underset{\_}{4.93\times 10^{-13}M}$.

The concentration of ${\text{Cu}}^{\text{2+}}$ is $\underset{\_}{1.77\times 10^{-17}M}$.

The equilibrium constant for equation (1) is calculated by the formula,

$K_1=\frac{\left[{\text{CuNH}}_{\text{3}}{}^{2+}\right]}{\left[{\text{Cu}}^{2+}\right]\left[{\text{NH}}_{\text{3}}\right]}$

Where,

• $K_1$ is the equilibrium constant of equation (1).

Substitute the value of $K_2$ and the concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{2+}$ and ${\text{NH}}_{\text{3}}$ in the above formula.

$\begin{array}{c}1.86\times {10}^4=\frac{\left[4.93\times {10}^{-13}\right]}{\left[{\text{Cu}}^{2+}\right]\left[1.50\right]}\\ \left[{\text{Cu}}^{2+}\right]=\underset{\_}{1.77\times 10^{-17}M}\end{array}$

Therefore, the concentration of ${\text{Cu}}^{\text{2+}}$ is $\underset{\_}{1.77\times 10^{-17}M}$.

Conclusion

Conclusion

The concentration of ${\text{NH}}_{\text{3}}$ is $\underset{\_}{1.50M}$.

The concentration of ${\text{Cu}}^{\text{2+}}$ is $\underset{\_}{1.77\times 10^{-17}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}^{\text{2+}}$ is $\underset{\_}{4.93\times 10^{-13}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_2{}^{\text{2+}}$ is $\underset{\_}{2.866\times 10^{-9}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_3{}^{\text{2+}}$ is $\underset{\_}{4.30\times 10^{-6}M}$.

The concentration of $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_4{}^{\text{2+}}$ is $\underset{\_}{1\times 10^{-3}M}$.