Chapter 17, Problem 17.31QP

(a)

Interpretation Introduction

To determine: The initial concentration of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ in the solution.

Interpretation: The initial concentration of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ in the solution is to be calculated.

Concept introduction: The concentration is calculated by the formula,

$\text{Concentration }=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}}$

The concentration is expressed in $\text{mol/L}$.

Answer to Problem 17.31QP

The initial concentration of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ in the solution is $\underset{\_}{0.00100M}$.

Explanation of Solution

The total volume of the solution is $\text{1}\text{L}$.

The number of moles of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $\text{0}\text{.00100}\text{mol}$.

The number of moles of ${\text{NH}}_{\text{3}}$ is $\text{0}\text{.500}\text{mol}$.

A solution is prepared from $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ and ${\text{NH}}_{\text{3}}$.

The initial concentration of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ in the solution is calculated by the formula,

$\text{Concentration}\text{of}\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2}{\text{Volume}\text{of}\text{solution}}$

The concentration is expressed in $\text{mol/L}$.

Substitute the concentration of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ and volume of solution in the above expression.

$\begin{array}{c}\left[\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2\right]=\frac{\text{0}\text{.00100}\text{mol}}{\text{1}\text{L}}\\ =\underset{\_}{0.00100M}\end{array}$

Therefore, the initial concentration of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ in the solution is $\underset{\_}{0.00100M}$.

(b)

Interpretation Introduction

To determine: The concentration of ${\text{Ni}}^{2+}$ in the solution at the equilibrium.

Interpretation: The concentration of ${\text{Ni}}^{2+}$ in the solution at the equilibrium is to be calculated.

Concept introduction: The constant for the complex ion formation in the equilibrium is known as the formation constant. It is denoted by $K_{\text{f}}$.

The concentration at the equilibrium is calculated by the expression,

$K_{\text{f}}=\frac{{\left[\text{Products}\right]}_{\text{equil}}}{{\left[\text{Reactants}\right]}_{{}_{\text{equil}}}}$

Answer to Problem 17.31QP

The concentration of ${\text{Ni}}^{2+}$ in the solution at the equilibrium is $\underset{\_}{1.25\times 10^{-10}M}$.

Explanation of Solution

The number of moles of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $\text{0}\text{.00100}\text{mol}$.

The number of moles of ${\text{NH}}_{\text{3}}$ is $\text{0}\text{.500}\text{mol}$.

The reaction of ${\text{Ni}}^{2+}$ and ${\text{NH}}_{\text{3}}$ is,

${\text{Ni}}^{2+}\left(aq\right)+6{\text{NH}}_{\text{3}}\left(g\right)\to {\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6\right]}^{2+}$ (1)

The ICE Table for the above reaction is below.

$\begin{array}{cccc}& \left[{\text{Ni}}^{2+}\right]\left(\text{M}\right)& \left[{\text{NH}}_{\text{3}}\right]\left(\text{M}\right)& {\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6\right]}^{2+}\left(\text{M}\right)\\ \text{Intial}& 0.00100& 0.500& 0\\ \text{Change}& 0.00100-x& 0.500-6\left(0.00100\right)& +0.00100\\ \text{Equilibrium}& \left(0.00100-x\right)\approx x& 0.494& 0.00100\end{array}$

The equilibrium constant for the equation (1) is,

$K_{\text{f}}=\frac{{\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6\right]}^{2+}}{\left[{\text{Ni}}^{2+}\right]{\left[{\text{NH}}_{\text{3}}\right]}^6}$

The value of equilibrium constant for ${\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6\right]}^{2+}$ is $5.5\times {10}^8\text{M}$.

Substitute the values of equilibrium concentrations from the above Table in the above expression.

$\begin{array}{c}5.5\times {10}^8\text{M}=\left(\frac{0.00100}{\left(x\right)\times {\left(0.494\right)}^6}\right)\text{M}\\ x=\frac{0.00100}{\left(5.5\times {10}^8\right)\times {\left(0.494\right)}^6}\\ =\underset{\_}{1.25\times 10^{-10}M}\end{array}$

Therefore, the concentration of ${\text{Ni}}^{2+}$ in the solution at the equilibrium is $\underset{\_}{1.25\times 10^{-10}M}$.

Conclusion

1. a. The initial concentration of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_2$ in the solution is $\underset{\_}{0.00100M}$.
2. b. The concentration of ${\text{Ni}}^{2+}$ in the solution at the equilibrium is $\underset{\_}{1.25\times 10^{-10}M}$.