Chapter 16, Problem 16.69QP

Interpretation Introduction

Interpretation: The concentration of nickel ion complex on dissolving the given moles of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ in the aqueous solution of given moles of ${\text{NH}}_{\text{3}}$ in the give volume of the solution is to be calculated.

Concept introduction: The complex ion formation constant is given by the formula,

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

To determine: The concentration of nickel ion complex in the aqueous solution of ammonia.

Answer to Problem 16.69QP

Solution:

The concentration of nickel ion complex in the aqueous solution of ammonia is $\underset{\_}{0.001M}$.

Explanation of Solution

Given

The number of moles of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $0.00100\text{mol}$.

The number of moles of ${\text{NH}}_{\text{3}}$ is $0.500\text{mol}$.

The total volume of the solution is $1.0\text{L}$.

The molar concentration is calculated by the formula,

$\text{Molar}\text{concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}}$ (1)

Substitute the value of number of moles of $\text{Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and the volume of the solution in equation (1).

$\begin{array}{c}\text{Molar}\text{concentration}=\frac{0.00100\text{mol}}{1.0\text{L}}\\ =0.001\text{M}\end{array}$

Substitute the value of number of moles of ${\text{NH}}_{\text{3}}$ and the volume of the solution in equation (1).

$\begin{array}{c}\text{Molar}\text{concentration}=\frac{0.500\text{mol}}{1.0\text{L}}\\ =0.5\text{M}\end{array}$

Nickel nitrate on dissolving in the aqueous solution of ammonia forma a complex ion with ammonia as,

${\text{Ni}}^{2+}+6{\text{NH}}_{\text{3}}\underset{}{\rightleftharpoons }\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6{}^{2+}$

The complex ion formation constant, $K_{\text{f}}$, of $\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6{}^{2+}$ is $5.5\times {10}^8$.

The concentration of formation of $\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6{}^{2+}$ complex is assumed to be $x$.

The ICE table for the formation of $\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6{}^{2+}$ is given as,

$\begin{array}{cccccc}& {\text{Ni}}^{2+}& +& 6{\text{NH}}_{\text{3}}& \underset{}{\rightleftharpoons }& \text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6{}^{2+}\\ \text{Initial}\left(\text{M}\right):& 0.001& & 0.5& & 0\\ \text{Change}\left(\text{M}\right):& -x& & -x& & +x\\ \text{Final}\left(\text{M}\right):& 0.001-x& & 0.5-x& & x\end{array}$

The complex ion formation constant is given by the formula,

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

Substitute the value of $K_{\text{f}}$, $\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6{}^{2+}\right]$, $\left[{\text{Ni}}^{2+}\right]$ and $\left[{\text{NH}}_{\text{3}}\right]$ in the above equation.

$5.5\times {10}^8=\frac{x}{0.001-x{\left(0.5-x\right)}^6}$

The value of $x$ is neglected since its value is very small compared to $0.5$.

Therefore, the above equation becomes,

$\begin{array}{c}5.5\times {10}^8=\frac{x}{0.001-x{\left(0.5\right)}^6}\\ 8.6\times {10}^6\times \left(0.001-x\right)=x\\ 8.6\times {10}^3-8.6\times {10}^{6}x=x\\ x=\underset{\_}{0.001M}\end{array}$

The concentration of $\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_6{}^{2+}$ is same as the concentration of $\text{Ni}{\left({\text{H}}_2\text{O}\right)}_6{}^{2+}$

Therefore, the concentration of $\text{Ni}{\left({\text{H}}_2\text{O}\right)}_6{}^{2+}$ is $\underset{\_}{0.001M}$.

Conclusion:

The concentration of in $\text{Ni}{\left({\text{H}}_2\text{O}\right)}_6{}^{2+}$ the aqueous solution of ammonia is $\underset{\_}{0.001M}$