Chapter 16, Problem 16.70QP

Interpretation Introduction

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

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

$K_{\text{f}}=\frac{\left[\text{Cu}{\left(\text{en}\right)}_2{}^{2+}\right]}{\left[{\text{Cu}}^{2+}\right]{\left[\text{en}\right]}^2}$

To determine: The concentration of copper ion complex in the aqueous solution of ethylenediamine.

Answer to Problem 16.70QP

Solution:

The concentration of copper ion complex in the aqueous solution of ethylenediamine is $\underset{\_}{5\times 10^{-5}M}$.

Explanation of Solution

Given

The molar concentration of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $5\times {10}^{-5}\text{M}$.

The molar concentration of ethylenediamine is $1\times {10}^{-3}\text{M}$.

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

Copper nitrate on dissolving in the aqueous solution of ethylenediamine forma a complex ion with ethylenediamine as,

${\text{Cu}}^{2+}+2\text{en}\underset{}{\rightleftharpoons }\text{Cu}{\left(\text{en}\right)}_2{}^{2+}$

The complex ion formation constant, $K_{\text{f}}$, of $\text{Cu}{\left(\text{en}\right)}_2{}^{2+}$ is $1\times {10}^{20}$.

The concentration of formation of $\text{Cu}{\left(\text{en}\right)}_2{}^{2+}$ complex is assumed to be $x$.

The ICE table for the formation of $\text{Cu}{\left(\text{en}\right)}_2{}^{2+}$ is given as,

$\begin{array}{cccccc}& {\text{Cu}}^{2+}& +& 2\left(\text{en}\right)& \underset{}{\rightleftharpoons }& \text{Cu}{\left(\text{en}\right)}_2{}^{2+}\\ \text{Initial}\left(\text{M}\right):& 5\times {10}^{-5}& & 1\times {10}^{-3}& & 0\\ \text{Change}\left(\text{M}\right):& -x& & -x& & +x\\ \text{Final}\left(\text{M}\right):& 5\times {10}^{-5}-x& & 1\times {10}^{-3}-x& & x\end{array}$

The complex ion formation constant is given by the formula,

$K_{\text{f}}=\frac{\left[\text{Cu}{\left(\text{en}\right)}_2{}^{2+}\right]}{\left[{\text{Cu}}^{2+}\right]{\left[\text{en}\right]}^2}$

Substitute the value of $K_{\text{f}}$, $\left[\text{Cu}{\left(\text{en}\right)}_2{}^{2+}\right]$, $\left[{\text{Cu}}^{2+}\right]$ and $\left[\text{en}\right]$ in the above equation.

$1\times {10}^{20}=\frac{x}{5\times {10}^{-5}-x{\left(1\times {10}^{-3}-x\right)}^2}$

The value of $x$ is neglected since its value is very small compared to $1\times {10}^{-3}$.

Therefore, the above equation becomes,

$\begin{array}{c}1\times {10}^{20}=\frac{x}{5\times {10}^{-5}-x{\left(1\times {10}^{-3}\right)}^2}\\ 1\times {10}^{20}\times \left(5\times {10}^{-11}-x\times {10}^{-6}\right)=x\\ 5\times {10}^9-x\times {10}^{14}=x\\ x=\underset{\_}{5\times 10^{-5}M}\end{array}$

The concentration of $\text{Cu}{\left(\text{en}\right)}_2{}^{2+}$ is same as the concentration of $\text{Cu}{\left({\text{H}}_2\text{O}\right)}_6{}^{2+}$

Therefore, the concentration of $\text{Cu}{\left({\text{H}}_2\text{O}\right)}_6{}^{2+}$ is $\underset{\_}{5\times 10^{-5}M}$.

Conclusion:

The concentration of in $\text{Cu}{\left({\text{H}}_2\text{O}\right)}_6{}^{2+}$ the aqueous solution of ethylenediamine is $\underset{\_}{5\times 10^{-5}M}$