Chapter 22, Problem 22.71QP

Interpretation Introduction

Interpretation:

The concentration of $C{u}^{2+}(aq)$, $N{H}_3(aq)$ and ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}(aq)$ at equilibrium has to be determined when $0.10\text{ mol}$ $C{u}^{2+}(aq)$ and $0.40\text{ mol}$ $N{H}_3(aq)$ are made up to $1.00\text{ L}$ of solution.

Concept Introduction:

Consider a ligand ‘X’ co-ordinates to central metal ion ‘A’, and the reaction is represented as,

$A\text{ + 6X }\stackrel{}{\to }{\left[A{X}_6\right]}^{n\pm }$

Equilibrium constant $K_c$ or dissociation constant, $K_d$ for formation of a complex compound by complexation reaction is represented as,

${\text{K}}_c{\text{ = K}}_d\text{ = }\frac{\left[A\right]{\left[X\right]}^6}{{\left[A{X}_6\right]}^{n\pm }}$

Answer to Problem 22.71QP

Concentration of $C{u}^{2+}(aq)$ at equilibrium is calculated as $6.1\times 10^{-4}M.$

Concentration of $N{H}_3(aq)$ at equilibrium is calculated as $2.4\times 10^{-3}M.$

Concentration of ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}(aq)$ at equilibrium is calculated as $0.10M.$

Explanation of Solution

The reaction between $C{u}^{2+}(aq)$, $N{H}_3(aq)$ and ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}(aq)$ can be represented as,

$C{u}^{2+}(aq)+4N{H}_3(aq)\stackrel{}{\rightleftharpoons }{\left[Cu{\left(N{H}_3\right)}_4\right]}^{+}(aq)$

Given that dissociation constant for the above reaction is $K_d\text{ = 2}\text{.1}\times {\text{10}}^{-13}$ also $K_c{\text{ = K}}_f\text{ = 2}\text{.1}\times {\text{10}}^{-13}$

The concentration of each reactant species and the product at equilibrium is given as,

$\begin{array}{l}\underset{\_}{Concentration(M){\text{ Cu}}^{2+}{\text{(aq) NH}}_3(aq){\left[Cu{\left(N{H}_3\right)}_4\right]}^{+}\text{(aq) }}\\ Initial\text{ 0}\text{.10 0}\text{.40 0}\\ Change\text{ -x -4x +x}\\ Equilibrium\text{ 0}\text{.10 - x 0}\text{.40-4x x}\end{array}$

  As we know,

$K_c{\text{ = K}}_d\text{ = 2}\text{.1}\times {\text{10}}^{-13}=\frac{\left[{\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}\right]}{\left[C{u}^{2+}\right]{\left[N{H}_3\right]}^4}$

Substitute the concentration $C{u}^{2+}(aq)$, $N{H}_3(aq)$ and ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}(aq)$ values in the above equation,

$K_c\text{ = 2}\text{.1}\times {\text{10}}^{-13}=\frac{\left(0.10-x\right){\left(4\times \left(0.10-x\right)\right)}^4}{x}$

To solve the above equation, let $y=0.10-x$ and then $x=0.10-y$.  Therefore the above equation becomes,

$2.1\times {10}^{-13}\text{ = }\frac{y{\left(4y\right)}^4}{\left(0.10-y\right)}$

Numerical value of ‘y’ is negligible as it is lesser than $0.10$. Hence $0.10-y\cong \text{0}\text{.10}$ Rearranging the above equation and solving for y,

$\begin{array}{l}2.1\times {10}^{-13}=\frac{y{\left(4y\right)}^4}{\left(0.10-y\right)}\\ y^5=\frac{\left(2.1\times {10}^{-13}\right)\left(0.10\right)}{4^4}\\ =\text{ 8}\text{.203}\times {\text{10}}^{-17}\\ y=\sqrt[5]{\text{8}\text{.203}\times {\text{10}}^{-17}}\\ =\text{ 6}\text{.064}\times {\text{10}}^{-4}\\ \end{array}$

Therefore, at equilibrium,

Concentration of $C{u}^{2+}(aq)$,

$\begin{array}{l}\left[C{u}^{2+}\right]\text{ = }\left(0.10\text{ M - x}\right)\text{= 6}\text{.064}\times {\text{10}}^{-4}\text{ M = 6}\text{.1}\times {\text{10}}^{-4}M\\ \end{array}$

Concentration of $N{H}_3(aq)$,

  $\begin{array}{l}\left[N{H}_3\right]\text{= 4}\left(0.10-x\right)\text{M }\\ \text{ = 4y = 4}\times \text{6}\text{.064}\times {\text{10}}^{-4}M\\ \text{= 2}\text{.4}\times {\text{10}}^{-3}\text{M}\end{array}$

Concentration of ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}=\text{ x = }\left(0.10-y\right)=0.10M$

Conclusion

Concentration of product and reactant species can be determined knowing the dissociation constant of the reaction.