Chapter 22, Problem 22.72QP

Interpretation Introduction

Interpretation:

The concentration of $A{g}^{+}(aq)$, $N{H}_3(aq)$ and ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}(aq)$ at equilibrium has to be determined when $0.10\text{ mol}$ $A{g}^{+}(aq)$ and $0.10\text{ mol}$ $N{H}_3(aq)$ 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,

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

The formation constant, $K_f$, for formation of a complex compound by complexation reaction is represented as,

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

Answer to Problem 22.72QP

Concentration of $A{g}^{+}(aq)$ at equilibrium is calculated as $0.050M.$

Concentration of $N{H}_3(aq)$ at equilibrium is calculated as $2.4\times 10^{-4}M.$

Concentration of ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}(aq)$ at equilibrium is calculated as $0.050M.$

Explanation of Solution

The reaction between $A{g}^{+}(aq)$, $N{H}_3(aq)$ and ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}(aq)$ can be represented as,

$A{g}^{+}(aq)+2N{H}_3(aq)\stackrel{}{\rightleftharpoons }{\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}(aq)$

From the figure 17.2 given in the text book is, the reaction constant for the above reaction is $K_d\text{ = 1}$ and the formation constant for the above reaction is $K_f\text{ = 1}\text{.7}\times {\text{10}}^7$

In this reaction ammonia is the limiting reagent and it is completely consumed during the formation of complex and the complex dissociates slightly later on.  The concentration of each reactant species and the product at equilibrium is given as,

$\begin{array}{l}\underset{\_}{Concentration(M){\text{ Ag}}^{+}{\text{(aq) NH}}_3(aq){\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}\text{(aq) }}\\ Initial\text{ 0}\text{.050 0}\text{.00 0}\text{.050}\\ Change\text{ +x +2x -x}\\ Equilibrium\text{ 0}\text{.050 + x 2x 0}\text{.050-x}\end{array}$

As we know,

$K_f\text{ = 1}\text{.7}\times {\text{10}}^7={\frac{\left[Ag{\left(N{H}_3\right)}_2\right]}{\left[A{g}^{+}\right]{\left[N{H}_3\right]}^2}}^{+}$

Substitute the concentration $A{g}^{+}(aq)$, $N{H}_3(aq)$ and ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}(aq)$ values in the above equation,

$K_f\text{ = 1}\text{.7}\times {\text{10}}^7=\frac{\left(0.050-x\right)}{\left(0.050+x\right){\left(2x\right)}^2}$

Numerical value of ‘x’ is negligible compared to $0.050\text{ M}$ and hence $0.050\text{ M}-x\text{ = 0}\text{.050 M}$ and $0.050\text{ M+}x\text{ = 0}\text{.050 M}$. Therefore the above equation becomes,

$1.7\times {10}^7\text{ = }\frac{1}{2x^2}$

Rearranging the above equation and solving for x,

$\begin{array}{l}2x^2=\frac{1}{1.7\times {10}^7}\\ x=\frac{1}{2}\sqrt{\frac{1}{1.7\times {10}^7}}\\ x=\text{ 1}\text{.212}\times {\text{10}}^{-4}M\\ \end{array}$

Therefore, at equilibrium,

Concentration of $A{g}^{+}(aq)$, $\left[A{g}^{+}\right]\text{ = }\left(0.050\text{ M + x}\right)\text{ = 0}\text{.050 M}$, as ‘x’ is negligible.

Concentration of $N{H}_3(aq)$, $\left[N{H}_3\right]\text{ = 2x = 2}\text{.42}\times {\text{10}}^{-4}\text{M }$

Concentration of ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}(aq)$, $\left[{\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}\right]\text{ = }\left(0.050\text{ M - x}\right)\text{ = 0}\text{.050 M}$, as ‘x’ is negligible.

Conclusion

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