Chapter 17, Problem 17.122QP

Interpretation Introduction

Interpretation:

The quantity of water and ammonia required to dissolve silver chloride has to be determined.

Concept introduction:

• The solubility product constant ( ${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion,  raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.
• Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

$\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging,

$\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

${\text{K}}_{\text{c}}$ is the equilibrium constant.

Answer to Problem 17.122QP

The quantity of water required to dissolve silver chloride is $\text{5}\text{.2}\text{×}{\text{10}}^{\text{3}}\text{L}$

The quantity of $\text{0}\text{.15M}$ ammonia required to dissolve silver chloride is $\text{9}\text{.4}\text{L}$

Explanation of Solution

To calculate: The quantity of water and ammonia required to dissolve silver chloride.

The number of moles of silver chloride is,

$\begin{array}{l}\text{mol AgCl =}\text{10}\text{.0 g AgCl}\text{×}\frac{\text{1 mol AgCl}}{\text{143}\text{.3 g AgCl}}\text{=}\text{0}\text{.06978 mol AgCl}\\ \\ {\text{mol AgCl = mol Ag}}^{\text{+}}{\text{ (In water) = mol Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{}^{\text{+}}{\text{ (In NH}}_{\text{3}}\text{)}\\ \\ \text{The volume required to dissolve AgCl}\text{is calculated using number of }\\ \text{moles of silver chloride and concentration of silver ion}\text{.}\\ \\ {\text{K}}_{\text{sp}}\text{ = 1}\text{.8 ×}{\text{10}}^{\text{-10}}\text{=}{\text{[Ag}}^{\text{+}}{\text{][Cl}}^{\text{-}}{\text{] = x}}^{\text{2}}\\ \\ {\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{x}\text{=}\sqrt{\text{1}\text{.8 ×}{\text{10}}^{\text{-10}}}\text{=}\text{1}\text{.34}\text{×}{\text{10}}^{\text{-5}}\text{M}\\ \\ \text{Volume of water}\text{=}\frac{\text{0}{\text{.06978 mol Ag}}^{\text{+}}}{\text{1}\text{.34}\text{×}{\text{10}}^{\text{-5}}{\text{mol Ag}}^{\text{+}}\text{/L}}\text{=}\text{5}\text{.2}\text{×}{\text{10}}^{\text{3}}\text{L}\end{array}$

Silver chloride reacts with ammonia to form silver-ammonia complex.  By combining the solubility equilibrium with complexation equilibrium to determine the total reaction.

When a chemical equation can be derived by taking sum of other equations, then the equilibrium constant of derived equation equals the product of the equilibrium constants of added equations.

$\begin{array}{l}{\text{AgCl}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{Ag}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{Cl}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{sp}}\text{ = 1}\text{.8}\text{×}{\text{10}}^{\text{-10}}\\ \\ {\text{Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{ + 2NH}}_{\text{3}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{{}^{\text{+}}}_{\text{(aq)}}{\text{K}}_{\text{f}}\text{ = 1}\text{.7}\text{×}{\text{10}}^{\text{7}}\\ \\ {\text{AgCl}}_{\text{(s)}}\text{+}{\text{2NH}}_{\text{3}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{{}^{\text{+}}}_{\text{(aq)}}\text{+}{\text{Cl}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{c}}\text{ = Ksp}\text{×}{\text{K}}_{\text{f}}\text{=}\text{3}\text{.06}\text{×}{\text{10}}^{\text{-3}}\end{array}$

$\begin{array}{l}{\text{AgCl}}_{\text{(s)}}\text{+}{\text{2NH}}_{\text{3}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{{}^{\text{+}}}_{\text{(aq)}}\text{+}{\text{Cl}}^{\text{-}}{}_{\text{(aq)}}\\ \text{Initial }\left(\text{M}\right)\text{: 0}\text{.15 }\text{0}\text{0}\\ \text{Change }\left(\text{M}\right)\text{: }\text{ -2x }\text{+x}\text{+x}\\ \text{Equilibrium }\left(\text{M}\right)\text{: }\text{0}\text{.15 - 2x}\text{x}\text{x}\\ \\ {\text{K}}_{\text{c}}\text{ =}\frac{{\text{[Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{}^{\text{+}}{\text{][Cl}}^{\text{-}}\text{]}}{{{\text{[NH}}_{\text{3}}\text{]}}^{\text{2}}}\text{=}\frac{{\text{(x)}}^{\text{2}}}{{\text{(0}\text{.15 - 2x)}}^{\text{2}}}\text{=}\text{3}\text{.06}\text{×}{\text{10}}^{\text{-3}}\\ \\ \text{Take}\text{square}\text{root}\text{on}\text{both}\text{sides}\text{we}\text{get,}\\ \\ \frac{\text{x}}{{\text{(0}\text{.15 - 2x)}}^{}}\text{=}\text{5}\text{.53}\text{×}{\text{10}}^{\text{-2}}\\ \\ \text{x}\text{=}\text{(0}\text{.15 - 2x)}\text{×}\text{5}\text{.53}\text{×}{\text{10}}^{\text{-2}}\text{=}\text{0}\text{.00829 - 0}\text{.1106x}\\ \\ \text{x}\text{=}\text{0}\text{.00746}\text{=}{\text{[Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{}^{\text{+}}\text{]}\\ \\ \text{The volume of ammonia required to dissolve AgCl}\text{is calculated using number of }\\ \text{moles of silver complex and concentration of silver complex}\text{.}\\ \\ \text{Volume of 0}\text{.15M ammonia}\text{=}\frac{\text{0}{\text{.06978 mol Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{}^{\text{+}}}{\text{0}{\text{.00746 mol Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{}^{\text{+}}\text{/L}}\text{=}\text{9}\text{.4}\text{L}\end{array}$

Due to complex formation and stability of complex ${\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{}^{\text{+}}$ ion, the volume required to dissolve decreases from $\text{5000 L to under 10 L}$

Conclusion

By using the solubility product expression, the volume of water and ammonia required to dissolve silver chloride were determined.