Chapter 17, Problem 17.92QP

Interpretation Introduction

Interpretation:

From the given data, the molar solubility of $\text{AgI}$ in $2.2\text{M}$ ${\text{NH}}_{\text{3}}$ has to be calculated.

Concept introduction:

• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• Solubility of a compound is expressed as concentration of its ions in saturated solution.

Answer to Problem 17.92QP

The molar solubility of $\text{AgI}$ in $2.2\text{M}$ ${\text{NH}}_{\text{3}}$ is $\text{8}\text{.3}\text{×}{\text{10}}^{\text{-5}}\text{M}$

Explanation of Solution

To calculate: The molar solubility of $\text{AgI}$ in $2.2\text{M}$ ${\text{NH}}_{\text{3}}$.

Given,

The strength of ${\text{NH}}_{\text{3}}$ is $2.2\text{M}$.

The molar solubility of $\text{AgI}$ is calculated as follows,

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.

$\begin{array}{l}{\text{AgBr}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{Ag}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{I}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{sp}}\text{ = 8}\text{.3}\text{×}{\text{10}}^{\text{-17}}\\ \\ {\text{Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{ + 2NH}}_{\text{3}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{{}^{\text{+}}}_{\text{(aq)}}{\text{K}}_{\text{f}}\text{ = 1}\text{.7}\text{×}{\text{10}}^{\text{7}}\\ \\ {\text{AgI}}_{\text{(s)}}\text{+}{\text{2NH}}_{\text{3}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{{}^{\text{+}}}_{\text{(aq)}}\text{+}{\text{I}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{c}}{\text{ = K}}_{\text{sp}}\text{×}{\text{K}}_{\text{f}}\text{=}1.41\text{×}{\text{10}}^{\text{-9}}\end{array}$

Before addition of $\text{AgI}$ in ${\text{NH}}_{\text{3}}$, the solution contains $\text{2}\text{.2}\text{M}{\text{I}}^{\text{-}}$.  After addition, the solution is in equilibrium that is s mol of solid $\text{AgI}$ dissolves to form s mol of ${\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{}^{\text{+}}$ and s mol of ${\text{I}}^{\text{-}}$.

$\begin{array}{l}{\text{AgI}}_{\text{(s)}}\text{+}{\text{2NH}}_{\text{3}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{{}^{\text{+}}}_{\text{(aq)}}\text{+}{\text{I}}^{\text{-}}{}_{\text{(aq)}}\\ \text{Initial }\left(\text{M}\right)\text{: 2}\text{.2 }\text{0}\text{0}\\ \text{Change }\left(\text{M}\right)\text{: }\text{ +2x }\text{+x}\text{+x}\\ \text{Equilibrium }\left(\text{M}\right)\text{: }2.2\text{ - 2x}\text{x}\text{x}\\ \\ \text{x}\text{is}\text{molar}\text{solubility}\\ \\ {\text{K}}_{\text{c}}\text{ =}\frac{{\text{[Ag(NH}}_{\text{3}}{\text{)}}_{\text{2}}{}^{\text{+}}{\text{][I}}^{\text{-}}\text{]}}{{{\text{[NH}}_{\text{3}}\text{]}}^{\text{2}}}\text{=}\frac{{\text{(x)}}^{\text{2}}}{{\text{(}2.2\text{ - 2x)}}^{\text{2}}}\text{=}1.41\text{×}{\text{10}}^{\text{-9}}\\ \\ \text{Take}\text{square}\text{root}\text{on}\text{both}\text{sides}\text{we}\text{get,}\\ \\ \frac{\text{x}}{{\text{(2}\text{.2 - 2x)}}^{}}\text{=}\text{3}\text{.75}\text{×}{\text{10}}^{\text{-5}}\\ \\ \text{x + (7}\text{.5 ×}{\text{10}}^{\text{-5}}\text{)x}\text{=}\text{8}\text{.26}\text{×}{\text{10}}^{\text{-5}}\\ \\ \text{x}\text{=}\text{8}\text{.3}\text{×}{\text{10}}^{\text{-5}}\text{M}\end{array}$

The molar solubility of $\text{AgI}$ in $2.2\text{M}$ ${\text{NH}}_{\text{3}}$ is $\text{8}\text{.3}\text{×}{\text{10}}^{\text{-5}}\text{M}$.

Conclusion

By using the equilibrium constant of $\text{AgI}$,  the molar solubility of $\text{AgI}$ in $2.2\text{M}$ ${\text{NH}}_{\text{3}}$ was calculated.