Question

The ${\mathit{K}}_{\mathbf{s}\mathbf{p}}$for silver sulfate is $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$Calculate the solubility of silver sulfate in each of the following

1. Water
2. $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{ }\mathit{M}\mathbf{ }\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}$
3. $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{ }\mathit{M}\mathbf{ }{\mathit{K}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$

Short answer

1. The solubility of silver sulfate in water is$s=1.2\times {10}^{-3}mol/L$
2. The solubility of silver sulfate in $0.10 M AgN{O}_3$is $s=1.2\times {10}^{-3}mol/L$
3. The solubility of silver sulfate in $0.10 M K_{2}S{O}_4$is $s=3.9\times {10}^{-3}mol/L$

Step-by-step solution

Definition of solubility

The maximum amount of a substance that will dissolve in a given amount of solvent at a specified temperature is defined as solubility

Calculation of solubility

Let the equilibrium concentrations be

Substituting the values,

$\begin{array}{c}1.2\times {10}^{-5}=(0.10+2s{)}^2\left(s\right)\\ 1.2\times {10}^{-5}=(0.10{)}^2\left(s\right)\\ s=\frac{1.2\times {10}^{-5}}{0.01}\\ s=1.2\times {10}^{-3}mol/L\end{array}$

Let the equilibrium concentrations be

Substituting the values,

$\begin{array}{c}1.2\times {10}^{-5}=(0.10+2s{)}^2\left(s\right)\\ 1.2\times {10}^{-5}=(0.10{)}^2\left(s\right)\\ s=\frac{1.2\times {10}^{-5}}{0.01}\\ s=1.2\times {10}^{-3}mol/L\end{array}$

Let the equilibrium concentrations be

Substituting the values,

$\begin{array}{c}1.2\times {10}^{-5}=(2s{)}^2(0.20+s)\\ 1.2\times {10}^{-5}=(2s{)}^2(0.20)\\ s=\frac{1.2\times {10}^{-5}}{0.8}\\ s=3.9\times {10}^{-3}mol/L\end{array}$