Chapter 17, Problem 17.42QP

The solubility of silver sulfate, Ag₂SO₄, in water has been determined to be 8.0 g/L. What is the solubility in 0.45 M sodium sulfate, Na₂SO₄?

Interpretation Introduction

Interpretation:

From the given data, the solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$  in $\text{0}\text{.45M}$ ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ 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.
• 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 ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.

Answer to Problem 17.42QP

The solubility of  ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$  in $\text{0}\text{.45M}$ ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{1}\text{.9}\text{g/L}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$

Explanation of Solution

To calculate: The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$  in $\text{0}\text{.45M}$ ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Given,

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in water is $\text{8 g/L}$

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$  in $\text{0}\text{.45M}$ ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated as follows,

By using the solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in water, the solubility product constant ( ${\text{K}}_{\text{sp}}$) is calculated as follows,

$\frac{\text{8 g}}{\text{1}\text{L}\text{soln}}\text{×}\frac{{\text{1molAg}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{311}\text{.81 g}}\text{=}\text{0}\text{.02565 M}$

$\begin{array}{l}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftharpoons }{\text{2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{+ }{\text{SO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}\\ \text{Initial }\left(\text{M}\right)\text{: }\text{0 }\text{0}\text{.13}\\ \text{Change }\left(\text{M}\right)\text{: }\text{+2×0}\text{.02565 }\text{+0}\text{.02565}\\ \text{Equilibrium }\left(\text{M}\right)\text{: }\text{2×0}\text{.02565 }\text{0}\text{.02565}\\ \\ {\text{K}}_{\text{sp}}\text{ = }{\left[{\text{Ag}}^{\text{+}}\right]}^{\text{2}}\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]\\ {\text{K}}_{\text{sp}}\text{= }{\text{(2×0}}^{\text{.02565)}}\left(\text{0}\text{.02565}\right)\\ {\text{K}}_{\text{sp}}\text{=}\text{6}\text{.755 ×}{\text{10}}^{\text{-5}}\end{array}$

Before addition of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$  in ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$, the solution contains $\text{0}\text{.45M}{\text{SO}}_{\text{4}}{}^{\text{2-}}$.  After addition the solution is in equilibrium that is s mol of solid ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ dissolves to form s mol of ${\text{Ag}}^{\text{+}}$ and s mol of ${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{\text{-}}$.

$\begin{array}{l}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftharpoons }{\text{2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{+ }{\text{SO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}\\ \text{Initial }\left(\text{M}\right)\text{: }\text{0 }\text{0}\text{.45}\\ \text{Change }\left(\text{M}\right)\text{: }\text{+2s +s}\\ \text{Equilibrium }\left(\text{M}\right)\text{: }\text{2s}\left(\text{0}\text{.45+}\text{s}\right)\\ \\ \text{s}\text{is}\text{molar}\text{solubility}\\ \\ {\text{K}}_{\text{sp}}\text{ = }{\left[{\text{Ag}}^{\text{+}}\right]}^{\text{2}}\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]\\ \\ \text{6}\text{.755 ×}{\text{10}}^{\text{-5}}\text{= }{\text{(2s)}}^{\text{2}}\left(\text{0}\text{.45}\text{+}\text{s}\right)\\ \\ \text{s}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ \\ \text{s}\text{=}\sqrt{\frac{\text{6}\text{.755 ×}{\text{10}}^{\text{-5}}}{\text{(4}\text{×0}\text{.45)}}}\text{=}\text{6}\text{.125}{\text{×10}}^{\text{-}}{}^{\text{3}}\text{M}\\ \\ \text{The}\text{mass}\text{of}\text{silver}\text{sulfate}\text{is}\\ \\ \frac{\text{6}\text{.125}{\text{×10}}^{\text{-}}{}^{\text{3}}\text{mol}}{\text{1}\text{L}\text{soln}}\text{×}\frac{\text{311}\text{.81 g}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1}{\text{mol Ag}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{=}\text{1}\text{.9}\text{g/L}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

Conclusion

By using the solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in water,  the solubility product constant ( ${\text{K}}_{\text{sp}}$)  of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ was determined and  the solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$  in $\text{0}\text{.45M}$ ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ was calculated.