Chapter 17.5, Problem 10PPB

Arrange the following salts in order of increasing molar solubility in 0.0010 M AgNO₃: AgBr, Ag₂CO₃. AgCl, AgI, Ag₂S.

Interpretation Introduction

Interpretation:

From the given salts, the ascending order of molar solubility has to be determined.

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.

To calculate: The molar solubility of given solutions dissolved in ${\text{AgNO}}_{\text{3}}$ .

Answer to Problem 10PPB

The ascending order of molar solubility is ${\text{Ag}}_{\text{2}}{\text{S < AgI < AgBr < AgCl < Ag}}_{\text{2}}{\text{CO}}_{\text{3}}$

Explanation of Solution

The molar solubility of given solutions dissolved in ${\text{AgNO}}_{\text{3}}$ is calculated below.

$\begin{array}{l}{\text{AgBr}}_{\left(\text{s}\right)}\stackrel{}{\rightleftarrows }{\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{ + }{\text{Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration }\left(\text{M}\right)\text{: }0.0010\text{0}\\ \text{Change in concentration }\left(\text{M}\right)\text{: }\text{+s }\text{+s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }0.0010+\text{s}\text{s}\\ \\ {\text{K}}_{\text{sp}}\text{ of AgBr is 7}\text{.7}{\text{×10}}^{\text{-}}{}^{\text{13}}\\ \text{s}\text{is}\text{molar}\text{solubility}\\ {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Ag}}^{\text{+}}\right]\left[{\text{Br}}^{\text{-}}\right]\\ \text{7}\text{.7}{\text{×10}}^{\text{-}}{}^{\text{13}}\text{= }\text{(}0.0010+\text{s)(s)}\\ \text{s}\text{=}7.7\text{×}{\text{10}}^{\text{-10}}\text{M}\\ \\ \text{molar solubility}\text{=}7.7\text{×}{\text{10}}^{\text{-10}}\text{M or}7.7\text{×}{\text{10}}^{\text{-10}}\text{mol/L}\end{array}$

$\begin{array}{l}{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftarrows }2{\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{ + }{\text{CO}}_{\text{3}}{{}^2}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration }\left(\text{M}\right)\text{: }\text{0}\text{.0010 }0\\ \text{Change in concentration }\left(\text{M}\right)\text{: }\text{+s }\text{+s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }\text{0010}+\text{s}\text{s}\\ \\ {\text{K}}_{\text{sp}}{\text{ of Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ is }8.1{\text{×10}}^{\text{-}}{}^{\text{12}}\\ \text{s}\text{is}\text{molar}\text{solubility}\\ {\text{K}}_{\text{sp}}\text{ = }{\left[{\text{Ag}}^{\text{+}}\right]}^2\left[{\text{CO}}_{\text{3}}{{}^2}^{\text{-}}\right]\\ 8.1{\text{×10}}^{\text{-}}{}^{\text{12}}\text{= }{\text{(}}^{0.0010}\text{(s)}\\ \text{s}\text{=}8.1\text{×}{\text{10}}^{\text{-18}}\text{M}\\ \\ \text{molar solubility}\text{=}8.1\text{×}{\text{10}}^{\text{-18}}\text{M or}8.1\text{×}{\text{10}}^{\text{-18}}\text{mol/L}\end{array}$

$\begin{array}{l}{\text{AgCl}}_{\left(\text{s}\right)}\stackrel{}{\rightleftarrows }{\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{ + }{\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration }\left(\text{M}\right)\text{: }0.0010\text{0}\\ \text{Change in concentration }\left(\text{M}\right)\text{: }\text{+s }\text{+s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }0.0010+\text{s}\text{s}\\ \\ {\text{K}}_{\text{sp}}\text{ of AgCl is 1}\text{.6}{\text{×10}}^{\text{-}}{}^{\text{10}}\\ \text{s}\text{is}\text{molar}\text{solubility}\\ {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Ag}}^{\text{+}}\right]\left[{\text{Cl}}^{\text{-}}\right]\\ \text{1}\text{.6}{\text{×10}}^{\text{-}}{}^{\text{10}}\text{= }\text{(}0.0010+\text{s)(s)}\\ \text{s}\text{=}1.6\text{×}{\text{10}}^{\text{-7}}\text{M}\\ \\ \text{molar solubility}\text{=}1.6\text{×}{\text{10}}^{\text{-7}}\text{M or}1.6\text{×}{\text{10}}^{\text{-7}}\text{mol/L}\end{array}$

$\begin{array}{l}{\text{AgI}}_{\left(\text{s}\right)}\stackrel{}{\rightleftarrows }{\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{ + }{\text{I}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration }\left(\text{M}\right)\text{: }0.0010\text{0}\\ \text{Change in concentration }\left(\text{M}\right)\text{: }\text{+s }\text{+s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }0.0010+\text{s}\text{s}\\ \\ {\text{K}}_{\text{sp}}\text{ of AgI is 8}\text{.3}{\text{×10}}^{\text{-}}{}^{\text{17}}\\ \text{s}\text{is}\text{molar}\text{solubility}\\ {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Ag}}^{\text{+}}\right]\left[{\text{I}}^{\text{-}}\right]\\ \text{8}\text{.3}{\text{×10}}^{\text{-}}{}^{\text{17}}\text{= }\text{(}0.0010+\text{s)(s)}\\ \text{s}\text{=}\text{8}\text{.3}{\text{×10}}^{\text{-}}{}^{\text{14}}\text{M}\\ \\ \text{molar solubility}\text{=}\text{8}\text{.3}{\text{×10}}^{\text{-}}{}^{\text{14}}\text{M or}\text{8}\text{.3}{\text{×10}}^{\text{-}}{}^{\text{14}}\text{mol/L}\end{array}$

$\begin{array}{l}{\text{Ag}}_{\text{2}}{\text{S}}_{\left(\text{s}\right)}\stackrel{}{\rightleftarrows }2{\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{ + }{\text{s}}^{\text{2-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration }\left(\text{M}\right)\text{: }\text{0}\text{.0010 }0\\ \text{Change in concentration }\left(\text{M}\right)\text{: }\text{+s }\text{+s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }\text{0010}+\text{s}\text{s}\\ \\ {\text{K}}_{\text{sp}}{\text{ of Ag}}_{\text{2}}\text{S is 6}\text{.0}{\text{×10}}^{\text{-}}{}^{51}\\ \text{s}\text{is}\text{molar}\text{solubility}\\ {\text{K}}_{\text{sp}}\text{ = }{\left[{\text{Ag}}^{\text{+}}\right]}^{\text{2}}\left[{\text{S}}^{\text{2}}{}^{\text{-}}\right]\\ \text{6}\text{.0}{\text{×10}}^{\text{-}}{}^{51}\text{= }{\text{(}}^{0.0010}\text{(s)}\\ \text{s}\text{=}\text{6}\text{.0}{\text{×10}}^{\text{-}}{}^{57}\text{M}\\ \\ \text{molar solubility}\text{=}\text{6}\text{.0}{\text{×10}}^{\text{-}}{}^{57}\text{M or}\text{6}\text{.0}{\text{×10}}^{\text{-}}{}^{57}\text{mol/L}\end{array}$

Conclusion

From the given salts, the ascending order of molar solubility was determined.