Chapter 17, Problem 45QP

Write balanced equations and solubility product expressions for the solubility equilibria of the following compounds: $\left(\text{a}\right)\text{ CuBr, }\left(\text{b}\right){\text{ ZnC}}_{\text{2}}{\text{O}}_{\text{4}}\text{, }\left(\text{c}\right){\text{ Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{, }\left(\text{d}\right){\text{ Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{, }\left(\text{e}\right){\text{ AuCl}}_{\text{3}}\text{, }\left(\text{f}\right){\text{ Mn}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$.

Interpretation Introduction

Interpretation:

Thebalanced equation and the solubility product constant of the given compounds are to be calculated.

Concept introduction:

The amount of solute dissolved in a given volume of the solvent to form a saturated solution at a given temperatureis termed as the solubility of the solute in the solvent at that temperature.

The solubility product of a sparingly-soluble salt is given as the product of the concentration of the ions raised to the power equal to the number of times the ion occurs in the equation, after the dissociation of the electrolyte.

The number of moles of the solute dissolved per litre of the solution is called molar solubility.

For a general reaction: $\text{AB}\left(s\right)\rightleftharpoons {\text{A}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)$

The solubility product can be calculated by the expression as: $K_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{][B}}^{-}\text{]}$

Here, $K_{\text{sp}}$ is the solubility product constant and $sp$ standsfor solubility product.

Answer to Problem 45QP

Solution:

$\left(\text{a}\right)$

Balanced equation: $\text{CuBr}\left(s\right)\rightleftharpoons {\text{Cu}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)$

Solubility product: $K_{\text{sp}}{\text{= [Cu}}^{\text{+}}{\text{][Br}}^{-}\text{]}$

$\left(\text{b}\right)$

Balanced equation: ${\text{ZnC}}_2{\text{O}}_4\left(s\right)\rightleftharpoons {\text{Zn}}^{2+}\left(aq\right)+{\text{C}}_2{\text{O}}_4{}^{2-}\left(aq\right)$

Solubility product: $K_{\text{sp}}{\text{= [Zn}}^{\text{2+}}{\text{][C}}_2{\text{O}}_4{}^{2-}\text{]}$

$\left(\text{c}\right)$

Balanced equation: ${\text{Ag}}_2{\text{CrO}}_4\left(s\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(aq\right){\text{+ CrO}}_4{}^{2-}\left(aq\right)$

Solubility product: $K_{\text{sp}}{\text{= [Ag}}^{\text{+}}{\text{]}}^2{\text{[CrO}}_4{}^{2-}\text{]}$

$\left(\text{d}\right)$

Balanced equation: ${\text{Hg}}_2{\text{Cl}}_2\left(s\right)\rightleftharpoons {\text{Hg}}_2^{2+}\left(aq\right)+{\text{2Cl}}^{-}\left(aq\right)$

Solubility product: $K_{\text{sp}}={\text{[Hg}}_2^{2+}{\text{][Cl}}^{-}{\text{]}}^2$

$\left(\text{e}\right)$

Balanced equation: ${\text{AuCl}}_3\left(s\right)\rightleftharpoons {\text{Au}}^{3+}\left(aq\right)+{\text{3Cl}}^{-}\left(aq\right)$

Solubility product: ${\text{K}}_{\text{sp}}{\text{= [Au}}^{\text{3+}}{\text{][Cl}}^{-}{\text{]}}^3$

$\left(\text{f}\right)$

Balanced equation: ${\text{Mn}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)\rightleftharpoons 3{\text{Mn}}^{2+}\left(aq\right)+{\text{2PO}}_4{}^{3-}\left(aq\right)$

Solubility product: $K_{\text{sp}}{\text{= [Mn}}^{\text{2+}}{\text{]}}^3{{\text{[PO}}_4{}^{3-}\text{]}}^2$

Explanation of Solution

a) $\text{CuBr}\left(s\right)$

The balanced equation for the dissociation of $\text{CuBr}\left(s\right)$ is as follows:

$\text{CuBr}\left(s\right)\rightleftharpoons {\text{Cu}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)$

Let $s$ be the molar solubility.

The initial change equilibrium table for the ionisation of $\text{CuBr}\left(s\right)$ is as follows:

$\begin{array}{l}\text{CuBr}\left(s\right)\rightleftharpoons {\text{Cu}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -s& +s& +s\\ \text{Equilibrium}\left(\text{M}\right)& -& s& s\end{array}\end{array}$

The equilibrium expression for the reaction is written as follows:

$K_{\text{sp}}{\text{= [Cu}}^{\text{+}}{\text{][Br}}^{-}\text{]}$

Here, $K_{\text{sp}}$ is the solubility product constant, $\left[{\text{Cu}}^{+}\right]$ is the concentration of the copper ion, and $\left[{\text{Br}}^{-}\right]$ is the concentration of the bromine ion.

Substitute the values of $\left[{\text{Cu}}^{+}\right]$ and $\left[{\text{Br}}^{-}\right]$ in the above expression,

$\begin{array}{l}{K}_{\text{sp}}=\left(s\right)\left(s\right)\\ =s^2\end{array}$

Hence, the solubility product constant of $\text{CuBr}\left(s\right)$ is $K_{sp}=s^2$.

b) ${\text{ZnC}}_2{\text{O}}_4\left(s\right)$

The balanced equation for the dissociation of ${\text{ZnC}}_2{\text{O}}_4\left(s\right)$ is as follows:

${\text{ZnC}}_2{\text{O}}_4\left(s\right)\rightleftharpoons {\text{Zn}}^{2+}\left(aq\right)+{\text{C}}_2{\text{O}}_4{}^{2-}\left(aq\right)$

Let $s$ be the molar solubility.

The initial change equilibrium table for the ionisation of ${\text{ZnC}}_2{\text{O}}_4\left(s\right)$ is as follows:

$\begin{array}{l}{\text{ZnC}}_2{\text{O}}_4\left(s\right)\rightleftharpoons {\text{Zn}}^{2+}\left(aq\right)+{\text{ C}}_2{\text{O}}_4{}^{2-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -s& +s& +s\\ \text{Equilibrium}\left(\text{M}\right)& -& s& s\end{array}\end{array}$

The equilibrium expression for the reaction is written as follows:

$K_{\text{sp}}{\text{= [Zn}}^{\text{2+}}{\text{][C}}_2{\text{O}}_4{}^{2-}\text{]}$

Here, $K_{\text{sp}}$ is the solubility product constant, $\left[{\text{Zn}}^{2+}\right]$ is the concentration of the zinc ion, and $\left[{\text{C}}_2{\text{O}}_4{}^{2-}\right]$ is the concentration of the oxalate ion.

Substitute the values of $\left[{\text{Zn}}^{2+}\right]$ and $\left[{\text{C}}_2{\text{O}}_4{}^{2-}\right]$ in the above expression,

$\begin{array}{l}{K}_{\text{sp}}=\left(s\right)\left(s\right)\\ =s^2\end{array}$

Hence, the solubility product constant of ${\text{ZnC}}_2{\text{O}}_4\left(s\right)$ is $K_{sp}=s^2$.

c) ${\text{Ag}}_2{\text{CrO}}_4\left(s\right)$

The balanced equation for the dissociation of ${\text{Ag}}_2{\text{CrO}}_4\left(s\right)$ is as follows:

${\text{Ag}}_2{\text{CrO}}_4\rightleftharpoons 2{\text{Ag}}^{+}\left(aq\right)+{\text{CrO}}_4{}^{2-}\left(aq\right)$

Let $s$ be the molar solubility.

The initial change equilibrium table for theionization of ${\text{Ag}}_2{\text{CrO}}_4\left(s\right)$ is as follows:

$\begin{array}{l}{\text{Ag}}_2{\text{CrO}}_4\left(s\right)\rightleftharpoons 2{\text{Ag}}^{+}\left(aq\right)+{\text{CrO}}_4{}^{2-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -s& +2s& +s\\ \text{Equilibrium}\left(\text{M}\right)& -& 2s& s\end{array}\end{array}$

The equilibrium expression for the reaction is written as follows:

$K_{\text{sp}}{\text{= [Ag}}^{\text{+}}{\text{]}}^2{\text{[CrO}}_4{}^{2-}\text{]}$

Here, $K_{\text{sp}}$ is the solubility product constant, $\left[{\text{Ag}}^{+}\right]$ is the concentration of the silver ion, and $\left[{\text{CrO}}_4{}^{2-}\right]$ is the concentration of the chromate ion.

Substitute the values of $\left[{\text{Ag}}^{+}\right]$ and $\left[{\text{CrO}}_4{}^{2-}\right]$ in the above expression,

$\begin{array}{l}{K}_{\text{sp}}={\left(2s\right)}^2\left(s\right)\\ =4s^3\end{array}$

Hence, the solubility product constant of ${\text{Ag}}_2{\text{CrO}}_4\left(s\right)$ is $K_{sp}=\text{ 4}{s}^3$.

d) ${\text{Hg}}_2{\text{Cl}}_2\left(s\right)$

The balanced equation for the dissociation of ${\text{Hg}}_2{\text{Cl}}_2\left(s\right)$ is as follows:

${\text{Hg}}_2{\text{Cl}}_2\left(s\right)\rightleftharpoons {\text{Hg}}_2^{2+}\left(aq\right)+{\text{2Cl}}^{-}\left(aq\right)$

Let $s$ be the molar solubility.

The initial change equilibrium table for the ionisation of ${\text{Hg}}_2{\text{Cl}}_2\left(s\right)$ is as follows:

$\begin{array}{l}{\text{Hg}}_2{\text{Cl}}_2\left(s\right)\rightleftharpoons {\text{Hg}}_2^{2+}\left(aq\right)+{\text{ 2Cl}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -s& +s& +2s\\ \text{Equilibrium}\left(\text{M}\right)& -& s& 2s\end{array}\end{array}$

The equilibrium expression for the reaction is written as follows:

$K_{\text{sp}}{\text{= [Hg}}_2^{2+}{\text{][Cl}}^{-}{\text{]}}^2$

Here, $K_{\text{sp}}$ is the solubility product constant, $\left[{\text{Hg}}_2^{2+}\right]$ is the concentration of $\text{Dimercury}\left(2+\right)$, and $\left[{\text{Cl}}^{-}\right]$ is the concentration of the chlorine ion.

Substitute the values of $\left[{\text{Hg}}_2^{2+}\right]$ and $\left[{\text{Cl}}^{-}\right]$ in the above expression,

$\begin{array}{l}{K}_{\text{sp}}=\left(s\right){\left(2s\right)}^2\\ =4s^3\end{array}$

Hence, the solubility product constant of ${\text{Hg}}_2{\text{Cl}}_2\left(s\right)$ is $K_{sp}=\text{ 4}{s}^3$.

e) ${\text{AuCl}}_3\left(s\right)$

The balanced equation for the dissociation of ${\text{AuCl}}_3\left(s\right)$ is as follows:

${\text{AuCl}}_3\left(s\right)\rightleftharpoons {\text{Au}}^{3+}\left(aq\right)+{\text{3Cl}}^{-}\left(aq\right)$

The ICE table for the ionisation for ${\text{AuCl}}_3\left(s\right)$ is as follows:

Let $s$ be the molar solubility.

$\begin{array}{l}{\text{AuCl}}_3\left(s\right)\rightleftharpoons {\text{Au}}^{3+}\left(aq\right)+{\text{ 3Cl}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -s& +s& +3s\\ \text{Equilibrium}\left(\text{M}\right)& -& s& 3s\end{array}\end{array}$

The equilibrium expression for the reaction is written as follows:

$K_{\text{sp}}{\text{= [Au}}^{\text{3+}}{\text{][Cl}}^{-}{\text{]}}^3$

Here, ${\text{K}}_{\text{sp}}$ is the solubility product constant, $\left[{\text{Au}}^{3+}\right]$ is the concentration of $\text{Gold}\left(3+\right)$, and $\left[{\text{Cl}}^{-}\right]$ is the concentration of the chlorine ion.

Substitute the values of $\left[{\text{Au}}^{3+}\right]$ and $\left[{\text{Cl}}^{-}\right]$ in the above expression,

$\begin{array}{l}{K}_{\text{sp}}=\left(s\right){\left(3s\right)}^3\\ =27s^4\end{array}$

Hence, the solubility product constant of ${\text{AuCl}}_3\left(s\right)$ is $K_{sp}=\text{ 27}{s}^4$.

f) ${\text{Mn}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$

The balanced equation for the dissociation of ${\text{Mn}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$ is as follows:

${\text{Mn}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)\rightleftharpoons 3{\text{Mn}}^{2+}\left(aq\right)+{\text{2PO}}_4{}^{3-}\left(aq\right)$

Let $s$ be the molar solubility.

The initial change equilibrium table for the ionisation of ${\text{Mn}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$ is as follows:

$\begin{array}{l}{\text{Mn}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)\rightleftharpoons 3{\text{Mn}}^{2+}\left(aq\right)+{\text{ 2PO}}_4{}^{2-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -s& +3s& +2s\\ \text{Equilibrium}\left(\text{M}\right)& -& 3s& 2s\end{array}\end{array}$

The equilibrium expression for the reaction is written as follows:

$K_{\text{sp}}{\text{= [Mn}}^{\text{2+}}{\text{]}}^3{{\text{[PO}}_4{}^{3-}\text{]}}^2$

Here, $K_{\text{sp}}$ is the solubility product constant, $\left[{\text{Mn}}^{2+}\right]$ is the concentration of the manganese $\left(2+\right)$ ion, and $\left[{\text{PO}}_4{}^{3-}\right]$ is the concentration of the phosphate $\left(3+\right)$ ion.

Substitute the values of $\left[{\text{Mn}}^{2+}\right]$ and $\left[{\text{PO}}_4{}^{3-}\right]$ in the above expression,

$\begin{array}{c}{K}_{\text{sp}}={\left(3s\right)}^3{\left(2s\right)}^2\\ =108s^5\end{array}$

Hence, the solubility product constant of ${\text{Mn}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)$ is $K_{sp}=\text{ 108}{s}^5$.