Question

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. ${\mathrm{FeC}}_2{\mathrm{O}}_4,K_{\mathrm{sp}}=2.1\times {10}^{-7}$, or $\mathrm{Cu}{\left({\mathrm{IO}}_4\right)}_2,K_{\mathrm{sp}}=1.4\times {10}^{-7}$ b. ${\mathrm{Ag}}_2{\mathrm{CO}}_3,K_{<\rho }=8.1\times {10}^{-12}$, or $\mathrm{Mn}(\mathrm{OH}{)}_2,K_{\mathrm{sp}}=2\times {10}^{-13}$

Short answer

a) $Cu{\left(I{O}_4\right)}_2$ with a molar solubility of $1.4\times {10}^{-7}$ has the smallest molar solubility compared to $Fe{C}_2{O}_4$. b) $Mn(OH{)}_2$ with a molar solubility of $2\times {10}^{-13}$ has the smallest molar solubility compared to $A{g}_{2}C{O}_3$.

Step-by-step solution

Write dissociation reactions and expressions for $K_{sp}$

For each of the two pairs of solids, write balanced dissociation reactions, and their corresponding expressions for the solubility product constant ($K_{sp}$). a) For ${\mathrm{FeC}}_2{\mathrm{O}}_4$: Dissociation reaction: ${\mathrm{FeC}}_2{\mathrm{O}}_4(s)\leftrightarrows {\mathrm{Fe}}^{2+}(aq)+2{\mathrm{C}}_2{\mathrm{O}}_4^{2–}(aq)$ Expression for $K_{sp}$: $K_{sp}=[{\mathrm{Fe}}^{2+}][{\mathrm{C}}_2{\mathrm{O}}_4^{2–}{]}^2$ For $\mathrm{Cu}{\left({\mathrm{IO}}_4\right)}_2$: Dissociation reaction: $\mathrm{Cu}{\left({\mathrm{IO}}_4\right)}_2(s)\leftrightarrows {\mathrm{Cu}}^{2+}(aq)+2{\mathrm{IO}}_4^{–}(aq)$ Expression for $K_{sp}$: $K_{sp}=[{\mathrm{Cu}}^{2+}][{\mathrm{IO}}_4^{–}{]}^2$ b) For ${\mathrm{Ag}}_2{\mathrm{CO}}_3$: Dissociation reaction: ${\mathrm{Ag}}_2{\mathrm{CO}}_3(s)\leftrightarrows 2{\mathrm{Ag}}^{+}(aq)+{\mathrm{CO}}_3^{2–}(aq)$ Expression for $K_{sp}$: $K_{sp}=[{\mathrm{Ag}}^{+}{]}^2[{\mathrm{CO}}_3^{2–}]$ For $\mathrm{Mn}(\mathrm{OH}{)}_2$: Dissociation reaction: $\mathrm{Mn}(\mathrm{OH}{)}_2(s)\leftrightarrows {\mathrm{Mn}}^{2+}(aq)+2{\mathrm{OH}}^{–}(aq)$ Expression for $K_{sp}$: $K_{sp}=[{\mathrm{Mn}}^{2+}][{\mathrm{OH}}^{–}{]}^2$

Calculate molar solubilities

For each solid, calculate its molar solubility ($S$) using the given $K_{sp}$ values and the expressions for the $K_{sp}$. a) ${\mathrm{FeC}}_2{\mathrm{O}}_4$: Let the molar solubility of ${\mathrm{Fe}}^{2+}$ and ${\mathrm{C}}_2{\mathrm{O}}_4^{2–}$ be $x$ and $2x$, respectively. $K_{sp}=2.1\times {10}^{-7}=x\times (2x{)}^2$ Solve for $x$ (molar solubility of ${\mathrm{FeC}}_2{\mathrm{O}}_4$) $\mathrm{Cu}{\left({\mathrm{IO}}_4\right)}_2$: Let the molar solubility of ${\mathrm{Cu}}^{2+}$ and ${\mathrm{IO}}_4^{–}$ be $x$ and $2x$, respectively. $K_{sp}=1.4\times {10}^{-7}=x\times (2x{)}^2$ Solve for $x$ (molar solubility of $\mathrm{Cu}{\left({\mathrm{IO}}_4\right)}_2$) b) ${\mathrm{Ag}}_2{\mathrm{CO}}_3$: Let the molar solubility of ${\mathrm{Ag}}^{+}$ and ${\mathrm{CO}}_3^{2–}$ be $2x$ and $x$, respectively. $K_{sp}=8.1\times {10}^{-12}=(2x{)}^2\times x$ Solve for $x$ (molar solubility of ${\mathrm{Ag}}_2{\mathrm{CO}}_3$) $\mathrm{Mn}(\mathrm{OH}{)}_2$: Let the molar solubility of ${\mathrm{Mn}}^{2+}$ and ${\mathrm{OH}}^{–}$ be $x$ and $2x$, respectively. $K_{sp}=2\times {10}^{-13}=x\times (2x{)}^2$ Solve for $x$ (molar solubility of $\mathrm{Mn}(\mathrm{OH}{)}_2$)

Compare molar solubilities

Compare the molar solubilities calculated in Step 2 to determine which solid in each pair has the smallest molar solubility. a) Comparing the molar solubilities of ${\mathrm{FeC}}_2{\mathrm{O}}_4$ and $\mathrm{Cu}{\left({\mathrm{IO}}_4\right)}_2$, the solid with the smallest molar solubility is ______. b) Comparing the molar solubilities of ${\mathrm{Ag}}_2{\mathrm{CO}}_3$ and $\mathrm{Mn}(\mathrm{OH}{)}_2$, the solid with the smallest molar solubility is ______.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, or $K_{sp}$, is a vital concept in understanding how substances dissolve in solutions. It refers to the equilibrium constant for a solid substance dissolving in an aqueous solution. Each solid has a specific $K_{sp}$ value, which quantitatively describes its solubility.
Small $K_{sp}$ values indicate a low solubility, meaning only a small amount of the solid can dissolve in water.
This constant is fundamental in predicting whether a precipitate will form when solutions are mixed.

• $K_{sp}$ is derived from the concentrations of the ions in a saturated solution.
• The formula generally takes the form $K_{sp}=[extion{]}^{n_1}[extion{]}^{n_2}$, involving the ions produced from dissociation.
• Knowing $K_{sp}$ helps chemists understand and control reactions where solubility is a factor.

In practical scenarios, students often calculate $K_{sp}$ to determine which of two solids has lower solubility, as it reflects how readily a substance can dissolve in a solution.

Dissociation Reactions

Dissociation reactions are equations that show how a compound breaks down into its constituent ions in a solution. These reactions play a key role in understanding the solubility and behavior of ionic compounds in water.
Each ionic solid has a characteristic way of dissociating.

• The balanced chemical equation for dissociation lays the foundation for forming the $K_{sp}$ expression.
• Dissociation is crucial for calculating the molar solubility, the amount of a substance that can dissolve to form a saturated solution.

By understanding dissociation, students can more accurately compute how much of a solid will actually dissolve.
For instance, the dissociation of ${\mathrm{FeC}}_2{\mathrm{O}}_4$ into ${\mathrm{Fe}}^{2+}$ and ${\mathrm{C}}_2{\mathrm{O}}_4^{2–}$ ions, and using the respective $K_{sp}$, allows one to determine its solubility accurately.

Chemical Equilibrium

Chemical equilibrium is a state where the concentrations of reactants and products remain constant over time. In the context of solubility and dissociation, it describes the balance between the dissolved ions and the undissolved solid in a saturated solution.
The concept of equilibrium is key in calculating and understanding $K_{sp}$.

• Equilibrium is reached when the rate of dissolution equals the rate of precipitation.
• At equilibrium, the system's $K_{sp}$ expression captures a precise relationship between ionic concentrations.

Understanding chemical equilibrium helps students predict and manipulate how changes in conditions affect solubility.
This knowledge is essential when determining which compound in a pair has a smaller molar solubility as it considers all dynamic interactions of ions within a solution.