Problem 36 - Chemistry An Atoms First Approach

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. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\mathrm{sp}}=8.1 \times 10^{-12},\) or \(\mathrm{Mn}(\mathrm{OH})_{2}\) \(K_{\mathrm{sp}}=2 \times 10^{-13}\)

Step-by-Step Solution

Answer

For the given pairs of solids, FeC2O4 has a smaller molar solubility than Cu(IO₄)₂, and Mn(OH)₂ has a smaller molar solubility than Ag₂CO₃. Therefore, the solids with the smallest molar solubilities are FeC2O4 and Mn(OH)₂.

Step 1: Balanced chemical equation for FeC2O4

$$\mathrm{FeC}_{2}\mathrm{O}_{4(s)} \rightleftharpoons \mathrm{Fe^{2+}_{(aq)}} + 2 \mathrm{C}_{2}\mathrm{O}_{4^{2-}_{(aq)}}$$ Step 2: Set up the Ksp expression

Step 2: Ksp expression for FeC2O4

Let the molar solubility of FeC2O4 be 's' Then, we have Ksp = [Fe²⁺][C₂O₄²⁻]² = (s)(2s)² Step 3: Calculate molar solubility 's'

Step 3: Calculate molar solubility of FeC2O4

Given, Ksp = 2.1 x 10⁻⁷ (2.1 x 10⁻⁷) = s*(2s)² Solve for 's' which is approximately 4.91 x 10⁻³ M 2) Cu(IO₄)₂ Step 1: Write the balanced chemical equation for the dissolution

Step 4: Balanced chemical equation for Cu(IO₄)₂

$$\mathrm{Cu}(\mathrm{IO}_{4})_{ 2(s)} \rightleftharpoons \mathrm{Cu^{2+}_{(aq)}} + 2 \mathrm{IO}_{4^{-}_{(aq)}}$$ Step 2: Set up the Ksp expression

Step 5: Ksp expression for Cu(IO₄)₂

Let the molar solubility of Cu(IO₄)₂ be 's' Then, we have Ksp = [Cu²⁺][IO₄⁻]² = (s)(2s)² Step 3: Calculate molar solubility 's'

Step 6: Calculate molar solubility of Cu(IO₄)₂

Given, Ksp = 1.4 x 10⁻⁷ (1.4 x 10⁻⁷) = s*(2s)² Solve for 's' which is approximately 5.04 x 10⁻³ M

Step 7: Compare the molar solubility of FeC2O4 and Cu(IO₄)₂

Molar solubility of FeC2O4: 4.91 x 10⁻³ M Molar solubility of Cu(IO₄)₂: 5.04 x 10⁻³ M FeC2O4 has a smaller molar solubility than Cu(IO₄)₂. b. We have two solids: 1) Ag₂CO₃ Step 1: Write the balanced chemical equation for the dissolution

Step 8: Balanced chemical equation for Ag₂CO₃

$$\mathrm{Ag}_{2}\mathrm{CO}_{3(s)} \rightleftharpoons 2 \mathrm{Ag^{+}_{(aq)}} + \mathrm{CO}_{3^{2-}_{(aq)}}$$ Step 2: Set up the Ksp expression

Step 9: Ksp expression for Ag₂CO₃

Let the molar solubility of Ag₂CO₃ be 's' Then, we have Ksp = [Ag⁺]²[CO₃²⁻] = (2s)²(s) Step 3: Calculate molar solubility 's'

Step 10: Calculate molar solubility of Ag₂CO₃

Given, Ksp = 8.1 x 10⁻¹² (8.1 x 10⁻¹²) = (2s)²(s) Solve for 's' which is approximately 9.45 x 10⁻⁵ M 2) Mn(OH)₂ Step 1: Write the balanced chemical equation for the dissolution

Step 11: Balanced chemical equation for Mn(OH)₂

$$\mathrm{Mn}(\mathrm{OH})_{2(s)} \rightleftharpoons \mathrm{Mn^{2+}_{(aq)}} + 2 \mathrm{OH^{-}_{(aq)}}$$ Step 2: Set up the Ksp expression

Step 12: Ksp expression for Mn(OH)₂

Let the molar solubility of Mn(OH)₂ be 's' Then, we have Ksp = [Mn²⁺][OH⁻]² = (s)(2s)² Step 3: Calculate molar solubility 's'

Step 13: Calculate molar solubility of Mn(OH)₂

Given, Ksp = 2 x 10⁻¹³ (2 x 10⁻¹³) = s*(2s)² Solve for 's' which is approximately 6.48 x 10⁻⁵ M

Step 14: Compare the molar solubility of Ag₂CO₃ and Mn(OH)₂

Molar solubility of Ag₂CO₃: 9.45 x 10⁻⁵ M Molar solubility of Mn(OH)₂: 6.48 x 10⁻⁵ M Mn(OH)₂ has a smaller molar solubility than Ag₂CO₃. So the solids with smaller molar solubilities are FeC2O4 and Mn(OH)₂.

Key Concepts

Ksp (Solubility Product Constant)

The solubility product constant, or Ksp, is a crucial concept in chemistry that measures the solubility of a sparingly soluble ionic compound. It is a specific type of equilibrium constant that applies to the dissolution of salts in water.

When a solid dissolves in a solvent, it dissociates into its constituent ions. The Ksp value represents the level at which the compound is in equilibrium with its ions in a saturated solution.

For a salt like \( ext{FeC}_2 ext{O}_4\), which dissociates into \( ext{Fe}^{2+}\) and \(2 ext{C}_2 ext{O}_4^{2-}\), we can write the equilibrium expression based on its ion concentrations:
\[K_{ ext{sp}} = [ ext{Fe}^{2+}][ ext{C}_2 ext{O}_4^{2-}]^2 \]

In this expression, \(s\) represents the molar solubility of the compound. If the Ksp value is known, it can be used to calculate \(s\), helping to determine how much of the compound will dissolve in water. Generally, a lower Ksp value indicates lower solubility.

Chemical Equilibrium

Chemical equilibrium is a state reached in a chemical reaction when the rates of the forward and reverse reactions are equal. At this point, the concentrations of reactants and products remain constant over time.

In the context of dissolution reactions, equilibrium describes the balance between the undissolved solid and its ions in solution. For example, for the dissolution of \( ext{Cu(IO}_4)_2\), the reaction can be shown as:
\( ext{Cu}( ext{IO}_4)_{2(s)} ightleftharpoons ext{Cu}^{2+}_{(aq)} + 2 ext{IO}_4^{-}_{(aq)}\)

At equilibrium, the ions in solution are in dynamic balance with the solid solute. The quantities involved do not change, but the reaction continuously occurs in both directions.

This equilibrium state is crucial when calculating solubility, as it ensures consistent concentrations of solute and solvent under given conditions. Understanding this concept allows chemists to predict how changes in conditions, such as temperature or pressure, will affect solubility and, subsequently, the Ksp.

Dissolution Equations

Dissolution equations describe how a solid substance dissolves in a solvent to form an aqueous solution of its ions. They involve writing balanced chemical equations that depict the breakup of a crystalline structure into ions.

These equations are essential for understanding solubility and for calculating Ksp expressions. For instance, when considering the dissolution of \( ext{Mn(OH)}_2\), we can write:
\( ext{Mn(OH)}_{2(s)} ightleftharpoons ext{Mn}^{2+}_{(aq)} + 2 ext{OH}^{-}_{(aq)}\)

Such equations help illustrate how the solid dissociates in water and assist in forming the basis for setting up Ksp expressions.

When writing dissolution equations, keep in mind:

• Balance the chemical equation correctly, ensuring equal numbers of each atom appear on both sides.
• Identify the stoichiometry, i.e., the ratio of ions formed from the solute.
• Use dissolution equations to assess relative solubility through the analysis of Ksp expressions.

This approach provides the foundational understanding necessary to explore concepts such as solubility, molar solubility, and chemical equilibrium.