Problem 32 - Chemistry

Question

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties. a. \(P b I_{2}, K_{s p}=1.4 \times 10^{-8}\) b. \(\operatorname{CdCO}_{3}, K_{s p}=5.2 \times 10^{-12}\) c. \(\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{s p}=1 \times 10^{-31}\)

Step-by-Step Solution

Answer

The solubility of the compounds are as follows: a. \(PbI_{2}\): \(7.1 \times 10^{-4}\) M b. \(CdCO_{3}\): \(2.3 \times 10^{-6}\) M c. \(Sr_3(PO_4)_2\): \(8.9 \times 10^{-11}\) M

Step 1: Write the dissolution equation for each compound

. For each of the three compounds, we will write a balanced chemical equation representing its dissolution in water. a. \(PbI_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)\) b. \(CdCO_{3}(s) \rightleftharpoons Cd^{2+}(aq) + CO_{3}^{2-}(aq)\) c. \(Sr_3(PO_4)_2(s) \rightleftharpoons 3Sr^{2+}(aq) + 2PO_{4}^{3-}(aq)\)

Step 2: Express the solubility in terms of variables

: For each compound, we will express the concentration of ions in terms of the solubility (S). a. \([Pb^{2+}] = S\) and \([I^-] = 2S\) b. \([Cd^{2+}] = S\) and \([CO_{3}^{2-}] = S\) c. \([Sr^{2+}] = 3S\) and \([PO_{4}^{3-}] = 2S\)

Step 3: Write the \(K_{sp}\) expression for each compound

: Using the solubility variables from Step 2, write the \(K_{sp}\) expressions for each of the compounds: a. \(K_{sp} = [Pb^{2+}][I^-]^2 = S(2S)^2\) b. \(K_{sp} = [Cd^{2+}][CO_{3}^{2-}] = S^2\) c. \(K_{sp} = [Sr^{2+}]^3[PO_{4}^{3-}]^2 = (3S)^3(2S)^2\)

Step 4: Solve for solubility (S) in each case using the given \(K_{sp}\) values

: Using the given \(K_{sp}\) values, solve for S: a. \(1.4 \times 10^{-8} = S(2S)^2 \Rightarrow S = 7.1 \times 10^{-4}\,\text{M}\) b. \(5.2 \times 10^{-12} = S^2 \Rightarrow S = 2.3 \times 10^{-6}\,\text{M}\) c. \(1 \times 10^{-31} = (3S)^3(2S)^2 \Rightarrow S = 8.9 \times 10^{-11}\,\text{M}\) Now we have the solubility of each compound in moles per liter: a. The solubility of \(PbI_{2}\) is \(7.1 \times 10^{-4}\) M. b. The solubility of \(CdCO_{3}\) is \(2.3 \times 10^{-6}\) M. c. The solubility of \(Sr_3(PO_4)_2\) is \(8.9 \times 10^{-11}\) M.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of the reactants and products remain constant, but not necessarily equal. In the context of dissolution reactions, equilibrium is established when the ionic compound dissolves in water, reaching a balance between the dissolved ions and the undissolved solid.
This equilibrium can be represented with a balanced chemical equation, showing the formation of ions from the solid compound, for example:

• For lead (II) iodide (\(PbI_2(s)\)) it dissociates into lead ions and iodide ions:\(PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)\).
• Equilibrium in dissolution indicates not all solid dissolves entirely but rather, some remains as solid while others are in ion form.
• The equilibrium position tells us how much solute dissolves, known as solubility.

Understanding chemical equilibrium concepts helps to predict the extent to which ionic compounds dissolve in water.

Ionic Compounds

Ionic compounds are made up of positive (called cations) and negative ions (called anions) held together by strong electrostatic forces. In solid form, these ions arrange in a fixed, repeating pattern, forming a crystal lattice.
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Integer felis turpis, tincidunt nec purus nec, lacinia scelerisque felis.Ionic bonds are strong contributing to characteristics like:

• High melting and boiling points.
• Brittleness.
• Electrical conductivity when molten or in solution.

When dissolved in water, these compounds dissociate into their component ions. This process enables the ionic compounds to conduct electricity. For applying solubility product constants, it’s necessary to know the stoichiometry of dissociation. This means recognizing how many ions form when one formula unit dissolves.For example:

• Cadmium carbonate (\(CdCO_3\)) breaks into cadmium ions and carbonate ions as:\(CdCO_3(s) \rightleftharpoons Cd^{2+}(aq) + CO_3^{2-}(aq)\).
• The stoichiometry is essential when calculating solubility using \(K_{sp}\).

This knowledge of ionic behavior allows for accurate predictions of solubility in solutions.

Dissolution Reactions

Dissolution reactions are processes where ionic compounds dissolve in a solvent, typically water, to form ions. These reactions are vital for understanding the solubility and behavior of substances in solution.
The dissolution of an ionic compound can be represented by an equation showing the solid turning into its ions.Key aspects to consider in a dissolution reaction:

• The extent of dissolution is quantified by its solubility, influencing how many ions form relative to the solid remaining.
• The solubility product constant (\(K_{sp}\)) is a measure of the solubility of an ionic compound, representing the product of the concentrations of its ions raised to their stoichiometric coefficients in equilibrium conditions.
• Finding solubility involves solving equilibrium expressions derived from the \(K_{sp}\) equation.

For example, in the dissolution of \(Sr_3(PO_4)_2\):

• The reaction is \(Sr_3(PO_4)_2(s) \rightleftharpoons 3Sr^{2+}(aq) + 2PO_4^{3-}(aq)\).
• The \K_{sp} expression is \((3S)^3(2S)^2\).
• This shows that for each mole of solute dissolving, 3 moles of \(Sr^{2+}\) and 2 moles of \(PO_4^{3-}\) ions form.

These concepts are foundational to solving dissolution problems and predicting how much of a compound will actually dissolve in a given solution.