Question

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) is \(4.8 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{BiI}_{3}\) is \(1.32 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\).

Short answer

The calculated \(K_{sp}\) values for each solid are: a. For \(\mathrm{CaC}_{2} \mathrm{O}_{4}\), the \(K_{sp}\) value is \(4.42 \times 10^{-13}\). b. For \(\mathrm{BiI}_{3}\), the \(K_{sp}\) value is \(6.50 \times 10^{-28}\).

Step-by-step solution

(Step 1: Write the balanced dissociation equation for each solid)

For \(\mathrm{CaC}_2\mathrm{O}_4\), the balanced equation is: \[ \mathrm{CaC}_2\mathrm{O}_4 (s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{C}\mathrm{O}^{2-}_4(aq) \] For \(\mathrm{BiI}_3\), the balanced equation is: \[ \mathrm{BiI}_3 (s) \rightleftharpoons \mathrm{Bi}^{3+}(aq) + 3\mathrm{I}^-(aq) \]

(Step 2: Determine the concentrations of ions using the solubility of the solids)

For \(\mathrm{CaC}_2\mathrm{O}_4\), we are given a solubility of \(4.8 \times 10^{-5}\ \mathrm{mol/L}\): \[ \begin{aligned} \mathrm{[Ca}^{2+}] &= 4.8 \times 10^{-5} \mathrm{M}\\ \mathrm{[C}\mathrm{O}^{2-}_4] &= 2 \times \mathrm{solubility} = 2 \times (4.8 \times 10^{-5}) \ \mathrm{M} = 9.6 \times 10^{-5} \mathrm{M} \end{aligned} \] For \(\mathrm{BiI}_3\), we are given a solubility of \(1.32 \times 10^{-5}\ \mathrm{mol/L}\): \[ \begin{aligned} \mathrm{[Bi}^{3+}] &= 1.32 \times 10^{-5} \mathrm{M}\\ \mathrm{[I}^-] &= 3 \times \mathrm{solubility} = 3 \times (1.32 \times 10^{-5}) \ \mathrm{M} = 3.96 \times 10^{-5} \mathrm{M} \end{aligned} \]

(Step 3: Calculate the \(K_{\mathrm{sp}}\) value for each solid)

For \(\mathrm{CaC}_2\mathrm{O}_4\), using the ion concentrations in the equilibrium expression for \(K_{\mathrm{sp}}\): \[ K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}][(\mathrm{C}\mathrm{O}^{2-}_4)]^2 = (4.8 \times 10^{-5})(9.6 \times 10^{-5})^2 = 4.42 \times 10^{-13} \] For \(\mathrm{BiI}_3\), using the ion concentrations in the equilibrium expression for \(K_{\mathrm{sp}}\): \[ K_{\mathrm{sp}} = [\mathrm{Bi}^{3+}][\mathrm{I}^-]^3 = (1.32 \times 10^{-5})(3.96 \times 10^{-5})^3 = 6.50 \times 10^{-28} \] The calculated \(K_{\mathrm{sp}}\) values for each solid are: a. For \(\mathrm{CaC}_{2} \mathrm{O}_{4}\), the \(K_{\mathrm{sp}}\) value is \(4.42 \times 10^{-13}\). b. For \(\mathrm{BiI}_{3}\), the \(K_{\mathrm{sp}}\) value is \(6.50 \times 10^{-28}\).

Key concepts

Ksp Calculation

The solubility product constant (\textbf{Ksp}) is a crucial concept in understanding solubility at a chemical level. It quantifies the maximum amount of a substance that can dissolve in a solution before reaching a saturation point, where no more of the substance can be dissolved.

Calculating \textbf{Ksp} involves determining the concentrations of the ions formed when a solid dissolves and reaches equilibrium in a solution. Each ion concentration is raised to the power of its coefficient in the balanced chemical equation for the dissolution process. These values are then multiplied together to give the \textbf{Ksp} value. For example, for calcium oxalate (\textbf{CaC}\(_2\)\textbf{O}\(_4\)), the concentration of \textbf{Ca}\(^{2+}\) ions is the solubility of the compound itself, while the concentration of \textbf{C}\(_2\)\textbf{O}\(_4^{2-}\) ions is twice the solubility due to the stoichiometry of the dissociation reaction:
\textbf{Ksp} = [Ca\(^{2+}\)][C\(^{2-}_{2}\)O\(_4\)]\(^2\). This calculation allows us to predict the behavior of sparingly soluble salts, crucial for applications in biochemistry, pharmacology, and industrial processes.

Chemical Equilibrium

Chemical equilibrium represents a state in which the forward and reverse reactions occur at the same rate, resulting in constant concentrations of reactants and products over time. In the context of solubility, once a solid starts dissolving in a solution, it will continue to do so until the \textbf{Ksp} is reached.At this point, any additional solid added will not dissolve because the product of the ion concentrations remains constant. Equilibrium can be represented with a balanced equation displaying the solid reactant on one side and its corresponding ions on the other, separated by a double arrow to represent the dynamic nature of the equilibrium state. For \textbf{BiI}\(_3\), equilibrium is shown as \textbf{BiI}\(_3\)\textbf{(s) \(\rightleftharpoons\) Bi}\(^{3+}\)\textbf{(aq) + 3I}\(^-\)(aq). Understanding this balance between solubility and precipitation is essential for predicting the outcome of reactions in various chemical systems.

Ion Concentration

Ion concentration refers to the amount of ions present in a given volume of solution, typically expressed in molarity (moles per liter, M). It's a fundamental concept when discussing solubility, as changes in ion concentration can shift a reaction's equilibrium position.In the examples provided, the solubility of the compounds directly relates to the concentration of ions produced. For instance, knowing the solubility of \textbf{CaC}\(_2\)\textbf{O}\(_4\) allows us to calculate not only the molarity of \textbf{Ca}\(^{2+}\) ions but also deduce that of \textbf{C}\(_2\)\textbf{O}\(_4^{2-}\) ions as being twice that due to the stoichiometry. Ion concentration plays a crucial role in the environment, biology, and chemistry fields; it helps determine pH levels, biochemical pathways, and titration outcomes.

Solubility of Compounds

Solubility of compounds is expressed as the maximum amount of a substance that can dissolve in a solvent at a given temperature and forms a significant basis for various scientific fields, including medicine, environmental science, and engineering.

The solubility can be affected by different factors, such as temperature, pressure, and the presence of other substances. For sparingly soluble salts like \textbf{CaC}\(_2\)\textbf{O}\(_4\) and \textbf{BiI}\(_3\), their low solubility means that these compounds dissolve to only a small extent before reaching equilibrium. The ability to predict a compound's solubility is essential for formulating medications, treating water, and synthesizing new materials. Furthermore, understanding the solubility principles aids in controlling crystal growth, purifying substances, and developing analytical methods for detecting various ions in solutions.