Question

Consider the following solubility data for calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right):\) $$ \begin{array}{l} K_{\mathrm{sp}} \text { at } 25^{\circ} \mathrm{C}=4 \times 10^{-9} \\ K_{\mathrm{sp}} \text { at } 95^{\circ} \mathrm{C}=1 \times 10^{-8} \end{array} $$ Five hundred \(\mathrm{mL}\) of a saturated solution are prepared at \(95^{\circ} \mathrm{C}\). How many milligrams of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) will precipitate when the solution is cooled to \(25^{\circ} \mathrm{C}\) ? (Assume that supersaturation does not take place.)

Short answer

When a 0.500 L solution of calcium oxalate is cooled from 95°C to 25°C, 0.74 milligrams of calcium oxalate will precipitate.

Step-by-step solution

Determine the ion concentrations at 𝟗𝟓°𝐂

In the saturated solution at \(95^{\circ} \mathrm{C}\), the solubility product constant (\(K_{sp}\)) is \(1\times10^{-8}\). For the reaction: $$ \mathrm{CaC}_{2} \mathrm{O}_{4} \left(\mathrm{solid}\right) \rightleftharpoons \mathrm{Ca}^{2+} \left(\mathrm{aq}\right) + \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \left(\mathrm{aq}\right) $$ Let the solubility of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) be \(s\) moles/L, then: $$ K_{sp} = [\mathrm{Ca}^{2+}] [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] = s^2 $$ Solve for \(s\): $$ s = \sqrt{K_{sp}} = \sqrt{1\times10^{-8}} = 3.16\times10^{-5}\; \text{moles/L} $$

Determine the ion concentrations at 25°𝐂

When the solution cools to \(25^{\circ} \mathrm{C}\), \(K_{sp}\) is \(4\times10^{-9}\). Using the same method as in Step 1, let the solubility be \(s^{\prime}\) moles/L, then the ion concentrations at \(25^{\circ} \mathrm{C}\) would be: $$ s^{\prime} = \sqrt{4\times10^{-9}}={2}\times10^{-5}\; \text{moles/L} $$

Determine the amount of precipitated calcium oxalate

Calculate the amount of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) that precipitates by finding the difference between the initial concentration (at \(95^{\circ}\mathrm{C}\)) and the final concentration (at \(25^{\circ}\mathrm{C}\)): $$ s_\text{precip} = 3.16\times10^{-5} - 2\times10^{-5} = 1.16\times10^{-5}\; \text{moles/L} $$ Multiply by the volume of the solution: $$ \text{moles of }\mathrm{CaC}_{2} \mathrm{O}_{4}\text{ precipitated} = 1.16\times10^{-5}\; \text{moles/L} \times 0.500\; \text{L} = 5.8\times10^{-6}\; \text{moles} $$

Convert moles of precipitated calcium oxalate to milligrams

To convert the amount of precipitated \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) from moles to milligrams, multiply by the molar mass of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) (1 mole of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) = 128 g): $$ \text{mass of }\mathrm{CaC}_{2} \mathrm{O}_{4}\text{ precipitated} = 5.8\times10^{-6}\; \text{moles} \times 128\; \frac{\text{g}}{\text{moles}} \times 1000\; \frac{\text{mg}}{\text{g}} = 0.74\; \text{mg} $$ Thus, 0.74 milligrams of calcium oxalate will precipitate when the solution is cooled from \(95^{\circ}\mathrm{C}\) to \(25^{\circ}\mathrm{C}\).

Key concepts

Ksp Calculations

The Solubility Product Constant, commonly abbreviated as Ksp, is fundamental in predicting the solubility of an ionic compound in water. In simple terms, the Ksp represents the maximum amount of solute that can dissolve in a solution before reaching a point where the solid will begin to precipitate.

Understanding how to perform Ksp calculations is crucial in many chemistry problems. To calculate Ksp, you must first write down the balanced chemical equation for the dissolution of the ionic compound. For instance, calcium oxalate \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) dissociates into calcium ions \(\mathrm{Ca}^{2+}\) and oxalate ions \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) in water. This equilibrium can be expressed as:
\(\mathrm{CaC}_{2} \mathrm{O}_{4} (\mathrm{solid}) \rightleftharpoons \mathrm{Ca}^{2+} (\mathrm{aq}) + \mathrm{C}_{2} \mathrm{O}_{4}^{2-} (\mathrm{aq})\).

For the solubility s in moles per liter, the Ksp expression becomes \(K_{sp} = [\mathrm{Ca}^{2+}] [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] = s^2\). Solving for s gives you the molar solubility of the compound. By undertaking this step, the maximum concentration of ions that can be present in the solution before precipitation is clear, an essential concept for anyone studying chemistry or working with solutions.

Temperature Effect on Solubility

Solubility is frequently dependent on temperature, and understanding this relationship is pivotal when studying or conducting experiments involving solutions. As a general rule, the solubility of solid substances tends to increase with rising temperature. As seen in the case of calcium oxalate, its solubility product constant (Ksp) at \(95^\circ\mathrm{C}\) is higher than that at \(25^\circ\mathrm{C}\), suggesting an increased solubility with temperature.

If we delve deeper, when the temperature increases, the kinetic energy of molecules escalates, enabling more solid to dissolve in the solvent. Conversely, when temperature decreases, the system releases energy, often leading to solid precipitation as solubility diminishes. This knowledge is tremendously useful, particularly in processes such as recrystallization, where temperature manipulation is a tool to purify substances or to induce crystallization of a dissolved solute.

In practical applications, such as in our exercise, the concept is essential for predicting the amount of substance that will precipitate upon cooling. For students and professionals alike, appreciating the temperature effect on solubility is essential for accurately controlling saturation and precipitation in chemical solutions.

Precipitation Calculations

Precipitation calculations form a cornerstone in analytical and environmental chemistry. Determining the amount of a substance that precipitates when conditions change is vital. As showcased in our calcium oxalate example, these calculations begin with understanding the difference in molar solubility at two temperatures and then applying stoichiometry to find the mass of the precipitate.

In the given exercise, we determine the concentration of ionic species at high temperature and then recalculate based on the lower temperature Ksp value. The difference in concentrations illustrates the amount that precipitates. Finally, knowing the volume of solution allows us to calculate the number of moles that precipitated. To complete the process, converting moles to mass involves using the molar mass of the compound, which for calcium oxalate is 128 g/mol, a crucial step for moving from theoretical to practical, measurable quantities.

This procedure is routine in industrial processes involving selective precipitation and in environmental science for understanding mineral deposition in nature. Approaching these calculations methodically allows students to gain proficiency in quantitative analysis and to make accurate predictions regarding the behavior of solutions in varying conditions.