Question

Consider the following solubility data for calcium oxalate \(\left(\mathrm{Ca} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) : $$ \begin{aligned} &K_{s p} \text { at } 25^{\circ} \mathrm{C}=4 \times 10^{-9} \\ &K_{s p} \text { at } 95^{\circ} \mathrm{C}=1 \times 10^{-8} \end{aligned} $$ Five hundred mL of a saturated solution is 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

Answer: In order to find the amount of CaC2O4 that will precipitate when cooled down to 25°C, we need to follow the steps which involve calculating molar solubility at both temperatures, finding the amount dissolved at each temperature, and then determining the difference between the two. Once we find the difference in moles, we can convert it to milligrams by using the molar mass of CaC2O4.

Step-by-step solution

Calculate molar solubility at 95°C

First, we need to calculate the molar solubility of CaC2O4 at 95°C. The solubility product constant, Ksp, at this temperature, is 1 × 10⁻⁸. Since the balanced dissolution reaction of CaC2O4 is: $$ \mathrm{CaC}_{2} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(aq) $$ Let the molar solubility of CaC2O4 at 95°C be S mol/L. Then, the Ksp expression becomes: $$ K_{s p}=[\mathrm{Ca}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]=S^{2} $$ Now, solve for S: $$ S=\sqrt{1 \times 10^{-8}} $$

Calculate the amount of CaC2O4 dissolved at 95°C

Using the molar solubility calculated in step 1, we can now find the amount of CaC2O4 that is dissolved in 500 mL of saturated solution at 95°C: $$ \text { Amount dissolved at } 95^{\circ} \mathrm{C}=\text { Molar solubility }\times \text { Volume of solution } $$ $$ =S \times 0.5\,\text{L} $$

Calculate molar solubility at 25°C

Next, we need to calculate the molar solubility of CaC2O4 at 25°C, using the given value of Ksp (4 × 10⁻⁹) in the same manner as in Step 1: $$ S=\sqrt{4 \times 10^{-9}} $$

Calculate the amount of CaC2O4 dissolved at 25°C

Using the molar solubility calculated in step 3, we can now find the amount of CaC2O4 that will be dissolved in 500 mL of solution at 25°C: $$ \text { Amount dissolved at } 25^{\circ} \mathrm{C}=\text { Molar solubility }\times \text { Volume of solution } $$ $$ =S \times 0.5\,\text{L} $$

Calculate the amount of CaC2O4 precipitated

To find the amount of CaC2O4 that will precipitate when the solution is cooled from 95°C to 25°C, subtract the amount dissolved at 25°C from the amount dissolved at 95°C: $$ \text { Amount precipitated }=\text { Amount dissolved at } 95^{\circ} \mathrm{C}-\text { Amount dissolved at } 25^{\circ} \mathrm{C} $$

Convert the amount precipitated to milligrams

The amount precipitated calculated in step 5 is in moles. We need to convert it to milligrams. To do this, multiply the amount precipitated by the molar mass of CaC2O4 and then by 1000 to convert grams to milligrams: $$ \text { Amount precipitated in mg }=\text { Amount precipitated in moles }\times \text { Molar mass of CaC }_{2} \mathrm{O}_{4}\times 1000 $$ Finally, the amount precipitated in milligrams is the answer we are looking for.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, or Ksp, is a vital concept in understanding how compounds dissolve in water. It represents the maximum amount of a solute that can dissolve in a solution at a given temperature before it becomes saturated. For a compound like calcium oxalate, represented by the dissolution equation:
\[\mathrm{CaC}_{2} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(aq)\]
Ksp is determined by the concentrations of the ions at equilibrium:
\[K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] = S^2\]Where \(S\) represents the molar solubility. The smaller the Ksp, the less soluble the compound is. In this problem, you can see that the Ksp changes with temperature, which affects how much calcium oxalate can dissolve.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve per liter of solution to reach saturation. This is directly related to Ksp. If you know the Ksp of a compound, you can calculate its molar solubility by solving for \(S\) in the expression:
\[S = \sqrt{K_{sp}}\]At 95°C, the Ksp of calcium oxalate is \(1 \times 10^{-8}\), leading to a specific molar solubility. Contrastingly, at 25°C, the Ksp is \(4 \times 10^{-9}\), resulting in a lower molar solubility. The difference in these solubilities helps determine how much of the compound will precipitate if the temperature changes.

Temperature Effect on Solubility

Temperature often plays a crucial role in the solubility of compounds. For calcium oxalate, as shown in this exercise, a higher temperature (95°C) results in a higher Ksp and greater solubility, while cooler temperatures (25°C) lead to lower solubility. This phenomenon is common as increased molecular movement at higher temperatures usually allows more solute to dissolve. In this exercise, understanding the change in solubility with temperature helps us calculate how much of the compound precipitates out when the solution is cooled from 95°C to 25°C. The precipitation can be calculated by finding the difference in the amount dissolved at the two temperatures. This is an essential consideration for practical applications, such as designing processes that require temperature control to manage solubility.