Question

A 725 mL sample of a saturated aqueous solution of calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4},\) at \(95^{\circ} \mathrm{C}\) is cooled to \(13^{\circ} \mathrm{C}\). How many milligrams of calcium oxalate will precipitate? For \(\mathrm{CaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=1.2 \times 10^{-8}\) at \(95^{\circ} \mathrm{C}\) and \(2.7 \times 10^{-9}\) at \(13^{\circ} \mathrm{C}\).

Short answer

To find the short answer, follow the calculations set out in the steps. The final result will be the amount in milligrams of calcium oxalate that precipitates when cooled to \(13^{\circ} C\).

Step-by-step solution

Calculate the molar solubility at \(95^{\circ} C\)

The solubility product constant, \(K_{sp}\), at \(95^{\circ} C\) is given as \(1.2 \times 10^{-8}\). Since the equation for calcium oxalate, \(CaC_{2}O_{4}\), can be written as \(CaC_{2}O_{4} ⇌ Ca^{2+} + C_{2}O_{4}^{2-}\), we know that the \(K_{sp}\) expression is \(K_{sp} = [Ca^{2+}][C_{2}O_{4}^{2-}]\). Because calcium oxalate is a 1:1 salt, the molar solubility at \(95^{\circ} C\) can be represented by \(s\) and so \(K_{sp}\) can be expressed as \(s^{2}\). Solving for \(s\), we get \(s = \sqrt{K_{sp}} = \sqrt{1.2 \times 10^{-8}}\).

Calculate the molar solubility at \(13^{\circ} C\)

We repeat the step for temperature \(13^{\circ} C\) where \(K_{sp}\) is given as \(2.7 \times 10^{-9}\). Thus, \(s = \sqrt{K_{sp}} = \sqrt{2.7 \times 10^{-9}}\).

Calculate the difference in molar solubility

The difference in molar solubility will give us the amount of \(CaC_{2}O_{4}\) that precipitates out when cooled to \(13^{\circ} C\). It's calculated as: \(\Delta s = s_{95} - s_{13}\).

Convert molar solubility to mass

Calculate the milligrams of \(CaC_{2}O_{4}\) that precipitates out by converting the molar solubility difference (\(\Delta s\)) into grams using its molar weight (one mole of \(CaC_{2}O_{4}\) has a weight of \(128.1g\)) and then into milligrams. Multiply this with the volume of the solution in litres to get the total milligrams that precipitate.

Key concepts

Understanding Calcium Oxalate

Calcium oxalate is a chemical compound with the formula \(\mathrm{CaC}_{2}\mathrm{O}_{4}\). It is often found in the form of a monohydrate and is known for its low solubility in water, which makes solubility a critical aspect.
This compound is commonly seen in both biological systems and industrial applications. For instance, calcium oxalate crystals are found in plant tissues and can form kidney stones in humans. These crystals form due to its low solubility and can cause issues depending on their concentration in bodily fluids.

• Formula: \(\mathrm{CaC}_{2}\mathrm{O}_{4}\)
• Molar Weight: Approximately 128.1 g/mol
• Low solubility in water, impacting its precipitation and solubility balance

Exploring Precipitation

Precipitation is a process where a substance in solution becomes a solid, forming crystals that settle out of the solution.
It happens commonly due to a change in conditions such as temperature or the concentration of ions. Understanding this process involves knowing that when the solubility product \(K_{sp}\) is surpassed, solids start to form.
To recognize a situation of precipitation, we look at the equilibrium of a solute. Taking our case of calcium oxalate, the equation can be written as:

• $\backslash (\backslash mathrm\{CaC\}\_\{2\}\backslash mathrm\{O\}\_\{4\}\; \backslash ;\backslash rightleftharpoons\backslash ;\; \backslash mathrm\{Ca\}^\{2+\}\; \backslash ;\; +\; \backslash ;\; \backslash mathrm\{C\}\_\{2\}\backslash mathrm\{O\}\_\{4\}^\{2-\}\backslash )$

In solution equilibrium, the product of the concentrations \([\mathrm{Ca}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]\) must equal \(K_{sp}\). If it exceeds this value due to cooling or other reasons, solid calcium oxalate will precipitate, lowering each ion's concentration until equilibrium is restored.

Temperature Effect on Solubility

Temperature significantly affects the solubility of many compounds, including calcium oxalate. Generally, solubility increases with temperature.
However, our compound presents a slight complexity—calcium oxalate becomes less soluble with a drop in temperature, as seen from the different values of \(K_{sp}\) at 95°C and 13°C.

• At \(95^{\circ}\mathrm{C},\;K_{sp} = 1.2 \times 10^{-8}\)
• At \(13^{\circ}\mathrm{C},\; K_{sp} = 2.7 \times 10^{-9}\)

This decrease in solubility with cooling means more calcium oxalate will precipitate out of the solution when cooled.
These ions reassociate into solid form, reducing solubility and thus increasing precipitation. Recognizing this helps in predicting how much of a compound will precipitate as the temperature changes.