Problem 110 - Chemistry

Question

Calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) is relatively insoluble in water \(\left(K_{\mathrm{sp}}=2 \times 10^{-9}\right) .\) However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in 0.10\(M \mathrm{H}^{+}\) than in pure water? In pure water, ignore the basic properties of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}.\)

Step-by-Step Solution

Answer

The solubility of calcium oxalate in pure water is \(1.41 \times 10^{-5} M\), calculated using the given \(K_{sp}\) value. The solubility in an acidic solution would be higher, as the oxalate ion reacts with the H+ ions. However, without more information about the solubility constant in acidic conditions (\(K_{solubility}\)), it is not possible to determine the exact difference in solubilities.

Step 1: Find the solubility of calcium oxalate in water

In water, the solubility equilibrium for calcium oxalate can be represented as: \[CaC_{2}O_{4}(s) \rightleftharpoons Ca^{2+}(aq) + C_{2}O_{4}^{2-}(aq)\] Let the solubility of calcium oxalate be represented by 's'. At equilibrium: \[K_{sp} = [Ca^{2+}][C_{2}O_{4}^{2-}] = s * s = s^2\] Given that, \(K_{sp}=2 \times 10^{-9}\) Now, solving for s: \[s = \sqrt{2 \times 10^{-9}}\] \[s = 1.41 \times 10^{-5}\: M\] The solubility of calcium oxalate in pure water is 1.41 x 10^-5 M.

Step 2: Determine the solubility of calcium oxalate in a 0.10 M H+ solution

In an acidic solution, the oxalate ion will react with the H+ ions: \[H^+ + C_{2}O_{4}^{2-} \rightleftharpoons HC_{2}O_{4}^-\] In 0.10 M H+ solution, the concentration of H+ ions will be much greater than the concentration of C2O4^2- ions, and the reaction will shift towards the products, which will increase the solubility of calcium oxalate. Now, let the solubility of calcium oxalate in the acidic solution be represented by 's_acid'. The concentration of H+ ions will be 0.10 M. At equilibrium: \[ s_{acid}[\mathrm{H^+}] = K_{solubility}\] \[ K_{solubility} = \frac{s_{acid} * 0.10\: M}{s}\] \[ s_{acid} = \frac{K_{solubility} * s}{0.10\: M}\]

Step 3: Calculate the difference in solubilities

To find how much more soluble calcium oxalate is in the acidic solution compared to pure water, subtract the solubility in water from the solubility in the acidic solution: \[Difference = s_{acid} - s = \frac{K_{solubility} * s}{0.10\: M} - s\] However, we are not given the K_solubility value. In this case, we can't solve the problem further without that information or any additional information.

Key Concepts

Calcium Oxalate

Calcium oxalate is a chemical compound consisting of calcium, carbon, and oxygen. It appears as a salt that is commonly found in plants like spinach and rhubarb, often appearing as needle-shaped crystals. Its chemical formula is \(\mathrm{CaC}_{2} \mathrm{O}_{4}\). This compound is infamous for its low solubility in water, which makes it less available for biological processes.
One of the reasons calcium oxalate is relatively insoluble is due to the ion pairs formed between \(\mathrm{Ca}^{2+}\) ions and \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) ions in solution, which settle out as a solid. This property has implications in both organic and inorganic chemistry, and plays a relevant role in various natural processes. Notably, calcium oxalate is responsible for certain types of kidney stones in humans, as excess calcium combines with oxalate in the kidneys.

Ksp (Solubility Product Constant)

In chemistry, the solubility of a compound is often described by its solubility product constant, or \(K_{sp}\). The \(K_{sp}\) is a measure of the extent to which a compound will dissolve in water. Specifically, \(K_{sp}\) indicates the concentrations of the ionic species when a chemical equilibrium is reached in a saturated solution.
For calcium oxalate, the solubility equilibrium can be represented by the equation:

• \[CaC_{2}O_{4}(s) \rightleftharpoons Ca^{2+}(aq) + C_{2}O_{4}^{2-}(aq)\]

The solubility product constant formula is \(K_{sp} = [Ca^{2+}][C_{2}O_{4}^{2-}]\).
If we consider calcium oxalate, with a \(K_{sp}\) of \(2 \times 10^{-9}\), we know it has low solubility in water. The smaller the \(K_{sp}\) value, the less soluble the compound is. Hence, calcium oxalate forms few ions in solution before reaching its saturation point.

Acidic Solution Chemistry

Acidic solutions can significantly affect the solubility of compounds, including calcium oxalate. In the presence of an acid, such as \(\mathrm{H}^+\), certain anions like oxalate can undergo protonation.
The reaction that illustrates this is:

• \[H^+ + C_{2}O_{4}^{2-} \rightleftharpoons HC_{2}O_{4}^-\]

This process consumes \(C_{2}O_{4}^{2-}\) ions, which drives the dissolution of more calcium oxalate because the equilibrium shifts to counteract the loss of \(C_{2}O_{4}^{2-}\) ions. Essentially, the presence of \(\mathrm{H}^+\) ions in an acidic solution increases the solubility of calcium oxalate by removing \(C_{2}O_{4}^{2-}\) ions from the equilibrium.
Consequently, the solubility of calcium oxalate in an acidic solution is higher compared to water, as the substance dissolves to alleviate the shift in chemical equilibrium caused by protonation.