Question

Solubility product of CaC₂O₄ at a given temperature in pure water is 4 × 10⁻⁹ mol²/L². Solubility of CaC₂O₄ at the same temperature is:

• A. 6.3×10−5mol/L
• B. 2×10−5mol/L
• C. 2×10−4mol/L
• D. 6.3×10−4mol/L

Hint:

The solubility product (Ksp) of CaC₂O₄ helps determine its solubility ().

Answer 1

The Correct Option is A

To solve this problem, we need to determine the solubility of calcium oxalate (\(CaC_2O_4\)) in pure water given its solubility product, \(K_{sp}\), at a certain temperature.

The solubility product expression for \(CaC_2O_4\) in water is given by:

\(K_{sp} = [Ca^{2+}][C_2O_4^{2-}]\) 

Since \(CaC_2O_4\) dissociates as follows:

\(CaC_2O_4(s) \rightleftharpoons Ca^{2+}(aq) + C_2O_4^{2-}(aq)\)

Let \(s\) be the solubility of \(CaC_2O_4\) in mol/L. Then, at equilibrium:

• \([Ca^{2+}] = s\)
• \([C_2O_4^{2-}] = s\)

Therefore, the expression for the solubility product becomes:

\(K_{sp} = s \times s = s^2\)

Given that \(K_{sp} = 4 \times 10^{-9} \, \text{mol}^2/\text{L}^2\), we find:

\(s^2 = 4 \times 10^{-9}\)

Solving for \(s\):

\(s = \sqrt{4 \times 10^{-9}} = 2 \times 10^{-5} \, \text{mol/L}\)

Upon reviewing the options provided, I realized there's a discrepancy with the given correct answer. The calculated solubility is \(2 \times 10^{-5} \, \text{mol/L}\), which matches the second option.

Therefore, the solubility of \(CaC_2O_4\) at this temperature is \(2 \times 10^{-5} \, \text{mol/L}\). The provided correct answer might need to be reviewed, as the calculated result is different.