If the K_spof calcium oxalate monohydrate at $42 ^ { \circ } \mathrm { C }$ is 6.45×10^-8, what is its solubility in water at the same temperature?

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Quizplus · Accepted Answer

The solubility product expression for calcium oxalate monohydrate is: $$ K_{sp} = [\ce{Ca^{2+}}][\ce{C2O4^{2-}}]\cdot[\ce{H2O}] $$ At $42 ^ { \circ } \mathrm { C }$, the ion product is given as: $$ IP = [\ce{Ca^{2+}}][\ce{C2O4^{2-}}]\cdot[\ce{H2O}] = 6.45\times10^{-8} $$ The molar mass of calcium oxalate monohydrate is 146.11 g/mol. Let x be the solubility of calcium oxalate monohydrate in g/100mL, then the concentration of calcium ions and oxalate ions in solution will be x/146.11 M. Since the solubility of calcium oxalate monohydrate dissociates into one calcium ion, one oxalate ion, and one water molecule, the concentration of water molecules in solution will be (100-x)/18 M. The ion product can now be expressed in terms of x: $$ IP = \frac{x}{146.11}\cdot\frac{x}{146.11}\cdot\frac{100-x}{18} = \frac{x^2(100-x)}{3885636}= 6.45\times10^{-8} $$ Solving for x gives a value of 3.71×10^-3 g/100mL, which matches option C.

If the Ksp of Calcium Oxalate Monohydrate At 42∘C42 ^ { \circ } \mathrm { C }42∘C

If the K_spof Calcium Oxalate Monohydrate At $42 ^ { \circ } \mathrm { C }$