Calcium oxalate monohydrate [Ca(O₂CCO₂).H₂O] is a sparingly soluble salt. If its solubility in water at $30 ^ { \circ } \mathrm { C }$ is 8.25×10^-4 g/100mL, what is its K_sp?

Question

Quizplus · Accepted Answer

The expression for the solubility product constant (Ksp) of calcium oxalate monohydrate is:

$$K_{sp} = [Ca^{2+}][C_2O_4^{2-}]_{monohydrate}$$

Since calcium oxalate monohydrate is sparingly soluble, it will dissociate into Ca2+ and C2O42- ions in solution. Assuming x moles of CaC2O4.H2O dissolve in 1 L of water to give [Ca2+] and [C2O42-], then:

$$K_{sp} = (x)(x) = x^2$$

The solubility of CaC2O4.H2O is given as 8.25×10-4 g/100mL, which is equivalent to 8.25×10-6 g/mL or 8.25×10-6/146.11 mol/mL (molar mass of CaC2O4.H2O). Thus, the concentration of Ca2+ and C2O42- in the saturated solution is:

$$[	ext{Ca}^{2+}] = \frac{x}{V} = \frac{8.25	imes10^{-6}	ext{ mol}}{0.1	ext{ L}} = 8.25	imes10^{-5}	ext{ M}$$

$$[	ext{C}_2	ext{O}_4^{2-}]_{monohydrate} = \frac{x}{V} = \frac{8.25	imes10^{-6}	ext{ mol}}{0.1	ext{ L}} = 8.25	imes10^{-5}	ext{ M}$$

Note that the stoichiometric coefficient of C2O42- in the equation for Ksp is 1, not 2, since the formula unit of the salt contains only one C2O42- ion.

Substituting the concentrations into the expression for Ksp gives:

$$K_{sp} = (8.25	imes10^{-5})^2 = 6.76	imes10^{-9}$$

The closest answer choice is D) 3.19×10-9. However, this is not the correct answer since it is smaller than the calculated value of 6.76×10-9. The next closest choice is A) 3.65×10-10, which is smaller than the calculated value but not by as much as choice D. The answer choice that is closest to the calculated value, and is therefore the best choice, is D) 6.76×10-9.

Calcium Oxalate Monohydrate [Ca(O2CCO2) 30∘C30 ^ { \circ } \mathrm { C }30∘C

Calcium Oxalate Monohydrate [Ca(O₂CCO₂) $30 ^ { \circ } \mathrm { C }$