.Calcium oxalate, CaC2O4 (128.1 g mol-1
), has a solubility of 6.7 x 10–4 g L–1
. What is its
Ksp?
The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble compound in water. It can be calculated using the concentration of the ions present in the solution.
The balanced equation for the dissolution of calcium oxalate is as follows:
CaC2O4(s) ⇌ Ca2+(aq) + C2O42-(aq)
From the equation, we can see that one mole of calcium oxalate dissolves to produce one mole of Ca2+ ions and one mole of C2O42- ions. Therefore, the concentration of Ca2+ and C2O42- ions in the solution will be equal to the solubility of calcium oxalate, which is given as 6.7 x 10^-4 g/L.
To calculate the concentration of Ca2+ and C2O42- ions in moles per liter (M), we need to convert the solubility from grams per liter (g/L) to moles per liter (mol/L).
The molar mass of calcium oxalate (CaC2O4) is 128.1 g/mol. Therefore, the solubility of calcium oxalate in moles per liter (mol/L) is:
solubility = 6.7 x 10^-4 g/L
molar mass = 128.1 g/mol
solubility (in mol/L) = (6.7 x 10^-4 g/L) / (128.1 g/mol)
solubility (in mol/L) = 5.225 x 10^-6 mol/L
Since the concentration of Ca2+ and C2O42- ions in the solution is the same as the solubility of calcium oxalate, the concentration of Ca2+ and C2O42- ions is 5.225 x 10^-6 mol/L.
The solubility product constant (Ksp) is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients. In this case, the stoichiometric coefficient of Ca2+ and C2O42- ions is 1.
Ksp = [Ca2+] * [C2O42-] = (5.225 x 10^-6 mol/L) * (5.225 x 10^-6 mol/L)
Ksp = 2.728 x 10^-11 mol^2/L^2
Therefore, the Ksp of calcium oxalate is 2.728 x 10^-11 mol^2/L^2.