Only 6.1 mg of calcium oxalate, Ca(C2O4) dissolves in 1.0 L of water at 25 Celsius. What is the solubility product of calcium oxalate?

Hi UOIT Students googling answers for chem!

To determine the solubility product of calcium oxalate, Ca(C2O4), we need to use the given information about its solubility in water.

The solubility product, also known as the solubility product constant (Ksp), is a mathematical expression that describes the equilibrium between a solid compound and its dissolved ions in a solution. For calcium oxalate, it can be represented by the following equilibrium reaction:

Ca(C2O4) (s) ⇌ Ca2+ (aq) + 2C2O42- (aq)

The Ksp expression for this equilibrium is written as:

Ksp = [Ca2+][C2O42-]^2

Where [Ca2+] represents the molar concentration of calcium ions and [C2O42-] represents the molar concentration of oxalate ions.

To find the solubility product (Ksp), we first need to determine the molar concentrations of calcium and oxalate ions in the solution using the given solubility information.

Given that 6.1 mg of calcium oxalate (Ca(C2O4)) dissolves in 1.0 L of water, we need to convert this mass into moles. The molar mass of calcium oxalate is calculated as:

Ca(C2O4) = 40.08 g/mol (calcium) + 2(12.01 g/mol) + 4(16.00 g/mol) = 128.08 g/mol

Converting the mass of calcium oxalate to moles:

6.1 mg * (1 g / 1000 mg) * (1 mol / 128.08 g) = 4.77 x 10^-5 mol

Since the reaction stoichiometry tells us that for every 1 mole of Ca(C2O4) that dissolves, 1 mole of Ca2+ ions and 2 moles of C2O42- ions are produced, the molar concentrations of the ions are as follows:

[Ca2+] = 4.77 x 10^-5 mol / 1 L = 4.77 x 10^-5 M
[C2O42-] = 2(4.77 x 10^-5 mol) / 1 L = 9.54 x 10^-5 M

Now that we have the molar concentrations of calcium and oxalate ions, we can substitute them into the Ksp expression to calculate the solubility product:

Ksp = [Ca2+][C2O42-]^2 = (4.77 x 10^-5 M)(9.54 x 10^-5 M)^2 = 4.33 x 10^-13

Therefore, the solubility product of calcium oxalate (Ca(C2O4)) is 4.33 x 10^-13.