A 1.00-L solution saturated at 25°C with calcium oxalate (CaC2O4) contains 0.0061 g of CaC2O4. The solubility constant (Ksp) for this salt is [CaC2O4=128.09 g/mol]
To find the solubility constant (Ksp) for calcium oxalate (CaC2O4), you need to use the given information about the saturation and concentration of the solution.
The solubility of a compound can be defined as the maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature. In this case, the problem states that the solution is saturated, which means that it contains the maximum amount of calcium oxalate that can dissolve at 25°C.
The solubility constant (Ksp) is a measure of how well a compound dissolves in a solvent. It represents the product of the concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient.
The chemical equation for the dissolution of calcium oxalate (CaC2O4) is:
CaC2O4(s) ⇌ Ca2+(aq) + C2O42-(aq)
First, we need to calculate the molar solubility of calcium oxalate, which is the concentration in moles per liter (mol/L). The molar mass of CaC2O4 is 128.09 g/mol, and the given mass of CaC2O4 is 0.0061 g. We can use these values to calculate the moles of CaC2O4:
moles of CaC2O4 = mass of CaC2O4 / molar mass of CaC2O4
= 0.0061 g / 128.09 g/mol
Next, we can calculate the molar solubility:
molar solubility = moles of CaC2O4 / volume of solution in liters
= moles of CaC2O4 / 1.00 L
Now, we have the concentration of Ca2+ and C2O42- ions in the saturated solution, which is equal to the molar solubility. Thus, we can substitute the concentration into the expression for Ksp:
Ksp = [Ca2+][C2O42-]
Remember that the concentration of Ca2+ and C2O42- ions in the saturated solution are equal, so we can use the molar solubility as the concentration for both ions. Therefore:
Ksp = (concentration of Ca2+)(concentration of C2O42-) = (molar solubility)^2
Substitute the calculated molar solubility into the expression for Ksp, and solve for Ksp.