a liter of a solution saturated at 25C with CaC2O4 is evaporated to dryness giving a 0.0061 gm residue of CaC2O4. Calculate the concentrations of the ions, and the molar solubility and the solubility product constant for this salt at 25C. Show the balanced reaction.
i get 4.76 x 10^-5 mol CaC2o4.. squaring the answer gives me a ksp= 2 x 10^-9
is this right?
Yes and no. 2 x 10^-9 is the right start. Why did you throw the other numbers away? It's hard to tell exactly how many significant figures you have (that could be 0.00610g and you just didn't type the last zero) but the 0.0061 give you at least two so you should have at least two in the answer. My calculator shows 2.265E-9. You should round that number to the correct number of s.f.
This could be a very complicated problem; I will assume it is to considered a simple one by ignoring the ionization of H2C2O4.
0.0061 g CaC2O4 = how many mols?
0.p0061/molar mass CaC2O4 = approximately ? = (Ca^2+) = (CO4^2-)
CaC2O4 ==> Ca^2+ + C2O4^2-
Ksp = (Ca^2+)(C2O4^2-) = ?
Substitute the molar solubilities of the ions into the Ksp expression and solve for Ksp.
To calculate the concentrations of the ions, molar solubility, and solubility product constant for CaC2O4, we can start by writing the balanced equation for the dissociation of CaC2O4 in water:
CaC2O4 (s) ⇌ Ca^2+ (aq) + C2O4^2- (aq)
From the equation, we can see that 1 mole of CaC2O4 produces 1 mole of Ca^2+ and 1 mole of C2O4^2-. Therefore, the concentration of Ca^2+ and C2O4^2- ions will be equal to the molar solubility of CaC2O4.
Given that the mass of the residue is 0.0061 g and the molar mass of CaC2O4 is 128 g/mol (40 g/mol for Ca + 12 g/mol for C + 2*16 g/mol for O), we can calculate the number of moles of CaC2O4:
Moles of CaC2O4 = mass / molar mass
= 0.0061 g / 128 g/mol
= 0.0000477 mol
Since the solution was initially 1 liter (or 1000 mL), the molar solubility (mol/L) is equal to the number of moles divided by the volume in liters:
Molar solubility of CaC2O4 = Moles of CaC2O4 / Volume of solution
= 0.0000477 mol / 1 L
= 0.0000477 M
Therefore, the concentration of Ca^2+ and C2O4^2- ions in the solution is 0.0000477 M.
To calculate the solubility product constant (Ksp), we need to assume that the solution is saturated, meaning that the concentrations of Ca^2+ and C2O4^2- ions are in equilibrium with the solid CaC2O4.
The solubility product constant (Ksp) is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients in the balanced equation:
Ksp = [Ca^2+] * [C2O4^2-]
Substituting the molar solubility of CaC2O4 for both [Ca^2+] and [C2O4^2-], we get:
Ksp = (0.0000477 M) * (0.0000477 M)
= 2.27 x 10^-9
Therefore, the solubility product constant for CaC2O4 at 25°C is 2.27 x 10^-9.
In summary:
- The concentrations of Ca^2+ and C2O4^2- ions in the solution are both 0.0000477 M.
- The molar solubility of CaC2O4 is 0.0000477 M.
- The solubility product constant (Ksp) for CaC2O4 at 25°C is 2.27 x 10^-9.