I have no idea how to solve this problem, please help me out.... thanks
a) A liter of a solution saturated at 25degree with calcium oxalate, CaC2O4, is evaporated to dryness, giving a 0.0061gm residue of CaC2O4. calculate the concentrations of the ions, and molar solubility and the solubility product constant for this salt at 25 degree C.
b)answer the above question, using a solution of 0.15M CaCl2 as the solvent instead of pure water.
So you have 0.0061 g CaC2O4 per liter of solution.
How many mols is that?
Isn't that also the molarity since it is in 1 L?
Now write the solubility equation.
CaC2O4 ==> Ca^+2 + C2O4^=
Ksp = (Ca^+)(C2O4^=) .
Plug in the values for the molarity and calculate Ksp. Post you work if you get stuck.
For part b, same thing EXCEPT you have more Ca^+ than in the saturated solution of water alone.
I know you're in Alpert's class
To solve part a), we need to calculate the concentration of ions and the molar solubility and solubility product constant for calcium oxalate (CaC2O4) at 25 degrees Celsius.
1. Start by converting the mass of the CaC2O4 residue into moles. The molar mass of CaC2O4 is 128 g/mol, so:
Moles of CaC2O4 = 0.0061 g / 128 g/mol = 4.77 x 10^-5 mol
2. Since 1 liter of the solution contains the same amount of moles, the concentration of CaC2O4 is also 4.77 x 10^-5 M.
3. Write the solubility equation for CaC2O4:
CaC2O4(s) ⇌ Ca^2+(aq) + C2O4^2-(aq)
4. The solubility product constant (Ksp) is given by the multiplication of the concentrations of Ca^2+ and C2O4^2- ions at equilibrium, raised to their stoichiometric coefficients:
Ksp = [Ca^2+][C2O4^2-]
5. Plug in the calculated concentration of CaC2O4 (4.77 x 10^-5 M) into the Ksp expression:
Ksp = (4.77 x 10^-5 M)(4.77 x 10^-5 M) = 2.28 x 10^-9
So, the molar solubility of CaC2O4 is 4.77 x 10^-5 M, and the solubility product constant (Ksp) for CaC2O4 at 25 degrees Celsius is 2.28 x 10^-9.
Now, for part b), we need to consider the presence of a solvent of 0.15 M CaCl2 instead of pure water.
1. The solubility equation remains the same:
CaC2O4(s) ⇌ Ca^2+(aq) + C2O4^2-(aq)
2. However, the concentration of Ca^2+ ions in the solution is different. Since there is 0.15 M CaCl2, the concentration of Ca^2+ ions is 0.15 M.
3. Plug in the concentration of Ca^2+ ions (0.15 M) into the Ksp expression:
Ksp = (0.15 M)(4.77 x 10^-5 M) = 7.15 x 10^-6
Thus, in part b), the solubility product constant (Ksp) for CaC2O4 at 25 degrees Celsius using a solution of 0.15 M CaCl2 as the solvent is 7.15 x 10^-6.
Remember to always double-check your calculations and units to ensure accuracy.