a 500 ml portion of .0020 M Na2C2O4 solution is able to dissolve .47g of MgC2O4. What is the Ksp for MgC2O4
The question is a little confusing by not telling us that the MgC2O4 "just" dissolves. My point is that perhaps the 500 mL of the Na2C2O4 could have dissolved more than 0.47 g MgC2O4. At any rate,
MgC2O4 ==> Mg^+ + C2O4^-2
Ksp = (Mg^+2)(C2O4^-2) = ??
You know (Mg^+2) = 0.47/molar mass MgC2O4 and that divided by 0.5 L.
You know C2O4 is the sum of the C2O4 from MgC2O4 and the C2O4 from Na2C2O4. Add those together to obtain the total C2O4, an solve for Ksp. I obtain an answer very close to my text book value for Ksp for MgC2O4.
To find the Ksp (solubility product constant) for MgC2O4, we can use the given information to calculate the concentrations of Mg2+ and C2O4^2- ions in the solution.
First, let's convert the mass of MgC2O4 dissolved to moles:
molar mass of MgC2O4 = (24.31 g/mol + 12.01 g/mol + (4 * 16.00 g/mol)) = 148.33 g/mol
moles of MgC2O4 dissolved = 0.47 g / 148.33 g/mol
Next, let's calculate the concentration of C2O4^2- ions in the solution:
volume of solution = 500 ml = 500/1000 L = 0.5 L
concentration of C2O4^2- ions = moles of C2O4^2- / volume of solution
Since MgC2O4 dissociates to produce one mole of C2O4^2- ions:
concentration of C2O4^2- ions = moles of MgC2O4 dissolved / volume of solution
Now, let's calculate the concentration of Mg2+ ions in the solution:
Since the stoichiometry of the dissociation of MgC2O4 is 1:1 for Mg2+ and C2O4^2-, the concentration of Mg2+ ions is equal to the concentration of C2O4^2- ions.
Now we can write the solubility product expression for MgC2O4:
Ksp = [Mg2+][C2O4^2-]
Since the concentrations of Mg2+ and C2O4^2- ions are equal, we can write:
Ksp = [Mg2+]^2
Substituting the value we obtained for the concentration:
Ksp = ([Mg2+])^2 = ([C2O4^2-])^2 = (moles of MgC2O4 dissolved / volume of solution)^2
Finally, substitute the given values in the equation:
Ksp = (0.47 g / 148.33 g/mol / 0.5 L)^2
Calculating it, we get:
Ksp = (0.003171 mol/L)^2 ≈ 1.004 × 10^-5
Therefore, the solubility product constant (Ksp) for MgC2O4 is approximately 1.004 × 10^-5.
To determine the Ksp (solubility product constant) for MgC2O4, we need to use the given information about the amount of solute dissolved in the solution.
First, we need to convert the given mass of MgC2O4 (0.47 g) into moles. This can be done using the molar mass of MgC2O4, which is 148.32 g/mol.
Number of moles of MgC2O4 = mass / molar mass = 0.47 g / 148.32 g/mol = 0.00317 mol
Next, we can use the stoichiometry of the balanced chemical equation for the dissolution of MgC2O4 to determine the number of moles of Mg2+ ions. The balanced equation is:
MgC2O4(s) -> Mg2+(aq) + 2C2O42-(aq)
Since the stoichiometric coefficient in front of Mg2+ is 1, the number of moles of Mg2+ ions will be the same as the number of moles of MgC2O4, which is 0.00317 mol.
Now, we can use the given volume of the Na2C2O4 solution (500 mL or 0.5 L) and its concentration (0.0020 M) to calculate the initial concentration of the Mg2+ ions.
Initial concentration of Mg2+ ions = 0.00317 mol / 0.5 L = 0.00634 M
Since the balanced chemical equation shows that 1 mole of Mg2+ ions is produced for every mole of MgC2O4, the initial concentration of Mg2+ ions is equal to the solubility of MgC2O4.
Therefore, the Ksp for MgC2O4 can be calculated as follows:
Ksp = [Mg2+][C2O42-]^2
Ksp = (0.00634 M)(0.00634 M)^2
Ksp = 2.55 x 10^-7
Therefore, the Ksp for MgC2O4 is 2.55 x 10^-7.