If 20.00mL of a solution that contains oxalate ion requires 10.00mL of 0.020M MnO4- solution in a titration, what was the molarity of the oxalate solution?
5C2O42- + 2MnO4- + 12H+ --> 2Mn2+ + 10CO2 + 8 H2O
Your equation isn't balanced. I think the 12 H^+ should be 16H^+. (H doesn't balance now and neither is the charge on the left the same as the charge on the right).
moles MnO4^- = M x L.
moles C2O4^-2 = moles MnO4^- x (5/2)
M of C2O4^- = moles/0.020
To find the molarity of the oxalate solution, we can use the stoichiometry of the balanced chemical equation and the volume and concentration of the MnO4- solution used in the titration.
First, let's determine the number of moles of MnO4- used in the titration. We know that the volume of the MnO4- solution used is 10.00 mL (0.01000 L) and the concentration of the MnO4- solution is 0.020 M. Multiplying these values together, we find:
moles of MnO4- used = 0.01000 L x 0.020 M = 0.00020 moles
According to the balanced equation, the stoichiometric ratio between MnO4- and oxalate ion (C2O42-) is 2:5. Therefore, the number of moles of oxalate ion in the solution can be determined using the following relationship:
moles of C2O42- = (moles of MnO4- used) x (5 moles of C2O42-/2 moles of MnO4-)
moles of C2O42- = 0.00020 moles x (5/2) = 0.00050 moles
Next, we can calculate the molarity (M) of the oxalate solution using the number of moles of oxalate ion and the volume of the solution used in the titration. We know that the volume of the oxalate solution used is 20.00 mL (0.02000 L). Dividing the number of moles of oxalate ion by the volume of the solution, we get:
Molarity of C2O42- = (moles of C2O42-) / (volume of oxalate solution)
Molarity of C2O42- = 0.00050 moles / 0.02000 L = 0.025 M
Therefore, the molarity of the oxalate solution is 0.025 M.