# 50.00 mL of a saturated solution of calcium oxalate is titrated with 0.0015 M potassium permanganate solution. If it takes 3.81 mL of the permanganate solution to titrate the oxalate, what is the...

50.00 mL of a saturated solution of calcium oxalate is titrated with 0.0015 M potassium permanganate solution. If it takes 3.81 mL of the permanganate solution to titrate the oxalate, what is the Ksp for calcium oxalate?

jerichorayel | Certified Educator

In this reaction, there are series of chemical equation to be considered.  First, the equilibrium of calcium oxalate. Second is the acidification of calcium oxalate. And last is the redox titration using potassium permanganate.

`CaC_2O_4 <=> Ca^(2+) + C_2O_4^(2-)`

`H^+ +CaC_2O_4 -> Ca^(2+) + H_2C_2O_4`

`5 H_2C_2O_4 + 2 MnO_4^- + 6H^+ -> 2Mn^(2+) + 10 CO_2 + 8H_2O`

Next thing to do is to get the moles of the oxalate ion from the third equation using the given amount of permanganate solution.

`mol es MnO_4^(-) = 0.0015 * (3.81/1000)L = 5.715x10^-6 mol es`

`5.715x10^-6 mol es MnO_4^(-) *(5 mol es C_2O_4^2-)/(2mol es MnO_4^-)`

`= 1.42875x10^-5 mol es C_2O_4`

To get the ksp, we know from the first equation that:

`CaC_2O_4 <=> Ca^(2+) + C_2O_4^(2-)`

Therefore:

`ksp = [Ca^(2+)][C_2O_4^(2-)]`

`[Ca^(2+)]=[C_2O_4^(2-)]`

`ksp =[C_2O_4^(2-)]^2`

`[C_2O_4^(2-)]= (1.42875x10^-5 mol es C_2O_4)/(0.05381 L)` **

`[C_2O_4^(2-)]= 2.6552x10^-4`

**Total volume = 50.00+ 3.81 = 53.81/1000 = 0.05381 L

`ksp =[C_2O_4^(2-)]^2`

`ksp =[2.6552x10^-4]^2`

ksp = 7.05x10^-8