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50.00 mL of a saturated solution of calcium oxalate is titrated with 0.0015 M potassium...

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macbeth110 | eNotes Newbie

Posted April 22, 2013 at 8:38 PM via web

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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?

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted April 23, 2013 at 1:47 AM (Answer #1)

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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

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