Consider a solution that is 1.5×10 2 M in Ba2+and 1.8×10 2 M in Ca2+. K sp(BaSO4

Question

Consider a solution that is 1.5×102 M in Ba2+and 1.8×102 M in Ca2+.
Ksp(BaSO4)=1.07×1010
Ksp(CaSO4)=7.10×105

Part A) If sodium sulfate is used to selectively precipitate one of the cations while leaving the other cation in solution, which cation will precipitate first? ANSWER Ba2+

Part B) What minimum concentration of Na2SO4 is required to cause the precipitation of the cation that precipitates first? ANSWER = 7.1*10-9

Part C) What is the remaining concentration of the cation that precipitates first, when the other cation just begins to precipitate? THIS IS WHAT I NEED HELP WITH

Explanation / Answer

Solution : -

What is the remaining concentration of the cation that precipitates first, when the other cation just begins to precipitate?

When the second cation starts to precipitate then concnetration of the sulfate ion needed is calculated as follows

CaSO4   ----- > Ca^2+ + SO4^2-

                         x             x

                      1.8*10^-2    x

ksp = [Ca^2+][SO4^2-]

7.10*10^-5 = [1.8*10^-2][x]

7.10*10^-5 / 1.8*10^-2 = x

0.00394 M = x

Now using this concentration of the SO4^2- lets calculate the concnetration of the Ba^2+ that is present when the CaSO4 starts to precipitate

BaSO4 ------- > Ba^2+    + SO4^2-

                     x                 0.00394 M

ksp = [Ba^2+ ] [SO4^2-]

1.07*10^-10 = [x] [0.00394]

1.07*10^-10 / 0.00394 = x

2.71*10^-8 M= [x]

So the concnetration of the first cation is 2.71*10^-8 when the second cation begins to precipitate.

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