Question

A solution contains 0.0350 M Ca2 and 0.0100 M Sr2 .What is the concentration of the first cation when the second starts to precipitate?

Answer #1

If the solution is precipitated by sulfate, then we need to
calculate [SO_{4}^{2-}]

For CaSO_{4}, K_{sp} = 2.4 x 10^{-5}

For SrSO_{4}, K_{sp} = 3.44 x
10^{-7}

Now, SrSO_{4} will precipitate when K_{sp} >
3.44 x 10^{-7}

Therefore, in a 0.0100 M Sr^{2+} (aq) solution
SrSO_{4} precipitates when

[SO_{4}^{2-}] > 3.44 x 10^{-7} /
0.0100
(as K_{sp} = [Sr^{2+}]
[SO_{4}^{2-}])

or [SO_{4}^{2-}] > 3.44 x 10^{-5}
M

Now, solubility product for Ca^{2+} at this point = 3.44
x 10^{-5} x 0.0350 = **0.12 x
10 ^{-5}**

However, K_{sp} for CaSO_{4} = 2.4 x
10^{-5}

thus solubility product for CaSO_{4} is exceeded, so the
**SrSO _{4} precipitates first**

At this point, [Ca^{2+}] = 2.4 x 10^{-5}/3.44 x
10^{-5} M

[Ca^{2+}] = **0.67 M**

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