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Question

A solution prepared by mixing 200 ml of 0.025 M CaCl2 and 400 ml of 0.15 M Na2SO4? Given Ksp of CaSO4 = 2.4 × 105 Then


A

will get precipitate

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B

No precipitate is formed

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C

get ready to precipitate

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D

NaCl will get precipitate

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Solution

The correct option is A

will get precipitate


The equation for the equilibrium is

CaSO4(s) Ca2+(aq) + SO24(aq)

and the solubility product expression is

Ksp = [Ca2+][SO24] = 2.4 × 105

If we assume that the volumes of the solutions that are mixed are additive, the final solution will have a volume of 600 ml. This total volume contains the equivalent of 200 ml of CaCl2, so the concentration of Ca2+ ions is

[Ca2+] = (200 ml CaCl2 solution600 ml total volume) × (0.15 M)

= 8.33 × 103 M

and the concentration of SO24 ion is:

[SO24] = (400 ml Na2SO4 solution600 ml total volume) × (0.15 M)

= 0.1 M

The ion product is

[Ca2+][SO24] = (8.33 × 103)(1.0 × 101)

= 8.33 × 104

Which is larger than Ksp, so CaSO4 should precipitate from the solution.


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