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 × 10−5 Then

A

will get precipitate

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B

No precipitate is formed

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C

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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) + SO2−4(aq) and the solubility product expression is Ksp = [Ca2+][SO2−4] = 2.4 × 10−5 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 × 10−3 M and the concentration of SO2−4 ion is: [SO2−4] = (400 ml Na2SO4 solution600 ml total volume) × (0.15 M) = 0.1 M The ion product is [Ca2+][SO2−4] = (8.33 × 10−3)(1.0 × 10−1) = 8.33 × 10−4 Which is larger than Ksp, so CaSO4 should precipitate from the solution.

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