Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g sample of CaC2 (s) (molar mass 64 g/mol) is used instead and all of it reacts, which of the following will occur and why? (the value of Ksp for Ca(OH)2 is 8.0 x 10-8)(A) Ca(OH)2 will precipitate because Q >K sp.(B) Ca(OH)2will precipitate because Q K sp.(D) Ca(OH)2 will not precipitate because Q

Ca(OH)2 will not precipitate because Q<Ksp

Explanation:

Ksp for Ca(OH)2 has already been stated in the question as 8.0 x 10-8mol2dm-6

The value of the reaction quotient depends heavily on the concentration of the reactants. As the initial concentration of the calcium carbide decreases considerably, the reaction quotient decreases until Q<Ksp hence the Ca(OH)2 will not precipitate from solution.

The reaction equation is:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

From

Ca(OH)2= Ca2+ + 2OH-

Concentration of solution= 0.064×1/64= 1×10-3

Since [Ca2+] = 1×10-3

[OH-]= (2×10-3)^2= 4×10^-6

Hence Q= 4×10^-9

This is less than the Ksp hence the answer.

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