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

(C) Ca (OH) 2 will not precipitate because Q >K sp.

(D) Ca (OH) 2 will not precipitate because Q

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

Explanation:

Here we have first a chemical reaction in which Ca (OH) ₂ is produced:

CaC₂ (s) + H₂O ⇒ Ca (OH) ₂ + C₂H₂

Ca (OH) ₂ is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product constant for Ca (OH) ₂ is:

Ca (OH) ₂ (s) ⇆ Ca²⁺ (aq) + 2OH⁻ (aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca (OH) ₂ = mol CaC₂ (reacted = 0.064 g / 64 g/mol = 0.001 mol Ca (OH) ₂

the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M = 0.002 M (From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q = [Ca²⁺][OH⁻]² = 0.001 x (0.002) ² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is option D