Answer:
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
