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23 June, 15:53

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

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  1. 23 June, 16:18
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    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
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