1.12 g of CaCl₂ is added to 4.80 L of a 5.00 x 10^-6 M solution of Na₂CO₃.

Will a precipitate of CaCO₃ (s) form?

Data: K_sp [CaCO₃ (s)] = 2.8 x 10^-9

Question

1.12 g of CaCl₂ is added to 4.80 L of a 5.00 x 10^-6 M solution of Na₂CO₃.

Will a precipitate of CaCO₃ (s) form?

Data: K_sp [CaCO₃ (s)] = 2.8 x 10^-9

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Accepted Answer

Dissolution reaction: CaCO₃ (s) $⇌$ Ca²⁺ (aq) + CO₃^2- (aq) [K_sp = 2.8 x 10^-9]

CaCO₃ (s) forms if Q_sp > K_sp

Let’s determine Q_sp:

Q_sp = [Ca²⁺]₀ [CO₃^2-]₀

CaCl₂ (s) → Ca²⁺ (aq) + 2 Cl^- (aq)

From the stoichiometry of this reaction: n_Ca2+,0 = n_CaCl2,0

[Ca²⁺]₀ = $\frac{n_{Ca 2 +, 0}}{V}$ = $\frac{n_{CaCl 2, 0}}{V}$ = $\frac{m_{CaCl 2, 0}}{M_{CaCl 2} \times V}$

[Ca²⁺]₀ = $\frac{1.12}{4.80 \times 110.98}$

[Ca²⁺]₀ = 2.10 x 10^-3 M

Na₂CO₃ (s) → 2 Na⁺ (aq) + CO₃^2- (aq)

From the stoichiometry of this reaction: n_CO32-,0 = n_Na2CO3,0

⇒ [CO₃^2-]₀ = [Na₂CO₃]₀ = 5.00 x 10^-6 M

Q_sp = [Ca²⁺]₀ [CO₃^2-]₀

Q_sp = 2.10 x 10^-3 x 5.00 x 10^-6

Q_sp = 1.05 x 10^-8 > K_sp

⇒ CaCO₃ (s) forms

Midterm 2 - Question 4 | General Chemistry 3

General Chemistry 3 - Midterm 2