Question

Sodium phosphate is added to a solution that contains 0.0094 M aluminum nitrate and 0.025 M calcium chloride. The concentration of the first ion to precipitate (either Al^3+ or Ca^2+) decreases as its precipitate forms. What is the concentration of this ion when the second ion begins to precipitate? Sulfur dioxide reacts with chlorine at 227 degree C: SO_2(g) +Cl_2(g) SO_2Cl_2(g) K_p for this reaction is 5.1 times 10^-2 atm-1. Initially, 1.00 g each of SO_2 and Cl_2 are placed in a 1.00 L reaction vessel. After 15 minutes, the concentration of SO_2Cl_2 is 45.5 pg/mL. You will determine if the system has reached equilibrium. First, what is K_c (in L/mol)? (A mu g is 10-6 g.)

Answer 1

14)

The two reaction that are happening here are

2Na₃PO₄ + 3CaCl₂ $\rightarrow$ 6NaCl + Ca₃(PO₄)₂

Na₃PO₄ + Al(NO₃)₃ $\rightarrow$ AlPO₄ + 3NaNO₃

Aluminum phosphate
AlPO₄(s) ⇄ Al³⁺(aq) + PO₄³⁻(aq)

Ksp = [Al³⁺]∙[PO₄³⁻]
with Ksp = 9.84×10⁻²¹

Calcium phosphate
Ca₃(PO₄)₂(s) ⇄ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)

Ksp = [Ca²⁺]³∙[PO₄³⁻]²
with Ksp = 2.07×10⁻³³
source: http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf

When you solve the equilibrium equation of each salt for [PO₄³⁻] you get the level of phosphate concentration at which the salt starts to precipitate:
- aluminum phosphate
[PO₄³⁻] = Ksp / [Al³⁺] = 9.84×10⁻²¹ / 0.0094 = 1.04×10⁻¹⁸ M
- calcium phosphate
[PO₄³⁻] = √(Ksp / [Ca²⁺]²) = √(2.07×10⁻³³ / 0.025³) = 1.15×10⁻¹⁴ M

Because [PO₄³⁻] is smaller for Al³⁺, aluminum phosphate precipitates first.

when calcium phosphate starts to precipitate the aluminum ion concentration is
[Al³⁺] = Ksp / [PO₄³⁻] = 9.84×10⁻²¹ / 1.15×10⁻¹⁴ = 8.55×10⁻⁷ M