14)
The two reaction that are happening here are
2Na3PO4 + 3CaCl2 6NaCl +
Ca3(PO4)2
Na3PO4 + Al(NO3)3
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
Sodium phosphate is added to a solution that contains 0.0094 M aluminum nitrate and 0.025 M...
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