If a solution is saturated in silver phosphate (Ksp = 1.8 x 10-18), what are the concentration of each aqueous ion and the solubility of the silver phosphate? If sodium phosphate is added to the solution such that it would be 0.001 M in Na3PO4, what would be the concentration of each aqueous ion in the solution?
I used Ksp = (xs)^x (ys)^y
I got s= 8.25*10^-4
I'm not really sure how to find the concentration from this, so I can't do the last part.
Solubility Question?
2016-08-05 5:25 am
回答 (2)
2016-08-05 6:48 am
Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq) .... Ksp = 1.8 × 10⁻¹⁸
At equilibrium :
Let [PO₄³⁻] = s M
Then, [Ag⁺] = 3s M
Ksp = [Ag⁺]³ [PO₄³⁻]
s (3s)³ = 1.8 × 10⁻¹⁸
s = ∜[(1.8 × 10⁻¹⁸) / 27] = 1.61 × 10⁻⁵
Hence,
[Ag⁺] = 3 × (1.61 × 10⁻⁵) = 4.83 × 10⁻⁵ M
[PO₄³⁻] = 1.61 × 10⁻⁵ M
====
Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq) .... Ksp = 1.8 × 10⁻¹⁸
Due to the fact that the solubility of Ag₃PO₄ is low and common ion effect, in 0.001 M Na₃PO₄ :
[PO₄³⁻] from the dissociation of Na₃PO₄ ≫ [PO₄³⁻] from the dissociation of Ag₃PO₄
Hence, [PO₄³⁻] in the solution ≈ 0.001 M
Ksp = [Ag⁺]³ [PO₄³⁻]
[Ag⁺]³ × 0.001 = 1.8 × 10⁻¹⁸
[Ag⁺] = ∛[(1.8 × 10⁻¹⁸) / 0.001] = 1.22 × 10⁻⁵ M
At equilibrium :
Let [PO₄³⁻] = s M
Then, [Ag⁺] = 3s M
Ksp = [Ag⁺]³ [PO₄³⁻]
s (3s)³ = 1.8 × 10⁻¹⁸
s = ∜[(1.8 × 10⁻¹⁸) / 27] = 1.61 × 10⁻⁵
Hence,
[Ag⁺] = 3 × (1.61 × 10⁻⁵) = 4.83 × 10⁻⁵ M
[PO₄³⁻] = 1.61 × 10⁻⁵ M
====
Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq) .... Ksp = 1.8 × 10⁻¹⁸
Due to the fact that the solubility of Ag₃PO₄ is low and common ion effect, in 0.001 M Na₃PO₄ :
[PO₄³⁻] from the dissociation of Na₃PO₄ ≫ [PO₄³⁻] from the dissociation of Ag₃PO₄
Hence, [PO₄³⁻] in the solution ≈ 0.001 M
Ksp = [Ag⁺]³ [PO₄³⁻]
[Ag⁺]³ × 0.001 = 1.8 × 10⁻¹⁸
[Ag⁺] = ∛[(1.8 × 10⁻¹⁸) / 0.001] = 1.22 × 10⁻⁵ M
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