Answer:
The answer to the question is;
The first ion to precipitate out is the Al³⁺ ion and the concentration of the Al³⁺ ion when the Ca²⁺ ion begins to precipitate is 1.12 × 10⁻⁵ M.
Explanation:
To solve the question, we note that
aluminum nitrate, Al(NO₃)₃ will dissociate as follows
Al(NO₃)₃ → Al³⁺ (aq) + 3NO₃⁻ (aq)
Therefore when sodium phosphate is added to a solution that contains aluminum nitrate we have the following system of aluminium phosphate which is
AlPO₄(s) ⇄ Al³⁺(aq) + PO₄³⁻(aq)
The solubility product for the above reaction is
Ksp = [Al³⁺][PO₄³⁻] = 9.84×10⁻²¹
The solubility product for calcium phosphate is expressed as
Ca₃(PO₄)₂(s) ⇄ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)
With Ksp = [Ca²⁺]³[PO₄³⁻]² = 2.07×10⁻³³
From the solubility product, we can find the concentration of [PO₄³⁻] at which precipitation starts as follows
The phosphate concretion for Al³⁺ when precipitation starts is
[PO₄³⁻] = ![\frac{K_{sp}}{[Al^{3+}]}](https://tex.z-dn.net/?f=%5Cfrac%7BK_%7Bsp%7D%7D%7B%5BAl%5E%7B3%2B%7D%5D%7D)
= 9.84×10⁻²¹ / 0.007 = 1.406×10⁻¹⁸ M
The phosphate concretion for Ca²⁺ when precipitation starts is
[PO₄³⁻] =![\sqrt{\frac{K_{sp}}{[Ca^{2+}]^2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7BK_%7Bsp%7D%7D%7B%5BCa%5E%7B2%2B%7D%5D%5E2%7D%7D)
= ![\sqrt{\frac{2.07\times10^{-33}}{[0.052]^2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2.07%5Ctimes10%5E%7B-33%7D%7D%7B%5B0.052%5D%5E2%7D%7D)
= 8.75×10⁻¹⁶ M
(Aluminium phosphate precipitates out first)
The reaction favors the precipitation of the aluminum phosphate first due to the lower concentration of the [PO₄³⁻] ions in the [Al³⁺][PO₄³⁻] system which is lower than the relative [PO₄³⁻] in the [Ca²⁺]³[PO₄³⁻]².
Therefore, the more sodium phosphate added serves to precipitate the remaining aluminium phosphate.
The process continues and the concentration of Al³⁺ decreases as more precipitates form. The process continues until the equilibrium conditions satisfies the precipitation threshold level for the calcium phosphate system concentration whereby the concentration of the Al³⁺ in the solution is given by.
[Al³⁺] = ![\frac{K_{sp}}{[PO_4^{3-}]} = \frac{9.84\times 10^{-21}}{8.75\times 10^{-16}}](https://tex.z-dn.net/?f=%5Cfrac%7BK_%7Bsp%7D%7D%7B%5BPO_4%5E%7B3-%7D%5D%7D%20%3D%20%5Cfrac%7B9.84%5Ctimes%2010%5E%7B-21%7D%7D%7B8.75%5Ctimes%2010%5E%7B-16%7D%7D)
= 1.12 × 10⁻⁵ M
Therefore the concentration of this aluminium ion when the calcium ion begins to precipitate = 1.12 × 10⁻⁵ M.
