Problem text:

Solid sodium phosphate is slowly added to $\pu{200 mL}$ of a solution containing $\pu{0.002 mol}$ of aluminum chloride and $\pu{0.001 mol}$ of calcium chloride (assuming no volume change occurs). How many grams of sodium phosphate must be added to precipitate as much of one ion as possible, while the other ion just does not precipitate? What percentage of the ion that will precipitate first is still left in the solution at that time?

My approach:

Two salts will be produced as the reaction goes: $\ce{AlPO4}$ and $\ce{Ca3(PO4)2}$

$K_\mathrm{sp} (\ce{AlPO4})=9.83\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\times 10^{-33}$

Therefore, I can conclude that $\ce{Ca3(PO4)2}$ will precipitate first.

I need to find concentration of $[\ce{PO4^3-}]$ via $$[\ce{Al^3+}][\ce{PO4^3-}]=K_\mathrm{sp}$$ $$[\ce{Al^3+}]=\frac{\pu{0.002 mol}}{\pu{200 mL}}=0.01M$$ $$[\ce{PO4^3-}]=\frac{9.83\times 10^{-21}}{\pu{0.01M}}=9.83\times 10^{-19}$$

Then I can find $m_\ce{Na3PO4}$ via finding number of moles first and multiplying by molar mass:

$$n_\ce{Na3PO4} = [\ce{PO4^3-}]\times V = 9.83\times 10^{-19}\times \pu{200 mL} = 1.966\times 10^{-19}$$ $$m_\ce{Na3PO4} = n_\ce{Na3PO4} \times \mathrm{Molar Mass} = \pu{3.22\times 10^{-17} g}$$

At this point, my result does not make much sense to me because the mass I obtained is extremely small. Unfortunately, I do not have the correct answers for that task, therefore, I would highly appreciate any help that can be useful in spotting mistakes I made in the solution.

  • $\begingroup$ Overall the problem is poorly setup considering the significant figures involved, and very poorly considering the real chemistry. // You calculated the $\ce{PO$^{3-}}$ remaining in solution correctly for the $\ce{Al^{3+}}$. However you need to do the same for $\ce{Ca^{2+}}$ to determine that $\ce{AlPO4}$ will precipitate first. // You calculated the amount of $\ce{PO$^{3-}}$ which will remain in solution, and did not add the amount needed to quantitatively precipitate the $\ce{AlPO4}$ too. $\endgroup$
    – MaxW
    Commented Jun 4, 2020 at 20:16

4 Answers 4


In an answer that is now hidden user Maurice pointed out the complexities of the chemistry. So let's assume:

  • All Al present in solution as $\ce{Al^{3+}}$ and any hydoxy complexes or precipitates will be ignored.
  • All Ca present in solution as $\ce{Ca^{2+}}$ and any hydoxy complexes will be ignored.
  • All $\ce{PO4^{3-}}$ added to solution stays as that ion and doesn't hydrolyze to $\ce{HPO4^{2-}}$
  • Two salts that will be produced are $\ce{AlPO4}$ and $\ce{Ca3(PO4)2}$
  • $K_\mathrm{sp} (\ce{AlPO4})=9.83\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\times 10^{-33}$

Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Al^{3+}}$ to precipitate: $$[\ce{Al^3+}]=\dfrac{\pu{0.002 mol}}{\pu{200 mL}}=0.01M$$ $$[\ce{PO4^3-}]_\ce{Al}=\dfrac{K_\mathrm{sp}}{\ce{[Al^{3+}]}} = \dfrac{9.83\times 10^{-21}}{\pu{0.01M}}=9.83\times 10^{-19}$$

Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Ca^{2+}}$ to precipitate: $$[\ce{Ca^2+}]=\dfrac{\pu{0.001 mol}}{\pu{200 mL}}=0.005M$$ $$[\ce{PO4^3-}]_\ce{Ca}=\sqrt{\dfrac{K_\mathrm{sp}}{\ce{[Ca^{2+}]^3}}} = \sqrt{\dfrac{2.07\times 10^{-33}}{(\pu{0.005 M})^3}}= 1.2890\times 10^{-13}$$

$\therefore \ce{AlPO4}$ will precipitate first.

The molecular mass of $\ce{Na3PO4}$ is 163.941 g/mol.

Note here that you can't just assume a stoichiometric amount of $\ce{Na3PO4}$ to say three significant figures. The solubility of the $\ce{Ca3(PO4)2}$ is so low that the $\ce{Na3PO4}$ would have to be weighed to an absolutely unrealistic precision.

So we'll go father down fantasy lane and suppose that:

  • There are $0.02\bar0$ moles of $\ce{Al^3+}$
  • That we are using God's balance which can weigh an infinitely precise amount of $\ce{Na3PO4}$.
  • That the molecular mass of $\ce{Na3PO4}$ is exactly 163.941 g/mol (this is so precise that isotopic ratios limit the value...)
  • $K_\mathrm{sp} (\ce{AlPO4})=9.83\bar0\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\bar0\times 10^{-33}$
  • That the volume changes in the solution when adding $\ce{Na3PO4}$ or precipitating $\ce{AlPO4}$ are insignificant.

So if the $[\ce{PO4^3-}] = 1.\bar0\times 10^{-13}$ then no $\ce{Ca3(PO4)2}$ will precipitate. At that concentration the amount of $\ce{Al^3+}$ left in solution will be:

$$\ce{[Al^{3+}]}=\dfrac{K_\mathrm{sp}}{\ce{[PO4^{3-}]}} = \dfrac{9.83\times 10^{-21}}{1.\bar0 \times \pu{10^{-13} M}} =9.83\times 10^{-8}$$

So $\pu{0.001999998034000 mol}$ of $\ce{AlPO4}$ will precipitate.

For 200 ml of a $1.\bar0 \times \pu{10^{-13} M}$ solution there are $2.\bar0 \times \pu{10^{-14} moles}$

So the total moles of $\ce{Na3PO4}$ needed is $\pu{0.001999998034020 mol}$ or ‭$\pu{0.327881677695273 g}$.

% Al left in solution $= \dfrac{0.002\bar0 - \pu{0.001999998034000 mol}}{0.002\bar0} =‭‭0.0000983‬\% $

Again the problem requires such a precise amount of $\ce{PO4^{3-}}$ to precipitate "all" the $\ce{Al^{3+}}$ yet leave "all" the $\ce{Ca^{2+}}$ in solution that it is totally unrealistic.


Followings can be calculated from given data:

$[\ce{AlCl3}] = \frac{\pu{0.002 mol}}{\pu{0.200 L}} = \pu{0.01 molL-1}$ and $[\ce{CaCl2}] = \frac{\pu{0.001 mol}}{\pu{0.200 L}} = \pu{0.005 molL-1}$.

Since both in same solution and dissolved, $[\ce{Al^3+}] = \pu{0.01 molL-1}$ and $[\ce{Ca^2+}] = \pu{0.005 molL-1}$.

If solid $\ce{Na3PO4}$ is added to the solution, two salts will be precipitated out: $\ce{AlPO4}$ with $K_\mathrm{sp}^\ce{AlPO4} = 9.83\times 10^{-21}$, and $\ce{Ca3(PO4)2}$ with $K_\mathrm{sp}^\ce{Ca3(PO4)2} = 2.07\times 10^{-33}$ (units ignored, and assume volume stays same during the process).

At first glance, since $K_\mathrm{sp}^\ce{Ca3(PO4)2} \lt \lt K_\mathrm{sp}^\ce{AlPO4} $, one can assume that $\ce{Ca3(PO4)2}$ would precipitate first. However, in reality, it is not the case:

Requirement for $\ce{Ca3(PO4)2}$ to precipitate: $$Q_\mathrm{sp}^\ce{Ca3(PO4)2} \ge K_\mathrm{sp}^\ce{Ca3(PO4)2} \Rightarrow (0.005)^3[\ce{PO4^3-}]^2 \ge 2.07\times 10^{-33}\\ \Rightarrow [\ce{PO4^3-}] \ge \sqrt{\frac{2.07\times 10^{-33}}{(0.005)^3}} \approx 1.29 \times 10^{-13}$$

Similarly, requirement for $\ce{AlPO4}$ to precipitate: $$Q_\mathrm{sp}^\ce{AlPO4} \ge K_\mathrm{sp}^\ce{AlPO4} \Rightarrow (0.010)[\ce{PO4^3-}] \ge 9.83\times 10^{-21}\\ \Rightarrow [\ce{PO4^3-}] \ge \frac{9.83\times 10^{-21}}{(0.010)} \approx 9.83\times 10^{-19}$$

Since, $9.83\times 10^{-19} \lt 1.29 \times 10^{-13}$, actually $\ce{AlPO4}$ precipitate first.

The requirement for $\ce{Ca3(PO4)2}$ to precipitate is: $[\ce{PO4^3-}] \ge 1.29 \times 10^{-13}$. Suppose, at this point $n$ amount of $\ce{AlPO4}$ been precipiteted. Thus, $[\ce{Al^3+}]$ is $\frac{0.002-n}{0.200}$. Hence:

$$\frac{0.002-n}{0.200} \times 1.29 \times 10^{-13} = K_\mathrm{sp}^\ce{AlPO4} = 9.83\times 10^{-21}$$

Now, you would be able to solve this equation for $n$. That is the amount of $\ce{Na3PO4}$ in $\pu{mols}$ you have have added to the solution before $\ce{Ca3(PO4)2}$ starts to precipitate. If you multiply this number by molar mass of $\ce{Na3PO4}$, you get your answer. I got $\pu{328 mg }$ for rough calculations. It's up to you.

  • $\begingroup$ There is a problem in that you can't just weigh out the $\ce{Na3PO4}$ "roughly." An extra milligram would be more than enough to cause some $\ce{Ca3(PO4)2}$ to precipitate. $\endgroup$
    – MaxW
    Commented Jun 4, 2020 at 19:25

The initial acidity due to $\ce{Al^{3+}}$ hydrolysis cannot be ignored. The initial concentration of $\ce{AlCl_3}$ is $0.02$ mol/$0.2$ L = $0.01$ M. In a $0.01$ M solution of aluminium chloride, the ion $\ce{Al^{3+}}$ is hydrolyzed according to :$$\ce{Al^{3+} + H_2O <-> [Al(OH)]^{2+} + H^+}......(1)$$ The $pK_a$ of this reaction is $4.9$. So in this solution, $\ce{[H^+] = \sqrt{K_a·[Al^{3+}]} = \sqrt{10^{-4.9}·0.01} = 3.55·10^{-4}}$. The corresponding $p$H is $-log[H^+] = 3.45$. At this $p$H, any addition of phosphate ion $\ce{PO_4^{3-}}$ will be transformed into monohydrogenophosphate $\ce{HPO_4^{2-}}$, and even dihydrogenophosphate $\ce{H_2PO_4^-}$ as the $pK_a$ of these equations are : $pK_{a2} = 12.3$ and $pK_{a3} = 7.2$ :$$\ce{PO_4^{3-} + H^+ -> HPO_4^{2-}}.........(2)$$ $$\ce{HPO_4^{2-} + H^+ -> H_2PO_4^-}........(3)$$ So if some sodium phosphate is added to such a solution ($p$H $3.45$), it will be immediately transformed into $\ce{H_2PO_4^-}$. The remaining concentration of $\ce{PO_4^{3-}}$ can be calculated .

Suppose we add $a$ mol $\ce{Na_3PO_{4}}$ to the $0.2$ L of this solution. The following concentrations can be calculated : $\ce{[H_2PO_4^-]}$ = $ a/{0.2} = 5$$a$ ;

$\ce{[HPO_4^{2-}] = K_{a3}·[H_2PO_4^-]/[H^+] = 10^{-7.2}·5$a$ /(3.55·10^{-4}) = 8.76·10^{-4}} a$

$\ce{[PO_4^{3-}] = K_{a2}·[HPO_4^-]/[H^+] = 10^{-12.3}·5$a$ /(3.55·10^{-4}) = 7.055·10^{-9}} a$

So the final solution has a concentration of phosphate ion $\ce{[PO_4^{3-}]}$ = $a·7.00·10^{-9}$ M. The solubility product for $\ce{AlPO_4}$ is obtained for a concentration of$\ce{[PO_4^{3-}]} = K_s/(0.01) = 9.83·10^{-21}/0.01 = 9.83·10^{-19} = a·7,00·10^{-9}$.

So : $a$ = $1,40·10^{-10}$ mol $\ce{Na_3PO_{4}}$

  • $\begingroup$ No, this isn't right. As you precipitate the $\ce{Al(OH)^2+}$ then $\ce{OH-}$ is released neutralizing the $\ce{H+}$ initially formed. In fact if the $\ce{Al(OH)^2+}$ is excatly precipitated then the pH should return to neutral. I have been looking somewhat the same notion, but wondering what pH you'd have to buffer the phosphate solution at. I think it would be between 4 and 5. You want a $\ce{[PO4^{3-}]} \approx 3.6\times10^{-16}$ $\endgroup$
    – MaxW
    Commented Jun 4, 2020 at 22:22
  • $\begingroup$ @ MaxW. $\ce{Al(OH)^{2+}}$ is not a precipitate. It is an ion. Ions are in solution. A precipitate has no charge. $\endgroup$
    – Maurice
    Commented Jun 5, 2020 at 6:55
  • $\begingroup$ Sorry I didn't express myself better. I was trying to infer as the $\ce{Al(OH)^2+}$ was precipitated as $\ce{AlPO4}$ then the $\ce{OH-}$ would be released. $\endgroup$
    – MaxW
    Commented Jun 5, 2020 at 8:15

I'm fiddling around with a different take on the problem. Let's assume that a buffered solution of phosphate species is used. What pH would the buffer have to be so that the $\ce{Al^3+}$ would be quantitatively precipitated but the $\ce{Ca^{2+}}$ would be left in solution?

(1) This gets around the problem of the hydroxy complexes of $\ce{Al^3+}$ and $\ce{Ca}$ since the solution will be acidic.

(2) Seems like a final concentration of $\ce{[PO4^{3-}]} = 3.6\times 10^{-16}$ would be a good target.

$$\log{(3.56\times 10^{-16})} \approx \dfrac{\log{(1.289\times10^{-13})} + \log{(9.83\times10^{-19}})}{2}$$

(3) Assume the total phosphate species is 1.00 molar. Then then the $\ce{AlPO4}$ precipitated can be assumed to cause a negligible change to the phosphate species at whatever pH the solution is buffered at.

  • Actually there will be some, but we can check to see that the affect is negligible.
  • We should really be using activities at this concentration, but we'll ignore that complication.

For a buffer with $\ce{[H3PO4] = [H2PO4-]}$ the pH will be 2.14.

For $\ce{H3PO4 <=> 3H+ + PO4^{3-}}$ the equilibrium expression is:

$$\ce{1.8208\times10^{-22} = \dfrac{\ce{[H+]^3[PO4^{3-}]}}{\ce{[H3PO4]}}}$$


$$\ce{[PO4^{3-}]} = \dfrac{(1.8208\times10^{-22})\times \ce{[H3PO4]}}{\ce{[H+]^3}} = \dfrac{(1.8208\times10^{-22})\times (0.5)}{(10^{-2.14})^3} = 2.4\times10^{-16}$$

which is just about perfect.


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