# The behaviour of Ag3PO4 in solution

To a $$\pu{1L}$$ flask, we add $$\pu{4.1900 g}$$ ($$\pu{0.01001 mol}$$) of $$\ce{Ag3PO4}$$, then fill it up with water. Knowing the following constants, we are asked to find the concentration of all species in solution:

$$K_\mathrm{s}(\ce{Ag3PO4})=2.7\cdot 10^{-18}$$, $$K_\mathrm{s}(\ce{AgOH})=4.0\cdot 10^{-16}$$, $$K_\mathrm{a1}=10^{-2}$$, $$K_\mathrm{a2}=10^{-7}$$, $$K_\mathrm{a3}=10^{-12}$$

$$\ce{Ag2O}$$ does not form.

By analyzing the problem, we notice that the formation of $$\ce{AgOH}$$ is more favorable:

$$\ce{Ag3PO4 + 3H2O <=> 3AgOH + H3PO4}, \\ K=\frac{K_\mathrm{s}(\ce{Ag3PO4})\cdot K_\mathrm{w}^3}{K_\mathrm{s}(\ce{AgOH})^3\cdot K_\mathrm{a1}\cdot K_\mathrm{a2}\cdot K_\mathrm{a3}} = 4.2\cdot10^7$$

Therefore, we start with $$[\ce{H3PO4}]_0 = \pu{0.01001 mol//dm^3}$$.

The equilibria we work with are:

\begin{align} \ce{AgOH &<=> Ag+ + HO-}\\ \ce{H3PO4 + 3HO- &-> PO4^{3-} + 3H2O}\\ \ce{PO4^{3-} + H2O &<=> HPO4^{2-} + HO-} \end{align}

The $$\ce{HO-}$$ formed from the insoluble salt is consumed in its entirety towards the formation of $$\ce{PO4^{3-}}$$, from which a fraction dissociates in water to form $$\ce{HPO4^{2-}}$$ and $$\ce{HO-}$$. Therefore, the principal conservation of mass law is:

$$[\ce{Ag+}]=3\ce{[PO4^{3-}]}+3\ce{[HPO4^{2-}]}$$

How do I tackle this problem in a more elegant manner? The equation on the top is solvable; but by looking at their results, we can see they do not fit at all:
$$[\ce{Ag+}]=6.5\cdot10^{-14}$$, $$[\ce{HO-}]=6.2\cdot10^{-3}$$, $$[\ce{PO4^{3-}}]=3.8\cdot10^{-3}$$

I'm thinking that, considering such a small concentration of $$\ce{Ag+}$$ compared to $$\ce{PO4^{3-}}$$, we might still have some $$\ce{Ag3PO4}$$ undissolved.

But this doesn't seem to make sense since the formation of $$\ce{AgOH}$$ is a lot more favorable, as seen from the value of $$K$$.