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:


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 $K$.


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