Question

The solubility product for Ag3PO4 is 2.8 × 10‑18. What is the solubility of silver phosphate in a solution which also contains 0.10 moles of silver nitrate per liter?

Answer 1

2.8x10⁻¹⁵ M.

Step-by-step explanation:

Hello,

In this case, the dissociation reaction for silver phosphate is:

`Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4(aq)`

Therefore, the equilibrium expression is:

`Ksp=[Ag^+]^3[PO_4^-]`

In such a way, since the initial solution contains an initial concentration of silver ions (from silver nitrate) of 0.10M, we can write the equilibrium expression in terms of the reaction extent
`x`:

`2.8x10^(-18)=(0.10+3x)^3*(x)`

Thus, solving for
`x` we have:

`x=2.8x10^(-15)M`

Thus, the molar solubility of silver phosphate is 2.8x10⁻¹⁵ M.

Regards.