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

Answer:

2.8x10⁻¹⁵ M.

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.