Question

For the following equilibrium, Ag3PO4(s)↽−−⇀3Ag+(aq)+PO3−4(aq) If Ksp=2.4×10−28, what is the molar solubility of Ag3PO4? Report your answer in scientific notation with the correct number of significant figures.

Answer 1

`\large \boxed{5.5* 10^(-8)\text{ mol/L }}`

Step-by-step explanation:

Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq); Ksp = 2.4 × 10⁻²⁸

3x x

`K_(sp) =\text{[Ag$^(+)$]$^(3)$[PO$_(4)^(3-)$]} = (3x)^(3)x = 2.4 * 10^(-28)\\27x^(4) = 2.4 * 10^(-28)\\x^(4) = 8.89 * 10^(-30)\\x = \sqrt[4]{8.89 * 10^(-30)}\\= \mathbf{5.5* 10^(-8)} \textbf{ mol/L}\\\text{The molar solubility of silver phosphate is $\large \boxed{\mathbf{5.5* 10^(-8)}\textbf{ mol/L }}$}`