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

Answer:

$\large \boxed{5.5\times 10^{-8}\text{ mol/L }}$

Explanation:

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

                          3x               x

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