Chapter 17.3, Problem 23PP

Interpretation Introduction

Interpretation:

The solubility of silver phosphate is to be calculated

Concept introduction

Solubility is the concentration of the solute dissolved in a given solvent. The equilibrium constant is the ratio of concentration of the products to the concentration of reactants.

${\text{K}}_{\text{sp}}\text{= }{\left[{\text{Ag}}^{\text{+1}}\right]}^{\text{3}}\text{×}\left[{\text{PO}}_{\text{4}}{}^{\text{3-}}\right]$

${\text{ [Ag}}_{\text{3}}{\text{PO}}_{\text{4}}\text{] }$= concentration of the product ${\left[{\text{Ag}}^{\text{+1}}\right]}^{\text{3}}\text{×}\left[{\text{PO}}_{\text{4}}{}^{\text{3+}}\right]$ = concentration of the reactants

Answer to Problem 23PP

The solubility of silver phosphate is $\text{1}{\text{.762×10}}^{\text{-5}}\text{mol/L}$

Explanation of Solution

Given information :

The solubility product is $\text{2}{\text{.6×10}}^{\text{-5}}\text{mol/L}$

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

${\text{AgPO}}_{\text{4}}{}_{}\underset{}{\rightleftharpoons }{\text{3Ag}}^{\text{1+}}{\text{+PO}}_{\text{4}}{}^{\text{3-}}$

$\text{I = 0 0 }$ ( initial concentration)

$\text{ C = +3x + x }$ ( change in concentration or change in amount from initial state to equilibrium)

${\text{ E= 27x}}^{\text{3}}\text{ x }$ (equilibrium concentration)

The solubility product is given as

$\text{Ksp = }{\left[{\text{Ag}}^{\text{1+}}\right]}^{\text{3}}\text{× }\left[{\text{PO}}_{\text{4}}{}^{\text{3-}}\right]$

${\text{= 27x}}^{\text{3}}\text{× x}$

$= 27\times X^4$

${}^{\text{4}}\sqrt{\left(\text{Ksp/27}\right)}\text{= X}$

Or, $\text{x = }\frac{{}^{\text{4}}\sqrt{\left(\text{2}{\text{.6 × 10}}^{\text{-18}}\right)}}{\text{27}}$

$\text{= 1}{\text{.762× 10}}^{\text{-5 }}\text{mol/L}$

Conclusion

The solubility of silver phosphate is $\text{1}{\text{.762× 10}}^{\text{-5 }}\text{mol/L}$