Chapter 18, Problem 104SAE

Interpretation Introduction

Interpretation:

The one which has the highest molar solubility should be determined.

Concept introduction:

$\begin{array}{l}\text{AB(s)}\to {\text{A}}^{\text{+}}(aq)+B^{-}(aq)\\ Solubility-xxx\\ K_{sp}=\left[{\text{A}}^{\text{+}}(aq)\right]\left[B^{-}(aq)\right]\\ K_{sp}=x^2\\ x=\sqrt{K_{sp}}\end{array}$

$\begin{array}{l}{\text{AB}}_{\text{2}}\text{(s)}\to {\text{A}}^{\text{+}}(aq)+2B^{-}(aq)\\ Solubility-xx{\left(2x\right)}^2\\ K_{sp}=\left[{\text{A}}^{\text{+}}(aq)\right]\left[B^{-}(aq)\right]\\ K_{sp}=x\times 4x^2\\ 4x^3=K_{sp}\\ x={\left(\frac{K_{sp}}{4}\right)}^{1/3}\end{array}$

$\begin{array}{l}{\text{A}}_{\text{3}}{\text{B}}_{\text{2}}\text{(s)}\to 3{\text{A}}^{\text{+}}(aq)+2B^{-}(aq)\\ Solubility-x3x2x\\ K_{sp}=\left[{\text{A}}^{\text{+}}(aq)\right]\left[B^{-}(aq)\right]\\ K_{sp}={\left(3x\right)}^3\times {\left(2x\right)}^2\\ 108x^5=K_{sp}\\ x={\left(\frac{K_{sp}}{108}\right)}^{1/5}\end{array}$

$\begin{array}{l}{\text{A}}_{\text{3}}\text{B}\text{(s)}\to 3{\text{A}}^{\text{+}}(aq)+B^{-}(aq)\\ Solubility-x3xx\\ K_{sp}=\left[{\text{A}}^{\text{+}}(aq)\right]\left[B^{-}(aq)\right]\\ K_{sp}={\left(3x\right)}^3\times x\\ 27x^4=K_{sp}\\ x={\left(\frac{K_{sp}}{27}\right)}^{1/4}\end{array}$