Chapter U6, Problem 9RE

Interpretation Introduction

(a)

Interpretation:

Which compound will have higher concentration of ions in solution at equilibrium needs to be determined.

Concept introduction:

The mathematical relationship between the products and reactants in an equilibrium is given as,

$\text{aA}\text{+}\text{bB}\rightleftharpoons \text{cC}\text{+ dD}$

$K=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$

Answer to Problem 9RE

${\text{PbSO}}_{\text{4}}$ have higher concentration of ions in solution at equilibrium.

Explanation of Solution

Given, solid ${\text{PbSO}}_{\text{4}}$ is dissolved in water to form a solution with excess undissolved solute at the bottom of the container.

Therefore, the chemical equation for the reversible change that is taking place is,

${\text{PbSO}}_{\text{4}}\left(\text{s}\right)\to {\text{Pb}}^{\text{2 +}}\left(\text{aq}\right)\text{+}{\text{SO}}_{\text{4}}{}^{-2}\left(\text{aq}\right)$

The product with larger $K_{sp}$ value is always favored. $K_{sp}$ value of ${\text{PbSO}}_{\text{4}}$ is large than $\text{CdS}$. Therefore, ${\text{PbSO}}_{\text{4}}$ is more soluble so ${\text{PbSO}}_{\text{4}}$ will have higher concentration of ionsin solution at equilibrium.

Interpretation Introduction

(b)

Interpretation:

The concentration of cadmium ions in solution at equilibrium needs to be calculated.

Concept introduction:

The mathematical relationship between the products and reactants in an equilibrium is given as,

$\text{aA}\text{+}\text{bB}\rightleftharpoons \text{cC}\text{+ dD}$

$K=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$

Answer to Problem 9RE

The concentration of cadmium ions in solution at equilibrium is ${\text{2×10 }}^{-\text{3}}\text{M}$

Explanation of Solution

Given, the $K_{sp}$ value for $\text{CdS}$ is $8.0\times {10}^{-27}$.

So, for the reaction,

$\text{CdS}\left(\text{s}\right)\rightleftharpoons {\text{Cd}}^{\text{2 +}}\left(\text{aq}\right)\text{+}{\text{S}}^{2-}\left(\text{aq}\right)$

$\begin{array}{l}{K}_{sp}=\left[{\text{Cd}}^{\text{2 +}}\right]\left[{\text{S}}^{2-}\right]\\ \left[{\text{Cd}}^{\text{2 +}}\right]=\left[{\text{S}}^{2-}\right]=\sqrt{K_{sp}}\\ \left[{\text{Cd}}^{\text{2 +}}\right]=\left[{\text{S}}^{2-}\right]=\sqrt{8\times {10}^{-27}}=2\times {10}^{-3}\text{M}\end{array}$