Chapter U6, Problem 8RE

Interpretation Introduction

(a)

Interpretation:

How can we tell that the given solution is saturated 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 8RE

As the excess undissolved solute settle at the bottom of the container, hence, we can say that the given solution is saturated.

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 and value of $K_{\text{sp}}$ for lead sulfate is $1.6\times {10}^{-8}$.

As it given that, excess undissolved solute settle at the bottom of the container, hence, we can conclude that the given solution is saturated.

Interpretation Introduction

(b)

Interpretation:

Is the 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 8RE

Yes, the given solution is at equilibrium

Explanation of Solution

A reaction is in equilibrium when the forward rate is equal to the reverse rate. The given solution is a saturated solution as excess ${\text{PbSO}}_{\text{4}}$ settle down at the bottom.

From this it can be concluded that the solution is at equilibrium between the dissolved and solid states as further addition of the solute with settle down as precipitate.

Interpretation Introduction

(c)

Interpretation:

The chemical equation for the reversible change that is taking place 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 8RE

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

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

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,

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

Interpretation Introduction

(d)

Interpretation:

The equilibrium-constant equation for the 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 8RE

The equilibrium-constant equation for the equilibrium is,

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

$\begin{array}{l}\text{K=}\left[\text{Products}\right]/\left[\text{Substrates}\right]\\ \text{K=}\left[{\text{Pb}}^{\text{+ 2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]/\left[{\text{PbSO}}_{\text{4}}\right]\end{array}$

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,

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

And, the equilibrium-constant equation for the equilibrium is

$\begin{array}{l}\text{K=}\left[\text{Products}\right]/\left[\text{Substrates}\right]\\ \text{K=}\left[{\text{Pb}}^{\text{+ 2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]/\left[{\text{PbSO}}_{\text{4}}\right]\end{array}$

Interpretation Introduction

(e)

Interpretation:

The equilibrium concentration of each ion 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 8RE

The equilibrium concentration of $\left[{\text{Pb}}^{\text{+ 2}}\right]=\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]=x=5.0\times {10}^{-4}$

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,

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

For this reaction,

$\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\text{2}{\text{.5×10}}^{\text{-8}}\text{=}\left[{\text{Pb}}^{\text{+ 2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]\\ {\text{x}}^{\text{2}}\text{=}\text{2}{\text{.5×10}}^{\text{-8}}\\ \left[{\text{Pb}}^{\text{+ 2}}\right]\text{=}\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]\text{=}\text{x}\text{=}\text{5}{\text{.0×10}}^{\text{-4}}\end{array}$

Interpretation Introduction

(f)

Interpretation:

On dissolving lead (II) sulfate shows an endothermic change. What will happen to the value of $\left[{\text{Pb}}^{\text{+ 2}}\right]$ and $\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]$ if the solution is heated 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 8RE

On heating, the equilibrium of the reaction will shift to make more products. The value or the concentration of $\left[{\text{Pb}}^{\text{+ 2}}\right]$ and $\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]$ if the solution on heating will increase.

Explanation of Solution

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

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

Given that, on dissolving lead (II) sulfate shows an endothermic change and the equilibrium of the reaction will shift to make more products.

So the concentration of $\left[{\text{Pb}}^{\text{+ 2}}\right]$ and $\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]$ if the solution will increaseon heating.

Interpretation Introduction

(g)

Interpretation:

What would happen if more sulfate ions is added to the solution 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 8RE

If more sulfate ions is added to the solution, ${\text{PbSO}}_{\text{4}}$ gets precipitated as the equilibrium shift to the left.

Explanation of Solution

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

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

On adding ${\text{SO}}_{\text{4}}{}^{\text{-2}}$ ions, the equilibrium shift to the left. Due to this, ${\text{PbSO}}_{\text{4}}$ gets precipitated.