Chapter 17, Problem 48A

(a)

Interpretation Introduction

Interpretation :

By balancing the given chemical equation, the solubility product constant expression should be written.

Concept Introduction :

When the number of atoms on the left side of the reaction is equal to the number of atoms on the product side, an equation is said to be a balanced chemical equation.

Answer to Problem 48A

The balanced chemical equation of silver chloride is given as

  $\text{AgCl}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{Cl}}^{\text{-}}\left(\text{aq}\right)$

This equilibrium is a heterogeneous equilibrium, so the concentration of silver chloride is one. Hence, the solubility product constant for the given reaction is;

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Ag}}^{\text{+}}\right]\left[{\text{Cl}}^{\text{-}}\right]$

Explanation of Solution

The balanced chemical equation of silver chloride is given as

  $\text{AgCl}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{Cl}}^{\text{-}}\left(\text{aq}\right)$

The Solubility expression for the given reaction is given as

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\frac{\left[{\text{Ag}}^{\text{+}}\right]\left[{\text{Cl}}^{\text{-}}\right]}{\left[\text{AgCl}\right]}\\ {\text{K}}_{\text{sp}}\left[\text{AgCl}\right]\text{=}\left[{\text{Ag}}^{\text{+}}\right]\left[{\text{Cl}}^{\text{-}}\right]\end{array}$

This equilibrium is a heterogeneous equilibrium, so the concentration of silver chloride is one.

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Ag}}^{\text{+}}\right]\left[{\text{Cl}}^{\text{-}}\right]$

(b)

Interpretation Introduction

Interpretation:

By balancing the given chemical equation, the solubility product constant expression should be written.

Concept Introduction:

When the number of atoms on the left side of the reaction is equal to the number of atoms on the product side, an equation is said to be a balanced chemical equation.

Answer to Problem 48A

The balanced chemical equation of dissociation of calcium sulphate in water is

  ${\text{CaSO}}_{\text{4}}\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{\text{+2}}\text{(aq)}\text{+}{\text{SO}}_{\text{4}}{}^{\text{-2}}\left(\text{aq}\right)$

This equilibrium is a heterogeneous equilibrium, so the concentration of calcium sulphate is one. Hence, the solubility product constant for the given reaction is;

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Ca}}^{\text{+2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]$

Explanation of Solution

The balanced chemical equation of calcium sulphate is given as

  ${\text{CaSO}}_{\text{4}}\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{SO}}_{\text{4}}{}^{\text{-2}}\left(\text{aq}\right)$

The Solubility expression for the given reaction is given as

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\frac{\left[{\text{Ca}}^{\text{+2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]}{\left[{\text{CaSO}}_{\text{4}}\right]}\\ {\text{K}}_{\text{sp}}\left[{\text{CaSO}}_{\text{4}}\right]\text{=}\left[{\text{Ca}}^{\text{+2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]\end{array}$

This equilibrium is a heterogeneous equilibrium, so the concentration of calcium sulphate is one.

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Ca}}^{\text{+2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]$

(c)

Interpretation Introduction

Interpretation :

By balancing the given chemical equation, the solubility product constant expression should be written.

Concept Introduction :

When the number of atoms on the left side of the reaction is equal to the number of atoms on the product side, an equation is said to be a balanced chemical equation.

Answer to Problem 48A

The balanced chemical equation of iron carbonate is given as

  ${\text{FeCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Fe}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{Co}}_{\text{3}}{}^{\text{-2}}\left(\text{aq}\right)$

This equilibrium is a heterogeneous equilibrium, so the concentration of iron carbonate is one. Hence, the solubility product constant for the given reaction is;

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Fe}}^{\text{+2}}\right]\left[{\text{CO}}_{\text{3}}{}^{\text{-2}}\right]$

Explanation of Solution

The balanced chemical equation of iron carbonate is given as

  ${\text{FeCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Fe}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{Co}}_{\text{3}}{}^{\text{-2}}\left(\text{aq}\right)$

The Solubility expression for the given reaction is given as

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\frac{\left[{\text{Fe}}^{\text{+2}}\right]\left[{\text{CO}}_{\text{3}}{}^{\text{-2}}\right]}{\left[{\text{FeCO}}_{\text{3}}\right]}\\ {\text{K}}_{\text{sp}}\left[{\text{FeCO}}_{\text{3}}\right]\text{=}\left[{\text{Fe}}^{\text{+2}}\right]\left[{\text{CO}}_{\text{3}}{}^{\text{-2}}\right]\end{array}$

This equilibrium is a heterogeneous equilibrium, so the concentration of iron carbonate is one.

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Fe}}^{\text{+2}}\right]\left[{\text{CO}}_{\text{3}}{}^{\text{-2}}\right]$

(d)

Interpretation Introduction

Interpretation:

By balancing the given chemical equation, the solubility product constant expression should be written.

Concept Introduction:

When the number of atoms on the left side of the reaction is equal to the number of atoms on the product side, an equation is said to be a balanced chemical equation.

Answer to Problem 48A

The balanced chemical equation of dissociation of lead chromate in water is

  ${\text{PbCrO}}_{\text{4}}\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{\text{+2}}\text{(aq)}\text{+}{\text{CrO}}_{\text{4}}{}^{\text{-2}}\left(\text{aq}\right)$

This equilibrium is a heterogeneous equilibrium, so the concentration of lead chromate is one. Hence, the solubility product constant for the given reaction is;

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Pb}}^{\text{+2}}\right]\left[{\text{CrO}}_{\text{4}}{}^{\text{-2}}\right]$

Explanation of Solution

The balanced chemical equation of lead chromate is given as

  ${\text{PbCrO}}_{\text{4}}\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{\text{+2}}\text{(aq)}\text{+}{\text{CrO}}_{\text{4}}{}^{\text{-2}}\left(\text{aq}\right)$

The Solubility expression for the given reaction is given as

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\frac{\left[{\text{Pb}}^{\text{+2}}\right]\left[{\text{CrO}}_{\text{4}}{}^{\text{-2}}\right]}{\left[{\text{PbCrO}}_{\text{4}}\right]}\\ {\text{K}}_{\text{sp}}\left[{\text{PbCrO}}_{\text{4}}\right]\text{=}\left[{\text{Pb}}^{\text{+2}}\right]\left[{\text{CrO}}_{\text{4}}{}^{\text{-2}}\right]\end{array}$

This equilibrium is a heterogeneous equilibrium, so the concentration of lead chromate is one.

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Pb}}^{\text{+2}}\right]\left[{\text{CrO}}_{\text{4}}{}^{\text{-2}}\right]$