Chapter 17, Problem 49A

(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 49A

The balanced chemical equation of magnesium fluoride is given as

  ${\text{MgF}}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Mg}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{2F}}^{\text{-}}\left(\text{aq}\right)$

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

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Mg}}^{\text{+2}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{-2}}$

Explanation of Solution

The balanced chemical equation of magnesium fluoride is given as

  ${\text{MgF}}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Mg}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{2F}}^{\text{-}}\left(\text{aq}\right)$

The Solubility expression for the given reaction is given as

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\frac{\left[{\text{Mg}}^{\text{+2}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{2}}}{\left[{\text{MgF}}_{\text{2}}\right]}\\ {\text{K}}_{\text{sp}}\left[{\text{MgF}}_{\text{2}}\right]\text{=}\left[{\text{Mg}}^{\text{+2}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{2}}\end{array}$

This equilibrium is a heterogeneous equilibrium, so the concentration of magnesium fluoride is one.

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Mg}}^{\text{+2}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{-2}}$

(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 49A

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

  ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{3Ca}}^{\text{+2}}\text{(aq)}\text{+}{\text{2PO}}_{\text{4}}{}^{\text{-3}}\left(\text{aq}\right)$

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

  ${\text{K}}_{\text{sp}}\text{=}{\left[{\text{Ca}}^{\text{+2}}\right]}^{\text{3}}{\left[{\text{PO}}_{\text{4}}{}^{\text{-3}}\right]}^{\text{2}}$

Explanation of Solution

The balanced chemical equation of calcium phosphate is given as

  ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{3Ca}}^{\text{+2}}\text{(aq)}\text{+}{\text{2PO}}_{\text{4}}{}^{\text{-3}}\left(\text{aq}\right)$

The Solubility expression for the given reaction is given as

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\frac{{\left[{\text{Ca}}^{\text{+2}}\right]}^{\text{3}}{\left[{\text{PO}}_{\text{4}}{}^{\text{-3}}\right]}^{\text{2}}}{\left[{\text{Ca}}_{\text{3}}\left({\text{PO}}_{\text{4}}\right)\right]}\\ {\text{K}}_{\text{sp}}\left[{\text{Ca}}_{\text{3}}\left({\text{PO}}_{\text{4}}\right)\right]\text{=}{\left[{\text{Ca}}^{\text{+2}}\right]}^{\text{3}}{\left[{\text{PO}}_{\text{4}}{}^{\text{-3}}\right]}^{\text{2}}\end{array}$

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

  ${\text{K}}_{\text{sp}}\text{=}{\left[{\text{Ca}}^{\text{+2}}\right]}^{\text{3}}{\left[{\text{PO}}_{\text{4}}{}^{\text{-3}}\right]}^{\text{2}}$

(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 49A

The balanced chemical equation of copper hydroxide is given as

  $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Cu}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{2OH}}^{\text{-1}}\left(\text{aq}\right)$

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

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Cu}}^{\text{+2}}\right]{\left[{\text{OH}}^{\text{-1}}\right]}^{\text{2}}$

Explanation of Solution

The balanced chemical equation of copper hydroxide is given as

  $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{Cu}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{2OH}}^{\text{-1}}\left(\text{aq}\right)$

The Solubility expression for the given reaction is given as

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\frac{\left[{\text{Cu}}^{\text{+2}}\right]{\left[{\text{OH}}^{\text{-1}}\right]}^{\text{2}}}{\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]}\\ {\text{K}}_{\text{sp}}\left[\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\right]\text{=}\left[{\text{Cu}}^{\text{+2}}\right]{\left[{\text{OH}}^{\text{-1}}\right]}^{\text{2}}\end{array}$

This equilibrium is a heterogeneous equilibrium, so the concentration of copper hydroxide is one.

  ${\text{K}}_{\text{sp}}\text{=}\left[{\text{Cu}}^{\text{+2}}\right]{\left[{\text{OH}}^{\text{-1}}\right]}^{\text{2}}$

(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 49A

The balanced chemical equation of dissociation of silver sulphate in water is ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{s}\right)\rightleftharpoons {\text{2Ag}}^{\text{+1}}\text{(aq)}\text{+}{\text{SO}}_{\text{4}}{}^{\text{-2}}\left(\text{aq}\right)$

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

  ${\text{K}}_{\text{sp}}\text{=}{\left[{\text{Ag}}^{\text{+1}}\right]}^{\text{2}}\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]$

Explanation of Solution

The balanced chemical equation of silver sulphate is given as

  ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{s}\right)\rightleftharpoons {\text{2Ag}}^{\text{+1}}\text{(aq)}\text{+}{\text{SO}}_{\text{4}}{}^{\text{-2}}\left(\text{aq}\right)$

The Solubility expression for the given reaction is given as

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\frac{{\left[{\text{Ag}}^{\text{+1}}\right]}^{\text{2}}\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]}{\left[{\text{AgSO}}_{\text{4}}\right]}\\ {\text{K}}_{\text{sp}}\left[{\text{AgSO}}_{\text{4}}\right]\text{=}{\left[{\text{Ag}}^{\text{+1}}\right]}^{\text{2}}\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]\end{array}$

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

  ${\text{K}}_{\text{sp}}\text{=}{\left[{\text{Ag}}^{\text{+1}}\right]}^{\text{2}}\left[{\text{SO}}_{\text{4}}{}^{\text{-2}}\right]$