Chapter 3, Problem 42QRT

(a)

Interpretation Introduction

Interpretation:

For ${\text{NaNO}}_2$, the acid and base from which it can be prepared have to be identified and the overall neutralization reaction in both complete and net ionic form has to be written.

Concept Introduction:

Acid - base reaction: Formation of the salt from the cation from the base and anion from the acid and formation of water is also the product.

Most of the ionic compounds are soluble in water, very few of the ionic compounds are sparingly soluble, and some of the ionic compounds are insoluble in water. When it is soluble in water ions gets separated in the solution.

Acids and bases:

Acid release hydrogen ion in water, base release hydroxide ions in water. An acid is a substance that produces hydronium ions, ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ when dissolved in water.

Explanation of Solution

The given compound is shown below,

  ${\text{NaNO}}_2$

${\text{NaNO}}_2$ dissociates in water (dissociation). Therefore ${\text{NaNO}}_2$ is formed from the neutralization of ${\text{HNO}}_2$ and $\text{NaOH}$. ${\text{HNO}}_2$ is weak acid and $\text{NaOH}$ is strong base.

The product are shown below,

  ${\text{HNO}}_{\text{2}}\text{(aq) + NaOH(aq) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{NaNO}}_{\text{2}}\text{(aq)}$

Nitrous acid (${\text{HNO}}_2$) reacts with sodium hydroxide ($\text{NaOH}$) gives sodium nitrite (${\text{NaNO}}_2$) and water.

Balance the equation,

    ${\text{HNO}}_{\text{2}}\text{(aq) + NaOH(aq) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{NaNO}}_{\text{2}}\text{(aq)}$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. The reaction is already balanced. Now, the balanced equation is given below.

  ${\text{HNO}}_{\text{2}}\text{(aq) + NaOH(aq) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{NaNO}}_{\text{2}}\text{(aq)}$

The complete ionic equation is given below,

    ${\text{HNO}}_{\text{2}}\text{(aq)}\text{+}{\text{Na}}^{\text{+}}{\text{(aq)+OH}}^{\text{-}}\text{(aq)}\to {\text{H}}_{\text{2}}\text{O(l)}\text{+}{\text{Na}}^{\text{+}}{\text{(aq)+NO}}_{\text{2}}^{\text{-}}\text{(aq)}$

Net ionic equation of the given reaction shown below

  ${\text{HNO}}_{\text{2}}\text{(aq)}\text{+}{\text{OH}}^{\text{-}}\text{(aq)}\to {\text{H}}_{\text{2}}\text{O(l)}\text{+}{\text{NO}}_{\text{2}}^{\text{-}}\text{(aq)}$

(b)

Interpretation Introduction

Interpretation:

For ${\text{CaSO}}_{\text{4}}$, the acid and base from which it can be prepared have to be identified and the overall neutralization reaction in both complete and net ionic form has to be written.

Concept introduction:

Refer to part (a)

Explanation of Solution

The given compound is shown below,

   ${\text{CaSO}}_4$

${\text{CaSO}}_4$ is formed from the neutralization of ${\text{H}}_{\text{2}}{\text{SO}}_4$ and ${\text{Ca(OH)}}_{\text{2}}$. ${\text{H}}_{\text{2}}{\text{SO}}_4$ is strong acid and ${\text{Ca(OH)}}_{\text{2}}$ is strong base.

The product are shown below,

   ${\text{H}}_{\text{2}}{\text{SO}}_4{\text{(aq) + Ca(OH)}}_{\text{2}}\text{(s) }\to {\text{ 2H}}_{\text{2}}\text{O (l)}\text{+}{\text{CaSO}}_4\text{(s)}$

Sulfuric acid (${\text{H}}_{\text{2}}{\text{SO}}_4$) reacts with calcium hydroxide (${\text{Ca(OH)}}_{\text{2}}$) gives calcium sulfate (${\text{CaSO}}_4$) and water.

Balance the equation,

    ${\text{H}}_{\text{2}}{\text{SO}}_4{\text{(aq) + Ca(OH)}}_{\text{2}}\text{(s) }\to {\text{ 2H}}_{\text{2}}\text{O (l)}\text{+}{\text{CaSO}}_4\text{(s)}$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. There are four hydrogen atoms in the left side and two hydrogen atoms in the right side. Therefore, two molecules of water molecule are added to right side of reaction. Now, the balanced equation is given below.

  ${\text{H}}_{\text{2}}{\text{SO}}_4{\text{(aq) + Ca(OH)}}_{\text{2}}\text{(s) }\to {\text{ 2H}}_{\text{2}}\text{O (l)}\text{+}{\text{CaSO}}_4\text{(s)}$

The complete ionic equation is given below,

    ${\text{H}}_{\text{2}}{\text{SO}}_4{\text{(aq) + Ca(OH)}}_{\text{2}}\text{(s) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{CaSO}}_4\text{(s)}$

Net ionic equation of the given reaction shown below

  ${\text{H}}^{\text{+}}{\text{(aq) + HSO}}_4^{-}{\text{+ Ca(OH)}}_{\text{2}}\text{(s) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{CaSO}}_4\text{(s)}$

(c)

Interpretation Introduction

Interpretation:

For $\text{NaI}$, the acid and base from which it can be prepared have to be identified and the overall neutralization reaction in both complete and net ionic form has to be written.

Concept introduction:

Refer to part (a)

Explanation of Solution

The given compound is shown below,

   $\text{NaI}$

$\text{NaI}$ is formed from the neutralization of $\text{HI}$ and $\text{NaOH}$. $\text{HI}$ is strong acid and $\text{NaOH}$ is strong base.

The product are shown below,

   $\text{HI(aq) + NaOH(aq) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}\text{NaI(aq)}$

Hydrogen iodide ($\text{HI}$) reacts with sodium hydroxide ($\text{NaOH}$) gives sodium iodide ($\text{NaI}$) and water.

Balance the equation,

    $\text{HI(aq) + NaOH(aq) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}\text{NaI(aq)}$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. The reaction is already balanced. Now, the balanced equation is given below.

  $\text{HI(aq) + NaOH(aq) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}\text{NaI(aq)}$

The complete ionic equation is given below,

    ${\text{H}}^{\text{+}}\text{(aq) +}{\text{I}}^{\text{-}}\text{(aq)+}{\text{Na}}^{\text{+}}{\text{(aq)+OH}}^{\text{-}}\text{(aq)}\to {\text{H}}_{\text{2}}\text{O(l)}\text{+}{\text{Na}}^{\text{+}}{\text{(aq)+ I}}^{-}\text{(aq)}$

Net ionic equation of the given reaction shown below

  ${\text{H}}^{\text{+}}\text{(aq) +}{\text{OH}}^{\text{-}}\text{(aq)}\to {\text{H}}_{\text{2}}\text{O(l)}$

(d)

Interpretation Introduction

Interpretation:

For ${\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$, the acid and base from which it can be prepared have to be identified and the overall neutralization reaction in both complete and net ionic form has to be written.

Concept introduction:

Refer to part (a)

Explanation of Solution

The given compound is shown below,

   ${\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$

${\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ is formed from the neutralization of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ and ${\text{Mg(OH)}}_{\text{2}}$. ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is weak acid and ${\text{Mg(OH)}}_{\text{2}}$ is weak base.

The product are shown below,

   ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + Mg(OH)}}_{\text{2}}\text{(s) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s)}$

Phosphoric acid (${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$) reacts with magnesium hydroxide (${\text{Mg(OH)}}_{\text{2}}$) gives magnesium phosphate (${\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$) and water.

Balance the equation,

    ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + Mg(OH)}}_{\text{2}}\text{(s) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s)}$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. There are three magnesium atoms in the right side and one magnesium atom in the left side. Therefore, three molecules of magnesium hydroxide (${\text{Mg(OH)}}_{\text{2}}$) are added to left side of reaction. Now, the balanced equation is given below.

  ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + 3Mg(OH)}}_{\text{2}}\text{(s) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s)}$

Now balance the phosphorus atom, there is one phosphorus atom in the left side and two phosphorus atoms in the right side. Therefore, two molecules of Phosphoric acid (${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$) are added to left side of reaction. Now, the balanced equation is given below.

  ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + 3Mg(OH)}}_{\text{2}}\text{(s) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s)}$

Now balance the hydrogen atom, there are two hydrogen atoms in the right side and twelve hydrogen atoms in the left side. Therefore, six molecules of water are added to right side of reaction. Now, the balanced equation is given below

  ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + 3Mg(OH)}}_{\text{2}}\text{(s) }\to {\text{ 6H}}_{\text{2}}\text{O (l)}\text{+}{\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(aq)}$

The complete ionic equation is given below,

    ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + 3Mg(OH)}}_{\text{2}}\text{(s) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s)}$

Net ionic equation of the given reaction shown below

  ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{(aq) + 3Mg(OH)}}_{\text{2}}\text{(s) }\to {\text{ H}}_{\text{2}}\text{O (l)}\text{+}{\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s)}$