Chapter 4, Problem 4.60QP

(a)

Interpretation Introduction

Interpretation: The balanced molecular equation and net ionic equation for the given reactions is to be stated.

Concept introduction: A balanced chemical equation is based on the law of conservation of mass for a closed system. According to this law, the mass of the products formed in a chemical reaction is equal to the mass of the reactants involved in the reaction.

In net ionic equations the same ions that are present on both the sides of reaction are cancelled out.

To determine: The overall ionic and net ionic equation for the given reaction.

Answer to Problem 4.60QP

Solution

The balanced reaction between $\text{HBr}$ and $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is,

$\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3\text{HBr}\left(aq\right)\to {\text{AlBr}}_3\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The net ionic equation is,

$\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+{\text{H}}^{+}\left(aq\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

Explanation

When an acid reacts with a base it forms a salt and water and this type of reaction is known as neutralization reaction.

According to the solubility rules, aluminum hydroxide act as a weak electrolyte as it is insoluble in water and only some ions go into solution.

The complete reaction between $\text{HBr}$ and $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is,

$\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+\text{HBr}\left(aq\right)\to {\text{AlBr}}_3\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

According to the law of conservation of mass, the mass of the reactants that take part in a chemical reaction is equal to the mass of the products formed in the reaction.

In the given reaction, the number of aluminum present on the reactant side is equal to their number on the product side. The number of bromine, oxygen and hydrogen atom present on the reactant side is not equal to their number on the product side. The coefficient $3$ is added to $\text{HBr}$ present on reactant side and coefficient $3$ is added to ${\text{H}}_{\text{2}}\text{O}$ present on product side. This balances the number of bromine, oxygen, and hydrogen atoms on either side of the reaction.

The balanced reaction between $\text{HBr}$ and $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is,

$\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3\text{HBr}\left(aq\right)\to {\text{AlBr}}_3\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The dissociation reaction of $\text{HBr}$ is,

$\text{HBr}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)$

In overall ionic reaction all the ions that are taking part in a reaction is represented in an equation.

Therefore, the overall ionic reaction is,

$\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+{\text{H}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{Br}}^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The ions that are present on both sides of the reaction are called spectator ions and are omitted from the reaction to represents the actual reactants and product.

Hence, the net ionic equation is,

$\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+{\text{H}}^{+}\left(aq\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(b)

Interpretation Introduction

To determine: The overall ionic and net ionic equation for the given reaction.

Answer to Problem 4.60QP

Solution

The balanced reaction between ${\text{H}}_2{\text{SO}}_4$ and ${\text{Na}}_2{\text{CO}}_3$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)+{\text{Na}}_2{\text{CO}}_3\left(aq\right)\to {\text{Na}}_2{\text{SO}}_4\left(aq\right)+{\text{H}}_{\text{2}}{\text{CO}}_3\left(aq\right)$

The net ionic equation is,

${\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)+{\text{CO}}_3{}^{2-}\left(aq\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Explanation of Solution

Explanation

When an acid reacts with a base it forms a salt and water and this type of reaction is known as neutralization reaction.

The complete reaction between ${\text{H}}_2{\text{SO}}_4$ and ${\text{Na}}_2{\text{CO}}_3$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)+{\text{Na}}_2{\text{CO}}_3\left(aq\right)\to {\text{Na}}_2{\text{SO}}_4\left(aq\right)+{\text{H}}_{\text{2}}{\text{CO}}_3\left(aq\right)$

According to the law of conservation of mass, the mass of the reactants that take part in a chemical reaction is equal to the mass of the products formed in the reaction.

In the given reaction, the number of sodium, sulphur, carbon, hydrogen and chlorine atoms present on the reactant side is equal to their number on the product side. Therefore the above reaction is a balanced reaction.

The complete reaction between ${\text{H}}_2{\text{SO}}_4$ and ${\text{Na}}_2{\text{CO}}_3$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)+{\text{Na}}_2{\text{CO}}_3\left(aq\right)\to {\text{Na}}_2{\text{SO}}_4\left(aq\right)+{\text{H}}_{\text{2}}{\text{CO}}_3\left(aq\right)$

In overall ionic reaction all the ions that are taking part in a reaction is represented in an equation.

The first dissociation reaction of ${\text{H}}_2{\text{SO}}_4$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)$

The second dissociation reaction of ${\text{H}}_2{\text{SO}}_4$ is,

${\text{HSO}}_4{}^{-}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)$

The dissociation of carbonic acid is,

${\text{H}}_{\text{2}}{\text{CO}}_3\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(aq\right)$

Therefore, the overall ionic reaction is,

$\begin{array}{c}{\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)+2{\text{Na}}^{+}+{\text{CO}}_3{}^{2-}\left(aq\right)\to 2{\text{Na}}^{+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ +{\text{CO}}_{\text{2}}\left(g\right)\end{array}$

The ions that are present on both sides of the reaction are called spectator ions and are omitted from the reaction to represents the actual reactants and product.

Hence, the net ionic equation is,

${\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)+{\text{CO}}_3{}^{2-}\left(aq\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

(c)

Interpretation Introduction

To determine: The overall ionic and net ionic equation for the given reaction.

Answer to Problem 4.60QP

Solution

The balanced reaction between ${\text{HNO}}_3$ and $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is,

${\text{HNO}}_3\left(aq\right)+2\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)\to \text{Ca}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

The net ionic equation is,

${\text{2H}}^{+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to 2{\text{H}}_2\text{O}\left(l\right)$

Explanation of Solution

Explanation

When an acid reacts with a base it forms a salt and water and this type of reaction is known as neutralization reaction.

According to the solubility rules, calcium hydroxide act as a weak electrolyte as it is insoluble in water and only some ions go into solution

The complete reaction between ${\text{HNO}}_3$ and $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is,

${\text{HNO}}_3\left(aq\right)+\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)\to \text{Ca}{\left({\text{NO}}_3\right)}_2\left(aq\right){\text{+H}}_2\text{O}\left(l\right)$

According to the law of conservation of mass, the mass of the reactants that take part in a chemical reaction is equal to the mass of the products formed in the reaction.

In the given reaction, the number of calcium present on the reactant side is equal to their number on the product side. The number of nitrogen, oxygen and hydrogen atom present on the reactant side is not equal to their number on the product side. The coefficient $2$ is added to ${\text{HNO}}_3$ present on reactant side and coefficient $2$ is added to ${\text{H}}_{\text{2}}\text{O}$ present on product side. This balances the number of nitrogen, oxygen, and hydrogen atoms on either side of the reaction.

The balanced reaction between ${\text{HNO}}_3$ and $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is,

${\text{HNO}}_3\left(aq\right)+2\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)\to \text{Ca}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

Therefore, the overall ionic reaction is,

${\text{2H}}^{+}\left(aq\right)+2{\text{NO}}_3{}^{-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{Ca}}^{2+}+2{\text{NO}}_3{}^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

The ions that are present on both sides of the reaction are called spectator ions and are omitted from the reaction to represents the actual reactants and product.

Hence, the net ionic equation is,

${\text{2H}}^{+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to 2{\text{H}}_2\text{O}\left(l\right)$

(d)

Interpretation Introduction

To determine: The overall ionic and net ionic equation for the given reaction.

Answer to Problem 4.60QP

Solution

The balanced between ${\text{H}}_2{\text{SO}}_4$ and $\text{KOH}$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)+\text{KOH}\left(aq\right)\to {\text{K}}_2{\text{SO}}_4\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The net ionic equation is,

${\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

Explanation

When an acid reacts with a base it forms a salt and water and this type of reaction is known as neutralization reaction.

The complete reaction between ${\text{H}}_{\text{2}}\text{O}$ and ${\text{K}}_2\text{O}$ is,

${\text{K}}_2\text{O}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2\text{KOH}\left(aq\right)$

The reaction between ${\text{H}}_2{\text{SO}}_4$ and $\text{KOH}$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)+\text{KOH}\left(aq\right)\to {\text{K}}_2{\text{SO}}_4\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

According to the law of conservation of mass, the mass of the reactants that take part in a chemical reaction is equal to the mass of the products formed in the reaction.

In the given reaction, the number of sulphur atoms present on the reactant side is equal to their number on the product side. The number of hydrogen, oxygen and potassium atom present on the reactant side is not equal to their number on the product side. The coefficient $2$ is added to $\text{KOH}$ present on reactant side and coefficient $2$ is added to ${\text{H}}_{\text{2}}\text{O}$ present on the product side. This balances the number of potassium, oxygen, and hydrogen atom on either side of the reaction.

The balanced between ${\text{H}}_2{\text{SO}}_4$ and $\text{KOH}$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)+2\text{KOH}\left(aq\right)\to {\text{K}}_2{\text{SO}}_4\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

In overall ionic reaction all the ions that are taking part in a reaction is represented in an equation.

The first dissociation reaction of ${\text{H}}_2{\text{SO}}_4$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)$

The second dissociation reaction of ${\text{H}}_2{\text{SO}}_4$ is,

${\text{HSO}}_4{}^{-}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)$

The dissociation reaction of $\text{KOH}$ is,

$\text{KOH}\left(aq\right)\rightleftharpoons {\text{K}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

Therefore, the overall ionic reaction is,

${\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)+2{\text{K}}^{+}+2{\text{OH}}^{-}\left(aq\right)\to 2{\text{K}}^{+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The ions that are present on both sides of the reaction are called spectator ions and are omitted from the reaction to represents the actual reactants and product.

Hence, the net ionic equation is,

${\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The balanced reaction between $\text{HBr}$ and $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is,

$\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3\text{HBr}\left(aq\right)\to {\text{AlBr}}_3\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The net ionic equation is,

$\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+{\text{H}}^{+}\left(aq\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The balanced reaction between ${\text{H}}_2{\text{SO}}_4$ and ${\text{Na}}_2{\text{CO}}_3$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)+{\text{Na}}_2{\text{CO}}_3\left(aq\right)\to {\text{Na}}_2{\text{SO}}_4\left(aq\right)+{\text{H}}_{\text{2}}{\text{CO}}_3\left(aq\right)$

The net ionic equation is,

${\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)+{\text{CO}}_3{}^{2-}\left(aq\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

The balanced reaction between ${\text{HNO}}_3$ and $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is,

${\text{HNO}}_3\left(aq\right)+2\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)\to \text{Ca}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

The net ionic equation is,

${\text{2H}}^{+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to 2{\text{H}}_2\text{O}\left(l\right)$

The balanced between ${\text{H}}_2{\text{SO}}_4$ and $\text{KOH}$ is,

${\text{H}}_2{\text{SO}}_4\left(aq\right)+\text{KOH}\left(aq\right)\to {\text{K}}_2{\text{SO}}_4\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The net ionic equation is, ${\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4{}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{SO}}_4{}^{2-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$