Chapter 18, Problem 64A

Interpretation Introduction

(a)

Interpretation:

The reaction between magnesium hydroxide and water is to be balanced.

Concept introduction:

To balance a reaction, the number of atoms of each element on both sides should be equal.

Answer to Problem 64A

$Mg{\left(OH\right)}_2\stackrel{H_{2}O}{\to }M{g}^{2+}+2O{H}^{-}$

Explanation of Solution

When magnesium hydroxide is dissolved in water it dissociates into its ions.

$Mg{\left(OH\right)}_2\stackrel{H_{2}O}{\to }M{g}^{2+}+2O{H}^{-}$

Interpretation Introduction

(b)

Interpretation:

The reaction between magnesium metal and hydrobromic acid is to be balanced.

Concept introduction:

To balance a reaction, the number of atoms of each element on both sides should be equal.

Answer to Problem 64A

$Mg+2HBr\to MgB{r}_2+H_2$

Explanation of Solution

When magnesium metal reacts with hydrobromic acid, magnesium bromide and hydrogen are formed.

$Mg+HBr\to MgB{r}_2+H_2$

Balancing the equation by making the number of each atom equal on both sides-

$Mg+2HBr\to MgB{r}_2+H_2$

Interpretation Introduction

(c)

Interpretation:

The reaction of propanoic acid in water is to be balanced.

Concept introduction:

To balance a reaction, the number of atoms of each element on both sides should be equal.

Answer to Problem 64A

$C_2{H}_{5}COOH\stackrel{H_{2}O}{\to }{C}_2{H}_{5}CO{O}^{-}+H^{+}$

Explanation of Solution

When magnesium hydroxide is dissolved in water it dissociates into its ions.

$C_2{H}_{5}COOH\stackrel{H_{2}O}{\to }{C}_2{H}_{5}CO{O}^{-}+H^{+}$

Interpretation Introduction

(d)

Interpretation:

The second ionization reaction of sulphuric acid in water is to be balanced.

Concept introduction:

To balance a reaction, the number of atoms of each element on both sides should be equal.

Answer to Problem 64A

$HS{O}_4{}^{-}\stackrel{H_{2}O}{\to }{H}^{+}+S{O}_4{}^{2-}$

Explanation of Solution

When $H_{2}S{O}_4$ is dissolved in water it dissociates into its ions as:

$\begin{array}{l}{H}_{2}S{O}_4\stackrel{H_{2}O}{\to }{H}^{+}+HS{O}_4{}^{-}\\ HS{O}_4{}^{-}\stackrel{H_{2}O}{\to }{H}^{+}+S{O}_4{}^{2-}\end{array}$

Thus, second ionization equation is:

$HS{O}_4{}^{-}\stackrel{H_{2}O}{\to }{H}^{+}+S{O}_4{}^{2-}$