Chapter 22, Problem 56QAP

Interpretation Introduction

(a)

Interpretation:

Balanced net ionic reaction should be written for the reaction of (CH₃ )₂ NH with HCl.

Concept introduction:

While writing the net ionic equation, an understanding of spectator ions is very important. The spectator ions are the ions which are found in same form in both the reactant and product sides. So, while writing the net ionic equation, these ions are not involved or simply not counted.

Answer to Problem 56QAP

${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{NH}(aq)+{\text{H}}^{+}(aq)\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{NH}}_2^{+}(aq)$

Explanation of Solution

The dissolved hydrochloric acid first dissociates into its ionic forms under the aqueous medium as follows:

$\text{HCl}(aq)\to {\text{H}}^{+}(aq)+{\text{Cl}}^{-}(aq)$

The lone pair of electron present at the ammine group is attacked by the proton dissociated through the acid.

The ionic reaction for the reaction occurs as

${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{NH}(aq)+{\text{H}}^{+}(aq)+{\text{Cl}}^{-}(aq)\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{NH}}_2^{+}(aq)+{\text{Cl}}^{-}(aq)$

Here, the Cl- ion is spectator ion and removing that will give the balanced net ionic equation as

${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{NH}(aq)+{\text{H}}^{+}(aq)\to {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{NH}}_2^{+}(aq)$

Interpretation Introduction

(b)

Interpretation:

Balanced net ionic reaction needs to be written for the reaction of acetic acid with a solution of barium hydroxide.

Concept introduction:

While writing the net ionic equation, an understanding of spectator ions is very important. The spectator ions are the ions which are found in same form in both the reactant and product sides. So, while writing the net ionic equation, these ions are not involved or simply not counted.

Answer to Problem 56QAP

${\text{CH}}_{\text{3}}\text{COOH}(aq)+{\text{OH}}^{-}(aq)\to {\text{CH}}_{\text{3}}{\text{COO}}^{-}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

Explanation of Solution

The dissolved barium hydroxide first ionizes into its ionic forms under the aqueous medium as follows:

${\text{Ba(OH)}}_{\text{2}}(aq)\to {\text{Ba}}^{+2}(aq)+2{\text{OH}}^{-}(aq)$

The acetic acid dissociates as follows:

${\text{CH}}_{\text{3}}\text{COOH}(aq)\to {\text{CH}}_{\text{3}}{\text{COO}}^{-}(aq)+{\text{H}}^{+}(aq)$

The Ba+2 ion reacts with two molecules of acetate ion to form the corresponding acetate and the water molecule is released.

The ionic reaction for the reaction occurs as

$2{\text{CH}}_{\text{3}}\text{COOH}(aq)+{\text{Ba}}^{2+}(aq)+2{\text{OH}}^{-}(aq)\to 2{\text{CH}}_{\text{3}}{\text{COO}}^{-}(aq)+{\text{Ba}}^{2+}(aq)+2{\text{H}}_{\text{2}}\text{O}(l)$

Here, the Ba+2 ion is spectator ion, and removing that will give the balanced net ionic equation as

$2{\text{CH}}_{\text{3}}\text{COOH}(aq)+2{\text{OH}}^{-}(aq)\to 2{\text{CH}}_{\text{3}}{\text{COO}}^{-}(aq)+2{\text{H}}_{\text{2}}\text{O}(l)$

It is reduced to simple mole ratio as

${\text{CH}}_{\text{3}}\text{COOH}(aq)+{\text{OH}}^{-}(aq)\to {\text{CH}}_{\text{3}}{\text{COO}}^{-}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

Interpretation Introduction

(c)

Interpretation:

Balanced net ionic reaction needs to be written for the reaction of 2-chloropropane with a solution of sodium hydroxide.

Concept introduction:

While writing the net ionic equation, an understanding of spectator ions is very important. The spectator ions are the ions which are found in same form in both the reactant and product sides. So, while writing the net ionic equation, these ions are not involved or simply not counted.

Answer to Problem 56QAP

${\text{CH}}_{\text{3}}{\text{-C(H)(Cl)-CH}}_{\text{3}}(l)+{\text{OH}}^{-}(aq)\to {\text{CH}}_{\text{3}}{\text{-C(H)(OH)-CH}}_{\text{3}}(aq)+{\text{Cl}}^{-}(aq)$

Explanation of Solution

The dissolved sodium hydroxide first ionizes into its ionic forms under the aqueous medium as follows:

$\text{NaOH}(aq)\to {\text{Na}}^{+}(aq)+{\text{OH}}^{-}(aq)$

The 2-chloropropane dissociates as follows:

${\text{CH}}_{\text{3}}{\text{-C(H)(Cl)-CH}}_{\text{3}}(l)\to {\text{CH}}_{\text{3}}{\text{-C}}^{\text{+}}{\text{(H)-CH}}_{\text{3}}(aq)+{\text{Cl}}^{-}(aq)$

The OH- ion is produced by sodium hydroxide adds to the carbocation and forms the corresponding alcohol releasing the sodium chloride molecule. Both the reactions take place together at once.

The ionic reaction for the reaction occurs as

In this reaction, the chlorine of 2-chloropropane is replaced by OH-group as follows:

${\text{CH}}_{\text{3}}{\text{-C(H)(Cl)-CH}}_{\text{3}}(l)+{\text{Na}}^{+}(aq)+{\text{OH}}^{-}(aq)\to {\text{CH}}_{\text{3}}{\text{-C(H)(OH)-CH}}_{\text{3}}(aq)+{\text{Cl}}^{-}(aq)+{\text{Na}}^{+}(aq)$

Here, the Na+ ion is spectator ions and removing that will give the balanced net ionic equation as

${\text{CH}}_{\text{3}}{\text{-C(H)(Cl)-CH}}_{\text{3}}(l)+{\text{OH}}^{-}(aq)\to {\text{CH}}_{\text{3}}{\text{-C(H)(OH)-CH}}_{\text{3}}(aq)+{\text{Cl}}^{-}(aq)$