Chapter 9, Problem 122A

Apply Write the chemical equations and net ionic equations for each of the following reactions that might occur in aqueous solutions Ifa reaction does not occur, write NR in place of the products. Magnesium phosphate precipitates in an aqueous solution.

1. ${\text{KNO}}_{\text{3}}+\text{CsCl}\to$

${\text{Ca(OH)}}_{\text{2}}+\text{KCN}\to$

${\text{LiPO}}_{\text{4}}+{\text{MgSO}}_{\text{4}}\to$

$\text{HBrO}+\text{NaOH}\to$

Interpretation Introduction

(a)

Interpretation:

The chemical reaction and net ionic equations needs to be determined.

${\text{KNO}}_{\text{3}}\text{+CsCl}\to$

Concept introduction:

Net ionic equations are ionic equations that include only the particles that participate in the reaction. Net ionic equations are written from complete ionic equations by removing all spectator ions. Ions that do not participate in a reaction are called spectator ions and are not usually shown in ionic equations.

Answer to Problem 122A

The reaction does not occur.

Explanation of Solution

In this reaction both hypothetical products are ionic and dissolved into ions in solution. Therefore, there is no reaction take place.

$\begin{array}{l}{\text{KNO}}_{\text{3}}\left(\text{aq}\right)\text{+CsCl}\left(\text{aq}\right)\to \text{KCl}\left(\text{aq}\right){\text{+CsNO}}_{\text{3}}\left(\text{aq}\right)\\ {\text{KNO}}_{\text{3}}\left(\text{aq}\right)\text{+CsCl}\left(\text{aq}\right)\to \text{NR}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The chemical reaction and net ionic equations needs to be determined.

$\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\text{+KCN}\to$

Concept introduction:

Net ionic equations are ionic equations that include only the particles that participate in the reaction. Net ionic equations are written from complete ionic equations by removing all spectator ions. Ions that do not participate in a reaction are called spectator ions and are not usually shown in ionic equations.

Answer to Problem 122A

The reaction does not occur.

Explanation of Solution

In this reaction both hypothetical products are ionic and dissolved into ions in solution. Therefore, there is no reaction take place.

$\begin{array}{l}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{aq}\right)\text{+2KCN}\left(\text{aq}\right)\to \text{Ca}{\left(\text{CN}\right)}_{\text{2}}\left(\text{aq}\right)\text{+2KOH}\left(\text{aq}\right)\\ \text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{aq}\right)\text{+KCN}\left(\text{aq}\right)\to \text{NR}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The chemical reaction and net ionic equations needs to be determined.

${\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+MgSO}}_4\to$

Concept introduction:

Net ionic equations are ionic equations that include only the particles that participate in the reaction. Net ionic equations are written from complete ionic equations by removing all spectator ions. Ions that do not participate in a reaction are called spectator ions and are not usually shown in ionic equations.

Answer to Problem 122A

The net ionic equation is as follows:

${\text{2PO}}_{\text{4}}{}^{3-}\left(\text{aq}\right){\text{+3Mg}}^{2+}\left(\text{aq}\right)\to \text{Mg}{}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)$

Explanation of Solution

In this reaction a solid compound is formed. So, the reaction is take place. Lithium has a single positive charge, phosphate has 3- charge, magnesium and sulfate has 2+ and 2- charge respectively. The solid product has 3 Mg atoms, so put 3 as the coefficient of second reactant. The solid product also has two ${\text{PO}}_{\text{4}}{}^{2-}$ so give coefficient 2 to the first reactant. Now, balance Li and ${\text{SO}}_{\text{4}}{}^{2-}$ by giving coefficeint 3 in front of first product. The balance reaction is as follows:

${\text{2Li}}_3{\text{PO}}_{\text{4}}\left(\text{aq}\right){\text{+3MgSO}}_4\left(\text{aq}\right)\to 3{\text{Li}}_2{\text{SO}}_{\text{4}}\left(\text{aq}\right)\text{+Mg}{}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)$

The ionic equation is as follows:

${\text{6Li}}^{+}\left(\text{aq}\right){\text{+2PO}}_{\text{4}}{}^{3-}\left(\text{aq}\right){\text{+3Mg}}^{2+}\left(\text{aq}\right){\text{+3SO}}_4{}^{2-}\left(\text{aq}\right)\to 6{\text{Li}}^{2+}{\text{+3SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)\text{+Mg}{}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)$

The net ionic equation is as follows:

${\text{2PO}}_{\text{4}}{}^{3-}\left(\text{aq}\right){\text{+3Mg}}^{2+}\left(\text{aq}\right)\to \text{Mg}{}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(\text{s}\right)$

Interpretation Introduction

(d)

Interpretation:

The chemical reaction and net ionic equations needs to be determined.

$\text{HBrO+NaOH}\to$

Concept introduction:

Net ionic equations are ionic equations that include only the particles that participate in the reaction. Net ionic equations are written from complete ionic equations by removing all spectator ions. Ions that do not participate in a reaction are called spectator ions and are not usually shown in ionic equations.

Answer to Problem 122A

The net ionic equation is as follows:

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

Explanation of Solution

In this reaction water is produced so, it does occur. All the cations have charge +1 and anions have charge -1. Therefore, the reaction is balanced.

$\text{HBrO}\left(\text{aq}\right)\text{+NaOH}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+NaBrO}\left(\text{aq}\right)$

The ionic equation is as follows:

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

The net ionic equation is as follows:

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