Chapter 10, Problem 10.51P

(a)

Interpretation Introduction

Interpretation:

The products of the reaction have to be predicted.  The complete and balanced equation along with corresponding net ionic equation has to be written.  If the reaction does not form precipitate, no reaction should be written.

Concept Introduction:

Double replacement reactions:  This type of reactions occurs when the cation and the anions switch between two reactants to form new products.  In order for the reaction to occur, one of the products is usually a solid precipitate, a gas or a molecular compound.

Solubility Rules:

1. 1) Most salts of metal alkali and ammonium salts are soluble.
2. 2) Most salts of nitrates, acetates and perchlorates are soluble.
3. 3) Most salts of silver, lead and mercury (I) are insoluble.
4. 4) Most salts of chlorides, bromides and iodides are soluble.
5. 5) Most salts of carbonates, chromates, sulphides, oxides, phosphates and hydroxide are insoluble (except for hydroxides of ${\text{Ba}}^{\text{2+}}{\text{,Ca}}^{\text{2+}}\text{and}{\text{Sr}}^{\text{2+}}$ are slightly soluble).
6. 6) Most salts of sulphates are soluble, except for calcium sulphate, barium sulphate and strontium sulphate that are insoluble.

Explanation of Solution

The product of the reaction between copper chloride and sodium sulphide is copper sulphide and sodium chloride.  By Rule 5, copper sulphide is insoluble in water.  The completed equation is,

  ${\text{CuCl}}_{2\left(\text{aq}\right)}{\text{+Na}}_2{\text{S}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{CuS}}_{\left(\text{s}\right)}{\text{+NaCl}}_{\left(\text{aq}\right)}$

The balanced equation is,

  ${\text{CuCl}}_{2\left(\text{aq}\right)}{\text{+Na}}_2{\text{S}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{CuS}}_{\left(\text{s}\right)}{\text{+2NaCl}}_{\left(\text{aq}\right)}$

The complete ionic equation is,

  ${\text{Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{CuS}}_{\left(\text{s}\right)}{\text{+2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The net ionic equation is,

  ${\text{Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{CuS}}_{\left(\text{s}\right)}$

(b)

Interpretation Introduction

Interpretation:

The products of the reaction have to be predicted.  The complete and balanced equation along with corresponding net ionic equation has to be written.  If the reaction does not form precipitate, no reaction should be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The product of the reaction between magnesium bromide and potassium carbonate is magnesium carbonate and potassium bromide.  By Rule 5, magnesium carbonate is insoluble in water.  The completed equation is,

  ${\text{MgBr}}_{\text{2}\left(\text{aq}\right)}{\text{+K}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{MgCO}}_{\text{3(s)}}{\text{+KBr}}_{\left(\text{aq}\right)}$

The balanced equation is,

  ${\text{MgBr}}_{\text{2}\left(\text{aq}\right)}{\text{+K}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{MgCO}}_{\text{3(s)}}{\text{+2KBr}}_{\left(\text{aq}\right)}$

The complete ionic equation is,

  ${\text{Mg}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Br}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{MgCO}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Br}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The net ionic equation is,

  ${\text{Mg}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{MgCO}}_{\text{3}}{}_{\left(\text{s}\right)}$

(c)

Interpretation Introduction

The products of the reaction have to be predicted.  The complete and balanced equation along with corresponding net ionic equation has to be written.  If the reaction does not form precipitate, no reaction should be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The product of the reaction between barium chloride and potassium sulphate is barium sulphate and potassium chloride.  By Rule 6, barium sulphate is insoluble in water.  The completed equation is,

  ${\text{BaCl}}_{\text{2}\left(\text{aq}\right)}{\text{+K}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{BaSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+KCl}}_{\left(\text{aq}\right)}$

The balanced equation is,

  ${\text{BaCl}}_{\text{2}\left(\text{aq}\right)}{\text{+K}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{BaSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2KCl}}_{\left(\text{aq}\right)}$

The complete ionic equation is,

  ${\text{Ba}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{BaSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$

The net ionic equation is,

  ${\text{Ba}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{BaSO}}_{\text{4}}{}_{\left(\text{s}\right)}$

(d)

Interpretation Introduction

The products of the reaction have to be predicted.  The complete and balanced equation along with corresponding net ionic equation has to be written.  If the reaction does not form precipitate, no reaction should be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The product of the reaction between mercury nitrate and potassium chloride is ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ and potassium nitrate.  By Rule (3), ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is insoluble in water.  The completed equation is,

  ${\text{Hg}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}\left(\text{aq}\right)}{\text{+KCl}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{KCl}}_{\text{(aq)}}{\text{+Hg}}_{\text{2}}{\text{Cl}}_{\text{2(s)}}$

The balanced equation is,

  ${\text{Hg}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}\left(\text{aq}\right)}{\text{+2KCl}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2KNO}}_{\text{3}}{}_{\text{(aq)}}{\text{+Hg}}_{\text{2}}{\text{Cl}}_{\text{2(s)}}$

The complete ionic equation is,

  ${\text{2Hg}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2NO}}^{\text{-}}{}_{\text{3(aq)}}{\text{+Hg}}_{\text{2}}{\text{Cl}}_{\text{2(s)}}$

The ionic equation is,

  ${\text{2Hg}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2(s)}}$