Chapter 10, Problem 10.52P

(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.  The reaction that does not form precipitate should be written as no reaction.

Concept Introduction:

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 sulphuric acid and calcium (II) perchlorate is calcium sulphate and perchloric acid.  By Rule 6, calcium sulphate is insoluble in water.  The completed equation is,

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\text{+Ca}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}\left(\text{aq}\right)}\stackrel{}{\to }{\text{HClO}}_{\text{4(aq)}}{\text{+CaSO}}_{\text{4(s)}}$

The balanced equation is,

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\text{+Ca}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}\left(\text{aq}\right)}\stackrel{}{\to }{\text{2HClO}}_{\text{4(aq)}}{\text{+CaSO}}_{\text{4(s)}}$

The complete ionic equation is,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{CaSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

The net ionic equation is,

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

(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.  The reaction that does not form precipitate should be written as no reaction.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The product of the reaction between silver nitrate and sodium perchlorate is sodium nitrate and silver perchlorate.  The completed equation is,

  ${\text{AgNO}}_{\text{3}\left(\text{aq}\right)}{\text{+NaClO}}_{\text{4}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{AgClO}}_{\text{4(aq)}}{\text{+NaNO}}_{\text{3(aq)}}$

The balanced equation is,

  ${\text{AgNO}}_{\text{3}\left(\text{aq}\right)}{\text{+NaClO}}_{\text{4}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{AgClO}}_{\text{4(aq)}}{\text{+NaNO}}_{\text{3(aq)}}$

The complete ionic equation is,

  ${\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+ClO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

No precipitate would form in the reaction between the silver nitrate and sodium perchlorate because both silver nitrate and sodium perchlorate are water-soluble (Rule 1 and Rule 2).  No precipitate occurs because there is no driving force for the reaction to take place.

  ${\text{AgNO}}_{\text{3}\left(\text{aq}\right)}{\text{+NaClO}}_{\text{4}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }\text{No reaction}$

(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.  The reaction that does not form precipitate should be written as no reaction.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The product of the reaction between mercury nitrate and sodium benzoate is mercury benzoate and sodium nitrate.  By Rule 3, mercury benzoate is insoluble in water.  The completed equation is,

  ${\text{Hg}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+NaC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2(aq)}}\stackrel{}{\to }{\text{Hg}}_{\text{2}}{\left({\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\right)}_{\text{2(s)}}{\text{+NaNO}}_{\text{3(aq)}}$

The balanced equation is,

  ${\text{Hg}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{+2NaC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2(aq)}}\stackrel{}{\to }{\text{Hg}}_{\text{2}}{\left({\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\right)}_{\text{2(s)}}{\text{+2NaNO}}_{\text{3(aq)}}$

The complete ionic equation is,

  ${\text{2Hg}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+Hg}}_{\text{2}}{\left({\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\right)}_{\text{2(s)}}$

The net ionic equation is,

  ${\text{2Hg}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Hg}}_{\text{2}}{\left({\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\right)}_{\text{2(s)}}$

(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.  The reaction that does not form precipitate should be written as no reaction.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The product of the reaction between lead acetate and potassium bromide is lead bromide and potassium acetate.  By Rule (3), lead bromide is insoluble.  The completed equation is,

  $\text{Pb}{\left({\text{CH}}_{\text{3}}\text{COO}\right)}_{\text{2(aq)}}{\text{+KBr}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbBr}}_{\text{2(s)}}{\text{+CH}}_{\text{3}}{\text{COOK}}_{\left(\text{aq}\right)}$

The balanced equation is,

  $\text{Pb}{\left({\text{CH}}_{\text{3}}\text{COO}\right)}_{\text{2(aq)}}{\text{+2KBr}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbBr}}_{\text{2(s)}}{\text{+2CH}}_{\text{3}}{\text{COOK}}_{\left(\text{aq}\right)}$

The complete ionic equation is,

  ${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbBr}}_{\text{2(s)}}{\text{+2CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+2K}}^{\text{+}}{}_{\left(\text{aq}\right)}$

The net ionic equation is,

  ${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbBr}}_{\text{2(s)}}$