Chapter 7, Problem 8CR

ummarize the simple solubility rules for ionic compounds. How do we use these rules in determining the identity of the solid formed in a precipitation reaction? Give examples including balanced complete and net ionic equations.

Interpretation Introduction

Interpretation:

Solubility rule of ionic compound. Examples of complete and net ionic equation if precipitation reaction.

Concept Introduction:

According to solubility rule precipitate formation depends upon how the two ions reacts.

Solubility also depends upon temperature and concentration.

Solubility Rule:

1. Elements present in group I are soluble.
2. Nitrate salt are soluble.
3. Salts with${\text{Cl}}^{\text{-}}{\text{ , Br}}^{\text{-}}{\text{ , I}}^{\text{-}}$ are soluble. But with${\text{Ag}}^{\text{+}}{\text{, Pb}}^{\text{2+}}{\text{, Hg}}_{\text{2}}^{\text{2+}}$ it is insoluble.
4. Mostly salts of silver are insoluble. Exception${\text{AgNO}}_{\text{3}}{\text{, Ag(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2)}}$ are only soluble.
5. Sulphate salts are soluble. Exception are lead sulfate, barium sulfate, strontium sulfate.
6. Elements which forms hydroxide with group I elements are soluble. Hydroxide formation with groupII are slightly soluble.
7. Sulfides salt are insoluble.
8. Chromates and phosphates salts are insoluble.

Precipitate is the solid mass which is floating on mixing of two or more compounds in a solutions. This reaction is called precipitation reaction. The remaining liquid on which precipitate is floating is called supernatant liquid. The precipitate and supernatant liquid can be separated by filtration, centrifugation or decantation methods.

Net ionic equation: gives the only ionic species which are involved in the chemical reaction.

Given below is the standard equation for precipitation or double displacement reaction.

$\text{AB}(aq)\text{+CD}(\text{a}\text{q}\text{)}\to \text{AD}(aq)\text{+CB}(s)$

Step one for ionic equation is to write separately each soluble reactants and products in to there cationic and anionic forms. Precipitate is insoluble in water. Resulting equation will be:${\text{A}}^{\text{+}}(aq){\text{+B}}^{\text{-}}(aq){\text{+C}}^{\text{+}}(aq){\text{+D}}^{-}(aq)\to {\text{A}}^{\text{+}}(aq){\text{+D}}^{-}(aq)+\text{CB}(s)$

Here in equation${\text{A}}^{\text{+}}$ &${\text{D}}^{-}$ ions are present on both sides of the equation. These ion are spectator ions because they remain unchanged throughout the reaction. These separator ions remains same, these can be eliminated to show the net ionic reaction.

${\text{C}}^{\text{+}}(aq){\text{+B}}^{\text{-}}(aq)\to \text{CB}(s)$

Molecular equation is also known as balanced complete reaction. In molecular equation all ionic compounds are represented as neutral compounds i.e is states of each compound is unknown.

Answer to Problem 8CR

Solubility rule of ionic compound are as follows:

1. Elements present in group I are soluble.
2. Nitrate salt are soluble.
3. Salts with${\text{Cl}}^{\text{-}}{\text{ , Br}}^{\text{-}}{\text{ , I}}^{\text{-}}$ are soluble. But with${\text{Ag}}^{\text{+}}{\text{, Pb}}^{\text{2+}}{\text{, Hg}}_{\text{2}}^{\text{2+}}$ it is insoluble.
4. Mostly salts of silver are insoluble. Exception${\text{AgNO}}_{\text{3}}{\text{, Ag(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2)}}$ are only soluble.
5. Sulphate salts are soluble. Exception are lead sulfate, barium sulfate, strontium sulfate.
6. Elements which forms hydroxide with group I elements are soluble. Hydroxide formation with groupII are slightly soluble.
7. Sulfides salt are insoluble.

Chromates and phosphates salts are insoluble Precipitation reaction are:

a)Molecular:${\text{2Na}}_{\text{(aq)}}^{\text{+}}{\text{+2OH}}_{\text{(aq)}}{\text{+Mg}}_{\text{(aq)}}^{\text{2+}}{\text{+2Cl}}_{\text{(aq)}}^{\text{-}}\to {\text{Mg(OH)}}_{\text{2}}{}_{\text{(s)}}{\text{+2Na}}_{\text{(aq)}}^{\text{+}}{\text{+2Cl}}_{\text{(aq)}}^{\text{-}}$

Ionic:${\text{Mg}}_{\text{(aq)}}^{\text{2+}}{\text{+2OH}}_{\text{(aq)}}\to {\text{Mg(OH)}}_{\text{2}}{}_{\text{(s)}}$

b) Molecular:${\text{Fe(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)+NaOH(aq)}\to {\text{Fe(OH)}}_{\text{3}}{\text{(s)+NaNO}}_{\text{3}}(aq)$

Ionic:${\text{Fe}}^{\text{3+}}{\text{(aq)+3OH}}^{\text{-}}\text{(aq)}\to {\text{Fe(OH)}}_{\text{3}}\text{(s)}$

c) Molecular: ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{(aq)+BaCl}}_{\text{2}}(aq)\to {\text{AlCl}}_{\text{3}}{\text{(aq)+BaSO}}_{\text{4}}\text{(s)}$

Ionic:${\text{Ba}}^{\text{2+}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}\to {\text{BaSO}}_{\text{4}}\text{(s)}$

d) Molecular:${\text{CaCl}}_{\text{2}}{\text{(aq)+NaPO}}_{\text{4}}\text{(aq)}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s)+NaCl(aq)}$

Ionic:${\text{Ca}}^{\text{2+}}{\text{(aq)+PO}}_{\text{4}}^{\text{3-}}\text{(aq)}\to {\text{Ca}}_3{(\text{P}{\text{O}}_4)}_2(s)$.

Explanation of Solution

Solubility rule of ionic compound are as follows:

1. Elements present in group I are soluble.
2. Nitrate salt are soluble.
3. Salts with${\text{Cl}}^{\text{-}}{\text{ , Br}}^{\text{-}}{\text{ , I}}^{\text{-}}$ are soluble. But with${\text{Ag}}^{\text{+}}{\text{, Pb}}^{\text{2+}}{\text{, Hg}}_{\text{2}}^{\text{2+}}$ it is insoluble.
4. Mostly salts of silver are insoluble. Exception${\text{AgNO}}_{\text{3}}{\text{, Ag(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2)}}$
5. are only soluble.
6. Sulphate salts are soluble. Exception are lead sulfate, barium sulfate, strontium sulfate.
7. Elements which forms hydroxide with group I elements are soluble. Hydroxide formation with groupII are slightly soluble.
8. Sulfides salt are insoluble.
9. Chromates and phosphates salts are insoluble.

Precipitation reaction occur by replacement of anion of cations. Example explains the reaction

$\text{AB + CD}\to \text{AD + CB}$

A combines with D

C combines with B.

Both the ion interchanges there companion according to their charges they have.

Below are some examples in which precipitation reaction helps in completing molecular equations and by solubility rule we can which reaction precipitate will form.

Examples of precipitation reaction are:

${\text{NaOH(aq)+MgCl}}_{\text{2}}\text{(aq)}\to$

Balances equation:${\text{2NaOH(aq)+MgCl}}_{\text{2}}\text{(aq)}\to 2{\text{NaCl+Mg(OH)}}_{\text{2}}$

Cations and anions exchanges there partner.

${\text{Na}}^{\text{+}}$ will combines with${\text{Cl}}^{\text{-}}$.

${\text{Mg}}^{\text{+}}$ combines with${\text{OH}}^{\text{-}}$ (According to double replacement reaction)

According to solubility rule, group I elements are soluble and Hydroxide are insoluble.

So, NaCl is aqueous and${\text{Mg(OH)}}_{\text{2}}$ is solid.

Spectator ions remain unchanged throughout the reaction. These separator ions remains same, these can be eliminated to show the net ionic reaction.

${\text{2Na}}_{\text{(aq)}}^{\text{+}}{\text{+2OH}}_{\text{(aq)}}{\text{+Mg}}_{\text{(aq)}}^{\text{2+}}{\text{+2Cl}}_{\text{(aq)}}^{\text{-}}\to {\text{Mg(OH)}}_{\text{2}}{}_{\text{(s)}}{\text{+2Na}}_{\text{(aq)}}^{\text{+}}{\text{+2Cl}}_{\text{(aq)}}^{\text{-}}$

Sodium and chloride are in aqueous state only. Hence final net ionic equation is :

${\text{Mg}}_{\text{(aq)}}^{\text{2+}}{\text{+2OH}}_{\text{(aq)}}\to {\text{Mg(OH)}}_{\text{2}}{}_{\text{(s)}}$

(b)

${\text{Fe(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)+NaOH(aq)}\to$

By double displacement reaction final equation of the reaction is:

${\text{Fe(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)+NaOH(aq)}\to {\text{Fe(OH)}}_{\text{3}}{\text{(s)+NaNO}}_{\text{3}}(aq)$

According to solubility rule${\text{NaNO}}_{\text{3}}$ is soluble as nitrates are soluble.${\text{Fe(OH)}}_{\text{3}}$ is insoluble because Fe is the ion which is soluble. The ionic equation is as follows:

${\text{Fe}}^{\text{3+}}{\text{(aq)+NO}}_{\text{3}}^{\text{-}}{\text{(aq)+Na}}^{\text{+}}{\text{(aq)+3OH}}^{\text{-}}\text{(aq)}\to {\text{Fe(OH)}}_{\text{3}}{\text{(s)+Na}}^{\text{+}}{\text{(aq)+NO}}_{\text{3}}^{\text{-}}$

Canceling out the separator ion forms the final equation:

${\text{Fe}}^{\text{3+}}{\text{(aq)+3OH}}^{\text{-}}\text{(aq)}\to {\text{Fe(OH)}}_{\text{3}}\text{(s)}$

(c)

${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{(aq)+BaCl}}_{\text{2}}(aq)\to$

By double displacement reaction final equation of the reaction is:

${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{(aq)+BaCl}}_{\text{2}}(aq)\to {\text{AlCl}}_{\text{3}}{\text{(aq)+BaSO}}_{\text{4}}\text{(s)}$

According to solubility rule point 5, barium sulfate is insoluble. Aluminum chloride are soluble as chlorides(halide) are soluble. The ionic equation is as follows:

${\text{2Al}}^{\text{3+}}{\text{(aq)+6Cl}}^{\text{-}}{\text{(aq)+3Ba}}^{\text{2+}}{\text{(aq)+3SO}}_{\text{4}}^{\text{2-}}\text{(aq)}\to {\text{2Al}}^{\text{3+}}{\text{(aq)+6Cl}}^{\text{-}}{\text{(aq)+3BaSO}}_{\text{4}}\text{(s)}$

Cancelling out the separator ions we get:${\text{Ba}}^{\text{2+}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}\to {\text{BaSO}}_{\text{4}}\text{(s)}$

(d)

${\text{CaCl}}_{\text{2}}{\text{(aq)+NaPO}}_{\text{4}}\text{(aq)}\to$

By double displacement reaction final equation of the reaction is:

${\text{CaCl}}_{\text{2}}{\text{(aq)+NaPO}}_{\text{4}}\text{(aq)}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s)+NaCl(aq)}$

According to solubility rule point 1, NaCl is soluble in (sodium is group I element.).Calcium phosphate is insoluble in water (point 5 in solubility rule.). The ionic equation is:${\text{Ca}}^{\text{2+}}{\text{(aq)+Cl}}^{\text{-}}{\text{(aq)+Na}}^{\text{+}}{\text{(aq)+PO}}_{\text{4}}^{\text{3-}}\text{(aq)}\to {\text{Ca}}_3{(\text{P}{\text{O}}_4)}_2(s)+{\text{Na}}^{\text{+}}\text{(aq)}+{\text{Cl}}^{\text{-}}\text{(aq)}$

Spectator ion are cancelled.

${\text{Ca}}^{\text{2+}}{\text{(aq)+PO}}_{\text{4}}^{\text{3-}}\text{(aq)}\to {\text{Ca}}_3{(\text{P}{\text{O}}_4)}_2(s)$.

Conclusion

Thus, solubility rule of ionic compound are as follows:

1. Elements present in group I are soluble.
2. Nitrate salt are soluble.
3. Salts with${\text{Cl}}^{\text{-}}{\text{ , Br}}^{\text{-}}{\text{ , I}}^{\text{-}}$ are soluble. But with${\text{Ag}}^{\text{+}}{\text{, Pb}}^{\text{2+}}{\text{, Hg}}_{\text{2}}^{\text{2+}}$ it is insoluble.
4. Mostly salts of silver are insoluble. Exception${\text{AgNO}}_{\text{3}}{\text{, Ag(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2)}}$
5. are only soluble.
6. Sulphate salts are soluble. Exception are lead sulfate, barium sulfate, strontium sulfate.
7. Elements which forms hydroxide with group I elements are soluble. Hydroxide formation with groupII are slightly soluble.
8. Sulfides salt are insoluble.

Chromates and phosphates salts are insoluble.

Molecular:${\text{2Na}}_{\text{(aq)}}^{\text{+}}{\text{+2OH}}_{\text{(aq)}}{\text{+Mg}}_{\text{(aq)}}^{\text{2+}}{\text{+2Cl}}_{\text{(aq)}}^{\text{-}}\to {\text{Mg(OH)}}_{\text{2}}{}_{\text{(s)}}{\text{+2Na}}_{\text{(aq)}}^{\text{+}}{\text{+2Cl}}_{\text{(aq)}}^{\text{-}}$

Ionic:${\text{Mg}}_{\text{(aq)}}^{\text{2+}}{\text{+2OH}}_{\text{(aq)}}\to {\text{Mg(OH)}}_{\text{2}}{}_{\text{(s)}}$

b) Molecular:${\text{Fe(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)+NaOH(aq)}\to {\text{Fe(OH)}}_{\text{3}}{\text{(s)+NaNO}}_{\text{3}}(aq)$

Ionic:${\text{Fe}}^{\text{3+}}{\text{(aq)+3OH}}^{\text{-}}\text{(aq)}\to {\text{Fe(OH)}}_{\text{3}}\text{(s)}$

c) Molecular: ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{(aq)+BaCl}}_{\text{2}}(aq)\to {\text{AlCl}}_{\text{3}}{\text{(aq)+BaSO}}_{\text{4}}\text{(s)}$

Ionic:${\text{Ba}}^{\text{2+}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}\to {\text{BaSO}}_{\text{4}}\text{(s)}$

d) Molecular:${\text{CaCl}}_{\text{2}}{\text{(aq)+NaPO}}_{\text{4}}\text{(aq)}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\text{(s)+NaCl(aq)}$

Ionic:${\text{Ca}}^{\text{2+}}{\text{(aq)+PO}}_{\text{4}}^{\text{3-}}\text{(aq)}\to {\text{Ca}}_3{(\text{P}{\text{O}}_4)}_2(s)$.