Chapter 7, Problem 6CR

xplain to your friend what chemists mean by a precipitation reaction. What is the driving force in a precipitation reaction? Using the information provided about solubility in these chapters, write balanced molecular and net ionic equations for five examples of precipitation reactions.

Interpretation Introduction

Interpretation:

Precipitation reaction and driving force for it.

Molecular and net ionic reaction for 5 precipitation reaction.

Concept Introduction:

Precipitate is the solid mass which is floating on mixing of two or more compounds in a solution. 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.

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 are insoluble.

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.

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

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 ions are spectator ions because they do not consume in the chemical reaction or remain the same. 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 6CR

Precipitate is the solid mass which is floating on mixing of two or more compounds in a solution. This reaction is called precipitation reaction.

The driving force for precipitation reaction is the formation of in soluble product.

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:${\text{CoCl}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{CoSO}}_{\text{4(aq)}}{\text{+2NaCl}}_{\text{(aq)}}$

Ionic:${\text{Co}}^{\text{2-}}{\text{(aq)+2Cl}}^{\text{-}}{\text{(aq)+2Na}}^{\text{+}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}\to {\text{Co}}^{\text{2-}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}+{\text{2Na}}^{\text{+}}\text{(aq)}+{\text{2Cl}}^{\text{-}}\text{(aq)}$

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

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

e) 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

Precipitation reaction occur between cation and anions in aqueous solution combines to form a solid mass which is ionic in nature and called precipitate

All reactions which are taking places in aqueous solution do not form precipitate. This depend upon the solubility of the compounds. Formation of precipitate depends upon whether the cation and anion present in the solution will combine with each other or not aqueous solution.

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.

The driving force for precipitation reaction is the formation of in soluble product.

Examples of precipitation reaction are:

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

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.

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

Now separate the reaction into there ionic forms which exits in aqueous solutions. Electrical charges and no. of atoms are same on both side

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

Element which occur on both the side of equation and are unchanged as they are soluble. Sodium and chloride are in aqueous state only. Hence final net ionic equation is :

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

${\text{CoCl}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to$

Balanced equation: ${\text{CoCl}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{CoSO}}_{\text{4(aq)}}{\text{+2NaCl}}_{\text{(aq)}}$

Cobalt combines with sulfate to form cobalt sulfate (${\text{CoSO}}_{\text{4}}$ ). Sodium combines with chloride ($\text{NaCl}$ ).

Separate the species into there ionic forms and balanced the equation:

${\text{Co}}^{\text{2-}}{\text{(aq)+2Cl}}^{\text{-}}{\text{(aq)+2Na}}^{\text{+}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}\to {\text{Co}}^{\text{2-}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}+{\text{2Na}}^{\text{+}}\text{(aq)}+{\text{2Cl}}^{\text{-}}\text{(aq)}$

Species on both side are same& canceled out of net ionic equation as all the species are in aqueous state. Therefore no precipitate is formed. This is no precipitate reaction.

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

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:

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

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

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

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

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.

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

Conclusion

Thus, precipitate is the solid mass which is floating on mixing of two or more compounds in a solution. This reaction is called precipitation reaction.

The driving force for precipitation reaction is the formation of in soluble product.

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:${\text{CoCl}}_{\text{2(aq)}}{\text{+Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{CoSO}}_{\text{4(aq)}}{\text{+2NaCl}}_{\text{(aq)}}$

Ionic:${\text{Co}}^{\text{2-}}{\text{(aq)+2Cl}}^{\text{-}}{\text{(aq)+2Na}}^{\text{+}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}\to {\text{Co}}^{\text{2-}}{\text{(aq)+SO}}_{\text{4}}^{\text{2-}}\text{(aq)}+{\text{2Na}}^{\text{+}}\text{(aq)}+{\text{2Cl}}^{\text{-}}\text{(aq)}$

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

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

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