Predicting precipitation reaction:Solved example


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(a) The formulas of the compounds are NaCl and Fe(NO3)2.Exchanging anions, you get sodium nitrate ,NaNO3 and iron (II) chloride,FeCl2.The equation for the exchange reaction is

NaCl + Fe(NO3)2 ------------> NaNO3 + FeCl2 ( not balanced)

note the NaCl are soluble as per rule 1 given in the basic solubility rule table in our download section at mychemistry  and NaNo3 are soluble as per rule 2 of the same table.FeCl2 is soluble as per rule 3 of this table there are some exceptions but that does not include FeCl2 .Also, Fe(NO3)2 is soluble as per rule 2 .

Since there is no precipitate,no reaction occurs.You obtain simply an aqueous solution of the four different ions (Na+,Cl-,Fe2+ and NO3- ) For the answer, we write

NaCl (aq) + Fe(NO3)2 (aq) ....................> NR

(B) The formula of the compound are Al2(SO4)2 AND NaOH .exchanging anions you get Al(OH)3 and Na2So4 .The equation for exchange reaction is.

Al2(SO4)3 + NaOH  ----------------> Al(OH)3 + Na2SO4  ( not balanced)

From table you see Al2(SO4)3 is soluble as per rule 4 ,NaOH and Na2SO4 as per rule 1 and  Al(OH)3 is insoluble as per rule 5 . so,

Al2(SO4)3 (aq) + 6NaOH (aq) ----------------> 2Al(OH)3 (s) + 3Na2SO4 (aq)

To get the ionic equation you write the strong electrolytes ( here soluble ionic compounds) as ions in aqueous solution and cancel spectator ions.

2Al3+(aq) + 3 SO42-(aq) + 6Na+(aq) + 6OH-(aq)  -------------> 2Al(OH)3 (s) + 6Na+(aq) +  3SO42-(aq)    

The net ionic equation is:

  Al3+(aq)  +  3OH-(aq)  ------------>  Al(OH)3 (s)