Question

Complete the equations for the formation of precipitates and write their net ionic equations. (Everything that was added to form the precipiates were aqueous solutions, so would that mean I would add water to some of the equations?)

Ag⁺ (aq) + HCl (aq) = AgCl (s)

Pb²⁺ (aq) + HCl (aq) = PbCl₂ (s)

Pb²⁺ (aq) + K₂CrO₄ (aq) = PbCrO₄ (s)

Ag(NH₃)₂⁺ (aq) + HNO₃ (aq) = AgCl (s)

Pb²⁺ (aq) + NH₃ (aq) = Pb(OH)₂ (s)

Al²⁺ (aq) + NH₃ (aq) = Al(OH)₂ (s)

Fe³⁺ (aq) + NH₃ (aq) = Fe(OH) (s)

Cr³⁺ (aq) NH₃ (aq) = Cr(OH) (s)

Pb(OH)₂ (s) + HSO₄ (aq) = PbSO₄ (s)

Al(OH)₄^- (aq) + NH₄NO₃ (aq) = Al(OH)₃ (s)

CrO₄^2- (aq) + Pb(NO₃)₂ (aq) = PbCrO₄ (s)

Answer 1

The equations are complete but we need to balance them

a) Ag + HCl === AgCl (s) + H⁺ (aq)

Ionic equation goes as follows:

Ag⁺ + H⁺ + Cl⁺ ==== AgCl(s) + H⁺, as you can see the hydrogen appears on both side of the equation, this is called an spectator ion it is there but does not take part on the reaction

net ionic equation is Ag⁺ + Cl^- === AgCl

b) Pb₂⁺ (aq) + 2 HCl (aq) = PbCl2 (s) + 2H⁺, this would be the balanced reaction

net ionic equation is

Pb₂⁺ (aq) + 2Cl^- (aq) = PbCl2 (s), if you make the same analysis made in part a you will find that the hydrogen is just an spectator ion, we just take it out of the equation

c) Pb₂⁺ (aq) + K₂CrO4 (aq) = PbCrO4 (s) + 2K⁺ (aq)

net ionic equation is

Pb₂⁺ (aq) + CrO₄^- (aq) = PbCrO4 (s), spectator ion is K⁺

d) Ag(NH3)₂+ (aq) + HNO₃ (aq) = AgCl (s), this equation is incomplete you should have a Cl on the reactants side in order to get AgCl so:

Ag(NH3)₂Cl (aq) + 2 HNO₃ (aq) = AgCl (s) + 2 NH₄NO₃ or

Ag(NH3)2⁺ + Cl^- (aq) + 2 H⁺ +  2NO₃^- (aq) = AgCl (s) + 2 NH4+   + 2 NO₃^-

net ionic equation is:

Ag⁺ + Cl^- ==== AgCl (s); this is the net ionic equation the other ones are just spectator ions