Chapter 4 Precipitation reactions

Net ionic reactions

General Chemistry IDr. Stone

If a compound is soluble, the solution is clear.

If a compound is insoluble, a solid will form and the solution will be cloudy.

Solid = precipitate Insoluble = precipitate

If a compound is insoluble, a solid will form and you can separate it from the solution by filtration.

Clear, soluble ions

Clear, soluble ions

Insoluble solid, precipitate

What ions are in solution?

A solution of sodium sulfate is mixed with a solution of barium nitrate.

Can an insoluble compound form from the ions in solution?

Na2SO4(aq) —> 2 Na+(aq) + SO4-2(aq)

Ba(NO3)2(aq) —> Ba+2(aq) + 2NO3-(aq)

Na2SO4(aq) —> 2 Na+(aq) + SO4-2(aq)

Ba(NO3)2(aq) —> Ba+2(aq) + 2NO3-(aq)

Total ionic Na2SO4(aq) + Ba(NO3)2(aq)—> ?

2 Na+(aq)+SO4-2(aq) + Ba+2(aq)+ 2NO3-(aq)—> BaSO4(s) +2 Na+(aq) + 2NO3-(aq)

Net ionic 2 Na+(aq)+SO4-2(aq) + Ba+2(aq)+ 2NO3-(aq)—> BaSO4(s) +2 Na+(aq) + 2NO3-(aq)

Remove spectators

SO4-2(aq) + Ba+2(aq)+—> BaSO4(s)

Barium sulfate

Complete the following table for precipitation reactions. If a precipitate forms, then write the correct formula, if no precipitate forms, write NR.

Ions Chloride Carbonate Sulfate Phosphate Nitrate Hydroxide

Potassium

Lead (II)

Barium

Silver(I)

Iron (II)

Magnesium

Solubility Table

AgOH(s)

2.Write the net ionic equation for the reaction that takes place between a solution of potassium chloride and a solution of lead nitrate.

a. 2KCl(aq) + Pb(NO3)2(aq) —> 2KNO3(aq) + PbCl2(aq) b. 2K+(aq) + 2Cl- (aq) 2Pb2+(aq) —> 2NO3-(aq) + 2KNO3(aq) + PbCl2(aq) c. 2K+(aq) + 2NO3-(aq) —>2KNO3(aq) d. 2K+(aq) + 2Cl- (aq) 2Pb2+(aq) —>2NO3-(aq) 2K+(aq) + 2NO3-(aq) + PbCl2(aq) e. Pb2+(aq) + 2Cl-(aq) —> PbCl2(s)

3. What mass of barium sulfate (molar mass = 233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of a 0.150 M solution of iron(III) sulfate?

3. What mass of barium sulfate (molar mass = 233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of a 0.150 M solution of iron(III) sulfate?

a. 7.59 g b. 3.65 g c. 13.1 g d. 18.8 g e. 4.37 g

Moles Ba+2 = 125 mL x 1 L. x 0.150 moles = 1.875 x 10-2 moles Ba+2 1000 mL L1.875 x 10-2 moles Ba+2 x 1 mole BaSO4 x 233 g = 4.37 g BaSO4

1 mole Ba+2 mole BaSO4

Moles SO4-2 = 125 mL x 1 L. x 0.150 moles x 3 moles SO4-2 = 5.625 x 10-2 moles 1000 mL L 1 mole Fe2(SO4)3

5.625 x 10-2 moles SO4-2 x 1 mole BaSO4 x 233 g = 13.1 g BaSO4

1 mole SO4-2 mole BaSO4

Limiting!

4. Which of the following compounds are soluble in water? PbCO3 , FeSO4 , AgNO3, NH4OH I II III IV

a. I and III c. II, III and IV e. All re soluble. b. II and IV d. I, II, and IV