chemistry precipitation reactions will help your teaching

1. Chapter 4 –
Precipitation Reactions

2. Types of Reactions – PPT
• Precipitation Reactions – mix two aqueous
solutions and one product is insoluble
• Precipitate – an insoluble solid compound
formed during a chemical reaction
• Use the solubility rules to determine if a
product is soluble or insoluble
– Yes, you must memorize these

3. The Regents Way….

4. Solubility Rules
Soluble:
1. Alkali metals (Group 1), ammonium ion (NH4
+),
Nitrates (NO3
-), Acetates (C2H3O2
-), hydrogen
carbonate (HCO3
-), chlorates (ClO3
-)
2. Chlorides, Bromides, and Iodides
exceptions: Pb2+, Ag+, and Hg2
2+ are insoluble
3. Sulfates (SO4
2-)
exceptions: Ca2+, Sr2+, Ba2+
, Pb2+, Ag+, and Hg2
2+
are insoluble

5. Solubility Rules
Insoluble: (except Group 1, NH4
+)
1. Carbonates (CO3
2-), Phosphates (PO4
3-),
and Sulfides (S2-)
2. Hydroxides (OH-)
exceptions: Ca2+, Sr2+, Ba2+
3. Chromates (CrO4
2-)
exceptions: Ca2+, Mg2+

6. Solubility Rules
Gas Formation:
With Acidic solutions:
o Any sulfide (S2-) compound and any acid forms H2S(g)
and a salt
o Any carbonate (CO3
2-) compound and any acid form
CO2(g), H2O(l) and a salt
o Any sulfite (SO3
2-) compound and any acid form SO2(g),
H2O and a salt
With Basic Solutions:
o Any ammonium (NH4
+) compound and any soluble
strong hydroxide form NH3(g), H2O and a salt

7. Memorize These Rules
Soluble:
1. Alkali metals (Group1), ammonium ion (NH4
+),
Nitrates (NO3
-),Acetates (C2H3O2
-), hydrogen
carbonate (HCO3
-), chlorates (ClO3
-)
1. Chlorides,Bromides,and Iodides
exceptions:Pb2+, Ag+,and Hg2
2+ are insoluble
1. Sulfates (SO4
2-)
exceptions:Sr2+, Ba2+, Pb2+,Ag+,Ca2+ and Hg2
2+ are
insoluble
Insoluble:(exceptions:Group1, NH4
+)
1. Carbonates (CO3
2-), Phosphates (PO4
3-), and
Sulfides(S2-)
1. Hydroxides(OH-)
exceptions: Ca2+, Sr2+, Ba2+ are soluble
3. Chromates (CrO4
2-)
exceptions: Ca2+, Mg2+ are soluble
GasFormingWithAcidicsolutions:
1. Any sulfide(S2-) compound and any acid forms H2S(g)
and a salt
1. Any carbonate (CO3
2-) compound and any acid form
CO2(g), H2O(l) and a salt
1. Any sulfite(SO3
2-) compound and any acid form SO2(g),
H2O and a salt
GasFormingWithBasicSolutions:
1. Any ammonium (NH4
+) compound and any
solublestrong hydroxide form NH3(g),H2O
and a salt

8. Solubility Rules
Determine whether the following compounds
are soluble or insoluble in water:
PbCl2
KI
Li2CO3
Lead (II) nitrate
Calcium carbonate
Silver (I) bromide

9. Using Solubility Rules to
Determine if a Reaction
Will Occur…

10. Predicting Precipitation Reactions:
1. Assign Oxidation states to ALL
elements on the reactant side

11. Assigning Oxidation States
1. Free element = 0
2. Total charge = 0 or the charge of the ion
3. Group 1 = +1; Group 2 = +2
4. Hydrogen = +1; Fluorine = -1
5. Oxygen = -2
6. Grp 17 = -1; Grp 16 = -2; Grp 15 = -3
Assign oxidation states to each element in:
K2Cr2O7 FeS2O3 Ni3P4 NaH

12. Predicting Precipitation Reactions:
2. Perform a double replacement
reaction to predict the possible
products by exchanging the cation
and anion partners

13. Predicting Precipitation Reactions:
3. Criss-Cross
oxidation states
to determine
formula of
possible products

14. Predicting Precipitation Reactions:
4. Determine if any
of the possible
products are
insoluble (solids
or liquids or
gases) using
solubility rules

15. Predicting Precipitation Reactions:
5. a. If neither product is
a solid or a liquid or a
gas, then “NR” no
reaction occurs
b. If one product is a
solid or a liquid or a
gas, then write a
complete ionic
equation

16. Predicting Precipitation Reactions:
6. Circle the solid product and the ions
that made the product – write them down
on a new line and BALANCE the net
ionic equation

17. Types of Reactions – PPT
• Molecular Equation – a chemical equation where
the reactants and products are written as if they
were molecular substances even though they
may actually exist in solution as ions
• Example of a molecular equation:
Ca(OH)2(aq) + Na2CO3(aq)  CaCO3(s) + 2 NaOH(aq)

18. Types of Reactions – PPT
• Complete Ionic Equation – a chemical equation
where aqueous ionic compounds are written as
separate ions in the solution
• Molecular Equation: Ca(OH)2(aq) + Na2CO3(aq)  CaCO3(s) + 2 NaOH(aq)
• Complete Ionic Equation:
Ca2+ + OH- + Na+ + CO3
2-  CaCO3(s) + Na+ + OH-

19. Types of Reactions – PPT
• Net Ionic Equation – an ionic equation from
which spectator ions have been omitted
• Spectator Ions – an ion that does not
participate in the reaction

20. Types of Reactions – PPT
• Molecular Equation: Ca(OH)2(aq) + Na2CO3(aq)  CaCO3(s) + 2 NaOH(aq)
• Complete Ionic Equation: Ca2+ + OH- + Na+ + CO3
2-  CaCO3(s) + Na+ + OH-
Spectator Ions = OH- and Na+ (cross out)
• Net Ionic Equation:
Ca2+ + CO3
2-  CaCO3(s) (Balanced!!)

21. Types of Reactions – PPT
Predict whether a reaction will occur in each of the
following cases. If a reaction does occur, write a net
ionic equation for the reaction:
1. NaOH(aq) + MgCl2(aq)  ?

22. Types of Reactions – PPT
2. (NH4)2SO4(aq) + ZnCl2(aq)  ?

23. Types of Reactions – PPT
• Stoichiometry of Precipitation Reactions –
works just like a normal stoichiometry
problem but you may need to determine
the “pieces” instead of the compounds

24. Types of Reactions – PPT
How many moles of chloride ions are needed to
precipitate 2.50 g of solid?
Pb(C2H3O2)2(aq) + NaCl(aq)  ?

25. Types of Reactions – PPT
What mass of solid is formed when 0.0100
moles of barium chloride and 0.0100 moles
of sodium hydroxide are mixed?

26. Calculations with Solutions
• Concentration – the quantity of solute in a
standard quantity of solution
• We have already used:
Mass percent = mass of solute/100g solution
• A new unit of concentration:
Molarity = moles of solute/1 L solution

27. Calculations with Solutions
Calculating Molarity:
A sample of NaCl weighing 0.0678 g is placed in a
25.0 mL volumetric flask. Enough water is
added to dissolve the NaCl, and then the flask is
filled to the mark with water. What is the
molarity of the resulting solution?

28. Calculations with Solutions
Molarity as a Conversion:
– Allows for conversion between moles of the
solute and liters of a solution
How many milliliters of 0.163 M NaOH are required
to give 0.0958 g of sodium hydroxide?

29. Calculations with Solutions
Molarity and Stoichiometry:
What volume of 0.650 M K2CrO4, in mL, is needed to
precipitate as Ag2CrO4(s) all the silver ions in 415 mL
of 0.186 M AgNO3?
2 AgNO3(aq) + K2CrO4(aq)  Ag2CrO4(s) + 2 KNO3(aq)

30. Calculations with Solutions
Molarity and Dilution: Mi x Vi = Mf x Vf
• Only use this equation for diluting samples
(because it is the same compound!)
A solution is 1.5 M H2SO4. How many milliliters of
this acid do you need to prepare 100.0 mL of
0.18 M H2SO4?

31. Calculations with Solutions
Molarity and Limiting Reagents:
75.0 mL of 0.250 M HCl is mixed with 225 mL of 0.055
M Ba(OH)2 . What mass of water is formed?

32. Calculations with Solutions
Making A Solution:
1. Calculate how many grams of the solute
are needed for the desired volume of
solution.
2. Weigh out the solid and put it in a
volumetric flask.
3. Add distilled water to the line on the
volumetric flask.
Prepare 25.0 mL of a 0.20 M solution of MgCl2