1. Aqueous solutions Types of reactions
Chapter 4
Aqueous solutions
Types of reactions

2. Parts of Solutions Solution- homogeneous mixture.
Solute- what gets dissolved.
Solvent- what does the dissolving.
Soluble- can be dissolved.
Miscible- liquids dissolve in each other.

3. Aqueous solutions Dissolved in water.
Water is a good solvent because the molecules are polar.
The oxygen atoms have a partial negative charge.
The hydrogen atoms have a partial positive charge.
The angle is 105º.

4. Hydration The process of breaking the ions of salts apart.
Ions have charges and attract the opposite charges on the water molecules.

5. Hydration H O H O H O H O H O H O H O H O H O

6. Solubility
How much of a substance will dissolve in a given amount of water.
Usually listed in g/100 mL
Solubilities vary greatly, but if they do dissolve the ions are separated, and they can move around.
Water can also dissolve non-ionic compounds if they have polar bonds.

7. Electrolytes Electricity is moving charges.
The ions that are dissolved can move.
Solutions of ionic compounds can conduct electricity and are therefore called “electrolytes.”

8. Types of solutions Solutions are classified three ways:
Strong electrolytes- completely dissociate (fall apart into ions).
Many ions- Conduct well.
Weak electrolytes- Partially fall apart into ions.
Few ions -Conduct electricity slightly.
Non-electrolytes- Don’t fall apart.
No ions- these sol’ns don’t conduct.

9. Types of solutions Acids- form H+ ions when dissolved.
Strong acids fall apart completely.
many ions
There are just 6 strong acids – MEMORIZE THEM!
Sulfuric acid: H2SO4
Nitric acid: HNO3
Hydrochloric acid: HCl
Hydrobromic acid: HBr
Hydroiodic acid: HI
Perchloric acid: HClO4
Weak acids- don’t dissociate completely.

10. Types of solutions Bases - form OH- ions when dissolved.
Strong bases- many ions. Most common:
Lithium hydroxide:  LiOH 
Sodium hydroxide:  NaOH 
Potassium hydroxide:  KOH 
Rubidium hydroxide:  RbOH 
Calcium hydroxide:  Ca(OH)2 
Strontium hydroxide:  Sr(OH)2 
Barium hydroxide:  Ba(OH)2

11. MOM to the rescue! Milk of magnesia = aqueous suspension of magnesium hydroxide
Is Mg(OH)2 a strong base?
Yes, technically… since all of the dissolved magnesium hydroxide dissociates.
However, it has a very low solubility in water.

12. Measuring Solutions Concentration- how much is dissolved.
1 M = 1 mol solute / 1 liter solution

13. What is this???
A one molar solution!!!
Ha ha ha ha ha ha!
1 L

14. Molarity Practice Problems
Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution.
How many grams of HCl would be required to make 50.0 mL of a 2.7 M solution?
What would the concentration be if you used 27g of CaCl2 to make 500. mL of solution?
What is the concentration of each ion?

15. Calculate the molarity of a solution with 34
Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution.

16. How many grams of HCl would be required to make 50. 0 mL of a 2
How many grams of HCl would be required to make 50.0 mL of a 2.7 M solution?

17. What would the concentration be if you used 27g of CaCl2 to make 500
What would the concentration be if you used 27g of CaCl2 to make 500. mL of solution? What is the concentration of each ion?
Why?
Because for every
one CaCl2 there
are two Cl- and
one Ca2+… thus for
every 1 mol CaCl2,
1 mole of Ca2+ and
2 moles of Cl- form.
[Ca2+]=0.49M
[Cl-]=0.98M

18. Molarity Practice Problems
Calculate the concentration of a solution made by dissolving 45.6 g of Fe2(SO4)3 to 475 mL.
What is the concentration of each ion?
This is because for every
1 Fe2(SO4)3 particle
there are 2 iron ions
and 3 sulfate ions
[Fe2+]=
2(.240 M)=.480 M
[SO42-]=
3(.240 M)=.720 M

19. Practice Problems – Making Solutions
Describe how to make mL of a 1.0 M K2Cr2O4 solution.
Answer: Measure out 24 g of K2Cr2O4 in a volumetric flask. Add enough
distilled water to dissolve the solid. Dilute to a total solution volume of mL.

20. Practice Problems – Making Solutions
Describe how to make 250. mL of an 2.0 M copper (II) sulfate dihydrate solution.
Answer: Measure out 98 g of CuSO4•2H2O in a volumetric flask. Add enough
distilled water to dissolve the solid. Dilute to a total solution volume of 250. mL.

21. M1 V1 = M2 V2 Dilution Adding more solvent to a known solution.
The moles of solute stay the same.
M1 V1 = M2 V2
Stock solution is a solution of known concentration used to make more dilute solutions

22. Dilution Practice Problems
What volume of a 1.7 M solutions is needed to make 250 mL of a 0.50 M solution?
M1V1 = M2V2
(1.7 M)(V1) = (0.50 M)(250 mL)
V1 = 74 mL

23. Dilution Practice Problems
18.5 mL of 2.3 M HCl is added to 250 mL of water. What is the concentration of the resulting solution?
M1V1 = M2V2
(2.3 M)(18.5 mL) = (M2)(250 mL+18.5 mL)
M2 = 0.16 M

24. Dilution Practice Problems
You have a 4.0 M stock solution. Describe how to make 1.0L of a .75 M solution.
M1V1 = M2V2
(4.0 M)(V1) = (0.75 M)(1.0 L)
V1 = 0.19 L = 190 mL
Measure out 190 mL of 4.0 M stock solution.
Dilute with enough water to make 1.0 L solution

25. Dilution Practice Problems
25 mL 0.67 M of H2SO4 is added to 35 mL of 0.40 M CaCl2 . What mass CaSO4 is formed?
H2SO4 + CaCl2  CaSO4 + 2HCl

26. Dilution Practice Problems
25 mL 0.67 M of H2SO4 is added to 35 mL of 0.40 M CaCl2 . What mass CaSO4 is formed? LR
H2SO4 + CaCl2  CaSO4 + 2HCl
0.017mol
0.014mol

27. Precipitation reactions
Types of Reactions
Precipitation reactions
When aqueous solutions of ionic compounds are poured together a solid forms.
A solid that forms from mixed solutions is a precipitate
If you’re not a part of the solution, your part of the precipitate! LOL! :-D

28. Precipitation reactions
Sample reaction…
Overall (molecular) eq’n:
3NaOH(aq) + FeCl3(aq) ® 3NaCl(aq) + Fe(OH)3(s)
The above rxn is really…
Total ionic eq’n:
3Na+(aq)+3OH-(aq)+Fe+3(aq) + 3Cl-(aq) ®3Na+(aq) + 3Cl-(aq) +Fe(OH)3(s)
So all that really happens is…
Net ionic eq’n:
3OH-(aq) + Fe+3(aq) ® Fe(OH)3(s)
(Double replacement reaction)

29. Precipitation reaction
We can predict the products
Can only be certain by experimenting
The anion and cation switch partners

30. Practice Problems – Predicting Products of Precipitation Rxns
AgNO3(aq) + KCl(aq) ®
Zn(NO3)2(aq) + BaCr2O7(aq) ®
CdCl2(aq) + Na2S(aq) ®
KNO3( ) + AgCl( )
Ag+ NO3-
K+ Cl-
Zn2+ NO3-
Ba2+ Cr2O72-
ZnCr2O7( ) + Ba(NO3)2( )
2NaCl( ) + CdS( )
Cd2+ Cl-
Na+ S2-

31. Precipitations Reactions
Only happen if one of the products is insoluble
Otherwise all the ions stay in solution- nothing has happened.
Need to memorize the rules for solubility!
See Solubility Rules Handout!

32. Solubility Rules All nitrates are soluble
Alkali metals ions and NH4+ ions are soluble
Halides are soluble except Ag+, Pb+2, and Hg2+2
Most sulfates are soluble, except Pb+2, Ba+2, Hg+2,and Ca+2
Most hydroxides are slightly soluble (insoluble) except NaOH and KOH
Sulfides, carbonates, chromates, and phosphates are insoluble
Lower number rules supersede so Na2S is soluble

33. Practice Problems – Predicting Products of Precipitation Rxns
AgNO3(aq) + KCl(aq) ®
Zn(NO3)2(aq) + BaCr2O7(aq) ®
CdCl2(aq) + Na2S(aq) ®
KNO3( ) + AgCl( ) aq s
Ag+ NO3-
K+ Cl-
Zn2+ NO3-
Ba2+ Cr2O72-
ZnCr2O7( ) + Ba(NO3)2( ) s aq
2NaCl( ) + CdS( ) aq s
Cd2+ Cl-
Na+ S2-

34. Three Types of Equations
Molecular Equation- written as whole formulas, not the ions.
K2CrO4(aq) + Ba(NO3)2(aq) ®BaCrO4(s) + 2KNO3(aq)
Complete Ionic Equation shows dissolved electrolytes as the ions.
2K+ + CrO4-2 + Ba NO3- ® BaCrO4(s) + 2K+ + 2 NO3-
Spectator ions are those that don’t react.
Net Ionic equation shows only those ions that react, not the spectator ions
Ba+2 + CrO4-2 ® BaCrO4(s)

35. Practice Problem
Write the three types of equations for the reactions when these solutions are mixed.
iron (III) sulfate and potassium sulfide
lead (II) nitrate and sulfuric acid

36. iron (III) sulfate and potassium sulfide
Fe2(SO4)3(aq) + 3K2S(aq)  Fe2S3(s) + 3K2SO4(aq)
2Fe3+(aq) + 3SO42-(aq) + 6K+(aq) + 3S2-(aq) 
Fe2S3(s) + 6K+(aq) + 3SO42-(aq)
2Fe3+(aq) + 3S2-(aq)  Fe2S3(s)

37. lead (II) nitrate and sulfuric acid
Pb(NO3)2(aq) + H2SO4(aq)  PbSO4(s) + 2HNO3(aq)
Pb2+(aq) + 2NO3-(aq) + 2H+ + SO42-(aq) 
PbSO4(s) + 2H+(aq) + 2NO3-(aq)
Pb2+(aq) + SO42-(aq)  PbSO4(s)

38. Stoichiometry of Precipitation- Practice Problems
Exactly the same, except you may have to figure out what the pieces are.
What mass of solid is formed when mL of M barium chloride is mixed with mL of M sodium hydroxide?
What volume of M HCl is needed to precipitate the silver from 50.ml of M silver nitrate solution ?

39. Stoichiometry of Precipitation- Practice Problems
What mass of solid is formed when mL of M barium chloride is mixed with mL of M sodium hydroxide?
BaCl2(aq) + 2NaOH(aq)  Ba(OH)2(s) + 2NaCl(aq)
0.01 mol
0.01 mol LR

40. Stoichiometry of Precipitation- Practice Problems
What volume of M HCl is needed to precipitate the silver from 50.ml of M silver nitrate solution ?
HCl(aq) + AgNO3(aq)  HNO3(aq) + AgCl(s)
? mL
mol

41. Types of Reactions Acid-Base
For our purposes an acid is a proton donor.
a base is a proton acceptor usually OH-
What is the net ionic equation for the reaction of HCl(aq) and KOH(aq)?
Acid + Base ® salt + water
H+ + OH- ® H2O

42. Acid - Base Reactions
Often called a neutralization reaction Because the acid neutralizes the base.
Often titrate to determine concentrations.
Solution of known concentration (titrant) is added to the unknown (analyte) until the equivalence point is reached where enough titrant has been added to neutralize it.

43. or Titration Where the indicator changes color is the endpoint.
Not always exactly at the equivalence point, but should pick an indicator that changes color very close to the endpoint.
# of H+ x MA x VA = # of OH- x MB x VB or
NA x VA = NB x VB

44. Titration Practice Problems
A mL sample of aqueous Ca(OH)2 requires mL of M Nitric acid for neutralization. What is [Ca(OH)2 ]?
NaVa = NbVb
(0.0980N)(34.66mL)=(Nb)(50.00mL)
Nb=
NCa(OH)2=(M)(2)
[Ca(OH)2]= M

45. Titration Practice Problems
75 mL of 0.25M HCl is mixed with 225 mL of M Ba(OH)2. What is the concentration of the excess H+ or OH-?
[H+] = 0.25 M
[OH-] = M
Excess [OH-] = mol – mol = mol

46. Types of Reactions Oxidation-Reduction (“Redox”)
Ionic compounds are formed through the transfer of electrons
An oxidation-reduction rxn involves the transfer of electrons
We need a way of keeping track – oxidation states

47. Oxidation States A way of keeping track of the electrons.
Not necessarily true of what is in nature, but it works.
need the rules for assigning
memorize these!

48. Rules for assigning oxidation states
The oxidation state of an atom in an element is zero (i.e. Na(s), O2(g), O3(g), Hg(l))
Oxidation state for monoatomic ions are the same as their charge. (i.e. Na+, Cl-)
Oxygen is assigned an oxidation state of -2 in its covalent compounds except as a peroxide (such as H2O2).
In compounds with nonmetals hydrogen is assigned the oxidation state +1.
In its compounds fluorine is always –1.
The sum of the oxidation states must be zero in compounds or equal the charge of the ion.

49. Oxidation States Practice Assign the oxidation states to each element in the following.
CO2
NO3-
H2SO4
Fe2O3
Fe3O4 +4 -2 +3 -2 +5 -2 -2
+8/3 -2 +1 +6 -2

50. Oxidation-Reduction
Electrons are transferred, so the oxidation states change.
2Na + Cl2 ® 2NaCl
CH4 + 2O2 ® CO2 + 2H2O
OIL RIG
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
LEO GER
Losing electrons - oxidation
Gaining electrons - reduction

51. Oxidation-Reduction
Oxidation means an increase in oxidation state - lose electrons.
Reduction means a decrease in oxidation state - gain electrons.
The substance that is oxidized is called the reducing agent.
The substance that is reduced is called the oxidizing agent.