Chapter 4
Aqueous solutions
Types of reactions
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.
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.
Hydration


The process of breaking the ions of salts
apart.
Ions have charges and attract the
opposite charges on the water
molecules.
Hydration
H
H
H
H
H
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.
Electrolytes



Electricity is moving charges.
The ions that are dissolved can move.
Solutions of ionic compounds can
conduct electricity and are therefore
called “electrolytes.”
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.
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.
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
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.
Measuring Solutions

Concentration- how much is dissolved.
mol solute
Molarity  M 
L sol'n

1 M = 1 mol solute / 1 liter solution
What is this???

A one molar
solution!!!

Ha ha ha
ha ha ha!
1L
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?
Calculate the molarity of a solution
with 34.6 g of NaCl dissolved in 125
mL of solution.
Molarity 
mol solute
L sol'n
 34.6g NaCl
1 mol 




58.44 g 

Molarity 
0.125L
Molarity  4.74 M
How many grams of HCl would be
required to make 50.0 mL of a 2.7 M
solution?
Molarity 
mol solute
L sol'n
mol HCl
2.7 M 
0.0500L
mol HCl  0.135 mol  36.46g HCl  4.92 g HCl
1 mol HCl
What would the concentration be if you used
27g of CaCl2 to make 500. mL of solution?
What is the concentration of each ion?
mol solute
Molarity 
L sol'n
Why?
 27 g CaCl2
1 mol 




110.98g 

Molarity 
0.500L
MolarityCaCl2  0.49 M
[Ca2+]=0.49M
[Cl-]=0.98M
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.
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?
 45.6g Fe2 (SO4 )3
1 mol 




399.88g 

Molarity of Fe2 (SO4 )3 
0.475L
[Fe2+]=
2(.240 M)=.480 M
[SO42-]= 3(.240 M)=.720 M
 0.240M
This is because for every
1 Fe2(SO4)3 particle
there are 2 iron ions
and 3 sulfate ions
Practice Problems –
Making Solutions

Describe how to make 100.0 mL of a 1.0
M K2Cr2O4 solution.
mol K2Cr2O4
1.0 M 
0.1000L
mol K2Cr2O4  0.10 mol 
246.2g
1 mol
 24 g
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 100.0 mL.
Practice Problems –
Making Solutions

Describe how to make 250. mL of an 2.0
M copper (II) sulfate dihydrate solution.
mol CuSO4  2H2O
2.0 M 
0.250L
195.61g
mol CuSO4  2H2O  0.50 mol 
1 mol
 98 g
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.
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
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
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
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
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
0.67mol
mol H2SO4 
 0.025L  0.017 mol
L
0.40mol
mol CaCl2 
 0.035L  0.014 mol
L
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
0.014 mol CaCl2 1 mol CaSO4 136.14 g CaSO4


 1.9 g CaSO4
1 mol CaCl2
1 mol CaSO4
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
Precipitation reactions
Sample reaction…
 Overall (molecular) eq’n:
3NaOH(aq) + FeCl3(aq) NaCl(aq) + Fe(OH)3(s)
The above rxn is really…
 Total ionic eq’n:
3Na+(aq)+3OH-(aq)+Fe+3(aq) + 3Cl-(aq) Na+(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)
Precipitation reaction



We can predict the products
Can only be certain by experimenting
The anion and cation switch partners
Practice Problems – Predicting
Products of Precipitation Rxns

AgNO3(aq) + KCl(aq)  KNO3(
Ag+

+ AgCl(
)
K+ Cl-
Zn(NO3)2(aq) + BaCr2O7(aq) 
Zn2+

NO3-
)
NO3-
Ba2+ Cr2O72-
ZnCr2O7(
CdCl2(aq) + Na2S(aq)  2NaCl(
Cd2+ Cl-
Na+ S2-
)
)
+ Ba(NO3)2(
+ CdS(
)
)
Precipitations Reactions



Only happen if one of the products is
insoluble
Otherwise all the ions stay in solutionnothing has happened.
Need to memorize the rules for solubility!
• See Solubility Rules Handout!
Solubility Rules
1) All nitrates are soluble
2) Alkali metals ions and NH4+ ions are soluble
3) Halides are soluble except Ag+, Pb+2, and Hg2+2
4) Most sulfates are soluble, except Pb+2, Ba+2, Hg+2,and Ca+2
5) Most hydroxides are slightly soluble (insoluble) except
NaOH and KOH
6) Sulfides, carbonates, chromates, and phosphates are
insoluble

Lower number rules supersede so Na2S is soluble
Practice Problems – Predicting
Products of Precipitation Rxns

AgNO3(aq) + KCl(aq)  KNO3(aq ) + AgCl( s
Ag+

K+ Cl-
Zn(NO3)2(aq) + BaCr2O7(aq) 
Zn2+

NO3-
)
NO3-
Ba2+ Cr2O72-
ZnCr2O7( s ) + Ba(NO3)2( aq)
CdCl2(aq) + Na2S(aq)  2NaCl(aq ) + CdS( s
Cd2+ Cl-
Na+ S2-
)
Three Types of Equations

Molecular Equation- written as whole formulas, not
the ions.
•

Complete Ionic Equation shows dissolved
electrolytes as the ions.
•
•

K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2KNO3(aq)
2K+ + CrO4-2 + Ba+2 + 2 NO3-  BaCrO4(s) + 2K+ + 2 NO3Spectator 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)
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
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)
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)
Stoichiometry of PrecipitationPractice Problems

Exactly the same, except you may have to
figure out what the pieces are.
• What mass of solid is formed when 100.00 mL
of 0.100 M barium chloride is mixed with 100.00
mL of 0.100 M sodium hydroxide?
• What volume of 0.204 M HCl is needed to
precipitate the silver from 50.ml of 0.0500 M
silver nitrate solution ?
Stoichiometry of PrecipitationPractice Problems


What mass of solid is formed when 100.00 mL of
0.100 M barium chloride is mixed with 100.00 mL
of 0.100 M sodium hydroxide?
BaCl2(aq) + 2NaOH(aq)  Ba(OH)2(s) + 2NaCl(aq)
0.01 mol
0.01 mol LR
0.01mol NaOH 1 mol Ba(OH)2 1 71 .35g Ba(OH)2


2 mol NaOH
1 mol Ba(OH)2
 0.857g Ba(OH)2
Stoichiometry of PrecipitationPractice Problems
• What volume of 0.204 M HCl is needed to
precipitate the silver from 50.ml of 0.0500 M silver
nitrate solution ?

HCl(aq) + AgNO3(aq)  HNO3(aq) + AgCl(s)
? mL
0.0025 mol
0.0025mol AgNO3
1 mol HCl

 0.0025 mol HCl
1 mol AgNO3
0.0025mol
0.204 M 
xL
volume  .012L  12mL
Types of Reactions
 Acid-Base





For our purposes an acid is a proton
donor.
a base is a proton acceptor usually OHWhat is the net ionic equation for the
reaction of HCl(aq) and KOH(aq)?
Acid + Base  salt + water
H+ + OH-  H2O
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.
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
Titration Practice Problems

A 50.00 mL sample of aqueous Ca(OH)2 requires
34.66 mL of 0.0980 M Nitric acid for
neutralization. What is [Ca(OH)2 ]?





NaVa = NbVb
(0.0980N)(34.66mL)=(Nb)(50.00mL)
Nb=0.06793
NCa(OH)2=(M)(2)
[Ca(OH)2]=0.03397M
Titration Practice Problems

75 mL of 0.25M HCl is mixed with 225 mL of 0.055 M
Ba(OH)2. What is the concentration of the excess H+ or OH-?
[H+] = 0.25 M
[OH-] = 0.110 M
0.25mol
 0.075L  0.01875mol H
L
0.110mol
 0.225L  0.02475mol OHL
Excess [OH-] = 0.02475 mol – 0.01875 mol = 0.00600 mol

[OH ]excess
0.00600mol
 0.000200 M

(.075L  .225L)
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
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!
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.
Oxidation States Practice
Assign the oxidation states to each element in the following.
• CO2
+4 -2
• NO3+5 -2
-2
• H2SO4
+1
+6 -2
• Fe2O3
+3
-2
• Fe3O4
+8/3
-2
Oxidation-Reduction

Electrons are transferred, so the oxidation
states change.
•
2Na + Cl2  2NaCl
•
CH4 + 2O2  CO2 + 2H2O
0
-4 +1


0
+1 -1
0
+4 -2
+1 -2
OIL RIG
•
•
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
LEO GER
•
•
Losing electrons - oxidation
Gaining electrons - reduction
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.