Chemical Reactions &
Solution Stoichiometry
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.

Hydration
The process of breaking the ions of salts
apart.
 Ions have charges and are attracted to the
opposite charges on the water molecules.

How Ionic solids dissolve
Click here for
Animation
H
H
H
H
H
Electrolytes
Electricity is moving charges.
 The ions that are dissolved can move.
 Solutions of ionic compounds can
conduct electricity.
 Electrolytes.
 Solutions are classified three ways.

Types of solutions

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- Don’t conduct.
Types of solutions
Acids- form H+ ions when dissolved.
 Strong acids fall apart completely, many ions
 Memorize this list
H2SO4 HNO3 HCl HBr HI HClO4
 Weak acids- don’t dissociate completely.
(everything else)
 Bases - form OH- ions when dissolved.
 Strong bases- completely ionize.
 Memorize LiOH, Ca(OH)2,NaOH, Sr(OH)2,
KOH, Ba(OH)2

Some Properties of Acids
þ Produce H+ ions in water (Ex. H3O+ the hydronium ion is a
hydrogen ion attached to a water molecule)
þ Taste sour
þ Corrode metals
þ Are electrolytes
þ React with bases to form a salt and water
þ pH is less than 7
þ Turns blue litmus paper red “Blue to Red A-CID”
Some Properties of Bases
 Produce OH- ions in water
 Taste bitter, chalky
 Are electrolytes
 Feel soapy, slippery
 React with acids to form salts and water
 pH greater than 7
 Turns red litmus paper to blue
“Basic Blue”
Acid/Base definitions
#1: Arrhenius (traditional)
contain H+ or
hydronium H3O+ ions as the only
positive ions
 Arrhenius Acids
Bases contain OH- ions
 problem: some bases don’t have
hydroxide ions
 Arrhenius
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
Acid/Base Definitions

Definition #2: Brønsted – Lowry
Acids – any substance that donates
a H+ ion (proton donor)
Bases – any substance that accepts
a H+ ion (proton acceptor)
A “proton” is really just a hydrogen
atom that has lost it’s electron!
ACID-BASE THEORIES
The Brønsted definition means NH3 is a
BASE in water — and water is itself
an ACID
NH3
Base
+
H2O
Acid
NH4+ + OHAcid
Base
A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor
base
acid
conjugate
acid
Conjugate what?
conjugate
base
Conjugate Pairs
When an acid loses a proton, the remaining
portion of the acid has an unshared pair of
electrons that can act as a base
 Ex. HCl  H+ + ClHCl donates a proton to Cl
The Cl ion is a base (proton acceptor)

When a pair of chemical formulas differ only by the
presence of H ion, they are a conjugate acidbase pair.
Conjugate Pairs
Acids & Base Definitions
Definition #3 – Lewis
Lewis acid - a substance
that accepts an
electron pair
Lewis base - a
substance that
donates an electron
pair
Lewis Acids & Bases
Formation of hydronium ion is an excellent
example.
H
+
ACID
•• ••
O—H
H
BASE
••
H O—H
H
•Electron pair of the new O-H bond
originates on the Lewis base.
Lewis Acid/Base Reaction
Measuring Solutions
Concentration- how much is dissolved.
 Molarity = Moles of solute
Liters of solution
 abbreviated M
 1 M = 1 mol solute / 1 liter solution

Molarity
Example Suppose you had 58.44 grams of NaCl and you
dissolved it in exactly 2.00 L of solution. What would be the
molarity of the solution?
 Two steps:
 Step One: convert grams to moles.
 Step Two: divide moles by liters to get molarity.
 In the above problem, 58.44 grams/mol is the molecular weight
of NaCl.
 Dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol.
 Then, dividing 1.00 mol by 2.00 L gives 0.50 mol/L (or 0.50 M).

Molarity
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?

Dilution
Adding more solvent to a known solution.
 The moles of solute stay the same.
 moles = M x L
 M1 V1 = M2 V2
 moles = moles
 Stock solution is a solution of known
concentration used to make more dilute
solutions

Dilution Example
53.4mL of a 1.50M solution of NaCl is on
hand, but you need some 0.800M
solution. How many mL of 0.800M can
you make?
 Find: V2 or mL of 0.800M solution
 Given: M1 = 1.50M, M2 = 0.800M
V1 = 53.4mL

Dilution Example
M1V1 = M2V2
 Rearrange our equation to get “find” by
itself
V2 = M1V1
M2
V2 = (1.50mol/L)(53.4ml) = 100mL
(0.800mol/L)
Dilution

What volume of a 1.7 M solutions is needed
to make 250 mL of a 0.50 M solution?
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

Precipitation reactions
 Formula Equation
NaOH(aq) + FeCl3(aq) NaCl(aq) + Fe(OH)3(s)
 is really

 Complete Ionic Equation

Na+(aq)+OH-(aq) + Fe+3 + Cl-(aq) Na+ (aq) + Cl- (aq) + Fe(OH)3(s)

So all that really happens is
 Net ionic Equation
OH-(aq) + Fe+3  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
 AgNO3(aq) + KCl(aq) 
 Zn(NO3)2(aq) + BaCr2O7(aq) 
 CdCl2(aq) + Na2S(aq) 

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 (pg 144)

Solubility Rules
1. All compounds that contain a group I element
(alkali metal) are soluble.
 2. All compounds that contain an ammonium ion
(NH4+) are soluble.
 3. All compounds that contain a nitrate ion (NO3-)
or acetate ion (C2H3O2-) are soluble.
 4. All compounds that contain a group VII element
(halogen) are soluble except those halides that
contain Ag+, Hg+, or Pb+2.
(This one is not perfect, check out the fluorides in the chart below.)

Solubility Rules
5. All compounds that contain a sulfate
ion (SO4-2) are soluble except that that
contain Ba+2, Sr+2, or Pb+2.
 6. All other compounds are
INSOLUBLE.

SOLUBILITY RULES-Chart Form

SOLUBLE COMPOUNDS

Almost all salts of Na+, K+, and NH4+

All salts of Cl-, Br-, I-

Salts of F-

Salts of: nitrates,
NO3chlorates,
ClO3perchlorates, ClO4acetates,
C2H3O2-

All salts of sulfates, SO4-2
COMBINATIONS THAT ARE
NOT SOLUBLE
Ag+, Hg2+2, Pb+2
Mg+2, Ca+2, Sr+2, Ba+2, Pb+2
Ba+2, Sr+2, Pb+2
Homework # 36
36.a.

b.

c.

d.

e.

f.

g.

h.

Soluble (Rule 3)
Soluble (Rule 1)
Inoluble (Rule 4)
Soluble (Rules 2 and 3)
Insoluble (Rule 6)
Insoluble (Rule 5)
Insoluble (Rule 6)
Soluble (Rule 2)
Three Types of Equations
Molecular Equation- written as whole formulas, not
the ions.
 K2CrO4(aq) + Ba(NO3)2(aq) 
 Complete Ionic equation show dissolved
electrolytes as the ions.


2K+ + CrO4-2 + Ba+2 + 2 NO3- BaCrO4(s) + 2K+ + 2 NO3-

Spectator ions are those that don’t react.
Three Type of Equations
Net Ionic equations show only those
ions that react, not the spectator ions
 Ba+2 + CrO4-2  BaCrO4(s)
 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.

Net Ionic Equations Guidelines:
1) Write the (balanced!) molecular equation first
- Reaction products: swap cations and anions
- Predict solubility (using Solubility rules)
2) Write the complete ionic equation next
- (s) compounds don’t ionize
- (aq) compounds do ionize
ion subscripts in the molecular equation become
coefficients in the complete ionic equation!
3) Write the net ionic equation next
- cancel spectator ions
The net ionic equation is a “simplified” form
of the complete ionic equation
Net Ionic Equations
Balanced Chemical Equation:
Pb(NO3)2(aq) + 2NaI(aq)  PbI2(s) + 2NaNO3(aq)
“Complete Ionic” Equation:
Pb2+(aq) + 2NO3-(aq) + 2Na+(aq)+ 2I-(aq)  PbI2(s) + 2Na+(aq) + 2NO3(aq)
Cancel the “spectator ions” that appear on both sides of the
arrow
Pb2+(aq) + 2NO3-(aq) + 2Na+(aq)+ 2I-(aq)  PbI2(s) + 2Na+(aq) + 2NO3(aq)
“Net Ionic” Equation:
Pb2+(aq) + 2I-(aq)  PbI2(s)
Example: Problem
Balanced Molecular Equation:
Ba(NO3)2 + NiSO4
Example: Problem
Balanced Molecular Equation:
Ba(NO3)2 + NiSO4
Ni(NO3)2 + BaSO4
Example: Problem
Balanced Molecular Equation:
Ba(NO3)2(aq)+ NiSO4(aq)
Ni(NO3)2 (aq)+ BaSO4(s)
Example: Problem
Balanced Molecular Equation:
Ba(NO3)2(aq)+ NiSO4(aq)
Ni(NO3)2 (aq)+ BaSO4(s)
Complete Ionic Equation:
Ba2+(aq) + 2NO3-(aq) + Ni2+(aq) + SO42-(aq)
Ni2+(aq) + 2NO3-(aq) + BaSO4 (s)
Example: Problem
Balanced Molecular Equation:
Ba(NO3)2(aq)+ NiSO4(aq)
Ni(NO3)2 (aq)+ BaSO4(s)
Complete Ionic Equation:
Ba2+(aq) + 2NO3-(aq) + Ni2+(aq) + SO42-(aq)
Ni2+(aq) + 2NO3-(aq) + BaSO4 (s)
Example: Problem
Balanced Molecular Equation:
Ba(NO3)2(aq)+ NiSO4(aq)
Ni(NO3)2 (aq)+ BaSO4(s)
Complete Ionic Equation:
Ba2+(aq) + 2NO3-(aq) + Ni2+(aq) + SO42-(aq)
Ni2+(aq) + 2NO3-(aq) + BaSO4 (s)
Net Ionic Equation:
Ba2+(aq) + SO42-(aq)
BaSO4 (s)
What is the Net Ionic Equation for the
reaction: HCl(aq) + NaOH(aq) ?
HCl(aq) + NaOH(aq)
H+(aq) + -OH(aq)
H2O(l) + NaCl(aq)
H2O(l)
Homework #44
44. a. CrCl3(aq) + 3 NaOH(aq) → Cr(OH)3(s) + 3 NaCl(aq)
Cr3+(aq) + 3 OH(aq) → Cr(OH)3(s)

b. 2 AgNO3(aq) + (NH4)2CO3(aq) → Ag2CO3(s) + 2 NH4NO3(aq)
2 Ag+(aq) + CO32(aq) → Ag2CO3(s)
c. CuSO4(aq) + Hg2(NO3)2(aq) → Cu(NO3)2(aq) + Hg2SO4(s)
Hg22+(aq) + SO42(aq) → Hg2SO4(s)
d. No reaction occurs because all possible products
(SrI2 and KNO3) are soluble.

Rules for solving Stoichiometry Problems
for Reactions in Solution
Identify what reaction occurs
 Write the net ionic equation
 Calculate moles of reactants
 Determine which is limiting
 Calculate moles of product
 Convert to units in the problem

Stoichiometry of Precipitation

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?
1. BaCl2 + 2NaOH  Ba(OH)2 + 2NaCl
 2.Ba+2(aq) +2OH-(aq)  Ba(OH)2(s)
 .1 L x .100mol Ba = .01 mol Ba = .01
1L
(1)
 .1L x .100 mol OH = .01 mol OH = .005 limiting
1L
(2)
 .01 mol OH x 1mol Ba(OH)2 x 171.3 g Ba(OH)2 = .8565 g
2 mol OH
1mol Ba(OH)2
Ba(OH)2

Homework #50

The balanced equation is: 3 BaCl2(aq) +
Fe2(SO4)3(aq) → 3 BaSO4(s) + 2 FeCl3(aq)
Solve for mols .01 mol BaCl2 and .01mol Fe2(SO4)3
 Find Limiting reactant

 .01 mol BaCl2
=.003
.01 mol Fe2(SO4)3 = .01
3

1
BaCl2 is the limiting reagent.
0.0100 mol BaCl2 x 3 mol BaSO4 x 234 g BaSO4= 2.33 g BaSO4
3 mol BaCl2
1mol BaCl2
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

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.

Acid-Base Reaction

75 mL of 0.25M HCl is mixed with 225 mL of 0.055
M Ba(OH)2 . What is the concentration (M) of the
excess H+ or OH- ?
2HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2H20(l)
 H+(aq) + OH-(aq)  H2O(l)



.075 L × .25mol HCl= 1.88 × 10-2mol HCl = 1.88 × 10-2mol H+ + 1.88 × 10-2mol Cl
1 L HCl
1
.225L × .055 mol Ba(OH)2 = 1.24 × mol Ba(OH)2
1L Ba(OH)2
= 1.24 × 10-2mol Ba2+ + 2.48 ×10-2 mol OH
1
75 mL of 0.25M HCl is mixed with 225 mL of 0.055 M Ba(OH)2 . What is the
concentration (M) of the excess H+ or OH- ?

Excess Concentration (M) Since 1.88 × 10-2 mol OH will be neutralized by
the H+, we have (2.48 - 1.88) × 10-2 = 0.60 ×10-2 mol OH remaining in
excess.
MOH = Moles excess = 6.0 x 10-3
Total Volume (.075 + .225)
= 2.0 × 10-2 M OH
Homework #62


HCl and HNO3 are strong acids; Ca(OH)2 and RbOH
are strong bases. The net ionic equation that occurs
is H+(aq) + OH(aq) → H2O(l).
0.100 mol HCl 1 mol H 

L
mol HCl
H+
mol
= 0.0500 L ×
+
1 mol H
0.1000 L × 0.200 molL HNO  mol
= 0.00500 + 0.0200 =
HNO
0.0250 mol H+

3
3

0.0100 mol Ca (OH) 2
2 mol OH 

L
mol Ca (OH) 2
0.100 mol RbOH 1 mol OH 

L
mol RbOH
mol OH = 0.5000 L ×
0.2000 L ×
+
= 0.0100 + 0.0200 =
0.0300 mol OH
#62 Continued

We have an excess of OH so the
solution is basic (not neutral). The mol of
excess OH = 0.0300 mol OH initially
 0.0250 mol OH reacted (with H+) =
0.0050 mol OH excess.
0.0050 mol OH 
0.0050 mol
 MOH = (0.0500  0.1000  0.5000  0.2000 ) L  0.8500 L
MOH = 5.9 × 10-3M
Redox Reactions
Types of Reaction
 Oxidation-Reduction called Redox
Ionic compounds are formed through the
transfer of electrons.
 An Oxidation-reduction reaction involves
the transfer of electrons.
 We need a way of keeping track.

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).
 The oxidation state of elements in their
standard states is zero.
 Oxidation state for monoatomic ions are
the same as their charge.

Oxidation states
 Oxygen is assigned an oxidation state of -2



in its covalent compounds except as a
peroxide.
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
Assign the oxidation states to each
element in the following.
 CO2
 NO3 H2SO4
 Fe2O3
 Fe3O4

Oxidation-Reduction

Transfer electrons, so the oxidation states
change.

Na + 2Cl2  2NaCl
CH4 + 2O2  CO2 + 2H2O
 Oxidation is the loss of electrons.
 Reduction is the gain of electrons.
 OIL RIG
 LEO the lion goes GER

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.

Agents
Oxidizing agent gets reduced.
 Gains electrons.
 More negative oxidation state.

Reducing agent gets oxidized.
 Loses electrons.
 More positive oxidation state.

Identify the
Substance oxidized
 Substance reduced
 Oxidizing agent
 Reducing agent
 in the following reactions
 Fe (s) + O2(g)  Fe2O3(s)
0
0
+3 -2


CH4 + H2O  CO + 3H
Half-Reactions
All redox reactions can be thought of as
happening in two halves.
 One produces electrons - Oxidation half.
 The other requires electrons - Reduction half.
 Write the half reactions for the following.
+
 Na + Cl2  Na + Cl

Na  Na+
Cl2  Cl-
 CH4
+ H2O  CO + 3H
Balancing Redox Equations
In aqueous solutions the key is the
number of electrons produced must be
the same as those required.
 For reactions in acidic solution an 8 step
procedure.
 Write separate half reactions
 For each half reaction balance all
reactants except H and O
 Balance O using H2O

Acidic Solution
 Balance H using H+
 Balance charge using e Multiply equations to make electrons
equal
 Add equations and cancel identical
species
 Check that charges and elements are
balanced.
Practice


The following reactions occur in aqueous solution. Balance them
MnO4- + Fe+2 Mn+2 + Fe+3
1. MnO4-  Mn+2
Fe+2  Fe+3
2. 8H+ + MnO4-  Mn+2 +4H2O
+7
+2
5. Equalize
5e-+8H+ + MnO4-  Mn+2 +4H2O
Fe+2  Fe+3 +e-
6. Multiply Equations and Add
5e-+8H+ + MnO4-  Mn+2 +4H2O
5 Fe+2  Fe+3 +e7. 5Fe+2 + 5e-+ 8H+ + MnO4-  Mn+2 +4H2O + 5Fe+3 +5e8. 5Fe+2 + 8H+ + MnO4-  Mn+2 +4H2O + 5Fe+3
You Try

CH3OH + Cr2O72 → CH2O + Cr3+
8 H+(aq) + 3 CH3OH(aq) + Cr2O72(aq) → 2 Cr3+(aq) + 3 CH2O(aq) + 7 H2O(l)
Basic Solution
Do everything you would with acid, but add
one more step.
 Add enough OH to both sides to neutralize
the H+, makes water
 Ag +CN +O2  Ag(CN)2

2 CrI3 + Cl2  CrO4
+ IO4 + Cl
 Fe(OH) + H O  Fe(OH)
2

2 2
Cr(OH)3 + OCl- + OH- 
CrO42- + Cl- + H2O
Basic Solution
 Basic Solution
Ag(s)+CN-(aq)+O2(g)Ag(CN)2-(aq)
1. Write and balance half reactions
oxidizing Ag(s)+CN-(aq) Ag(CN)2-(aq)
Ag(s)+2CN-(aq) Ag(CN)2 -(aq)+eReducing 4e-+O2(g)+4H+2H2O
2. Equalize electron tranfer
Mulitiply by 4 Ag(s)+2CN-(aq) Ag(CN)2 -(aq)+e4Ag(s)+8CN-(aq) 4Ag(CN)2 -(aq)+4e-
3. Add half reactions and cancel identical species
4Ag(s)+8CN-(aq)+O2(g)+4H+4Ag(CN)2 -(aq)+2H2O
Basic Solution
4. Add OH- ions to both sides of balanced
equation to cancel eliminate H+
4Ag(s)+8CN-(aq)+O2(g)+4H++4OH-4Ag(CN)2 -(aq)+2H2O+ 4OH-
5. Eliminate as many H2O molecule possible
4Ag(s)+8CN-(aq)+O2(g)+ 2H2O 4Ag(CN)2 -(aq)+ 4OH-
6. Lastly check your charges are balanced
You Try

Al + MnO4 → Al(OH)4 + MnO2

2 H2O(l) + Al(s) + MnO4(aq) → Al(OH)4(aq) + MnO2(s)
(basic Solution)