SOLUTIONS AND THEIR COMPONENTS
Definitions
solvent – bulk material used to dissolve substance
solute – material dissolved in solvent
Examples
1) salt water
solvent – water
solute – salt
2) dish water
solvent – water
solute – dish soap
3) engine coolant
solvent – ethylene glycol
solute – water
THE AQUEOUS PHASE
Q: What happens to salt when it is dissolved in water?
- It is a physical change, but how can that be?
A: i) Solid salt is a collection of ions, Na+ and Cl-, assembled in a lattice.
- Ions are surrounded by each other.
ii) When dissolved, the ions are surrounded by water.
- Thus, the ions themselves don’t change, only their environment.
+
-
+
-
-
+
-
+
H
H
O
H
H
O
+
-
+
+
O
-
H
H
-
+
-
O
+
H
ion surrounded by other ions
H
ion surrounded by water molecules
(solvent cage)
Ions (or sometimes molecules) that are surrounded by water are in the aqueous phase.
Ionic compounds which are soluble (that is, dissolve in water) dissociate (that is, break
apart) into dissolved cations and anions.
Examples: NaCl (aq) = Na+ (aq) + Cl- (aq)
K2O (aq) = 2 K+ (aq) + O2- (aq)
- note 2 K+ ions dissociate for every K2O unit.
Ca(NO3)2 (aq) = Ca2+ (aq) + 2 NO3- (aq)
- ions dissociate, but polyatomic ions remain covalently bonded.
Fe2(SO4)3 (aq) = 2 Fe3+ (aq) + 3 SO42- (aq)
Cs3PO4 (aq) = 3 Cs+ (aq) + PO43- (aq)
*Not all compounds dissociate into ions when in the aqueous phase, as we shall soon see.*
Aside: phases of reactants and products
(g) – gas
(l) – liquid
(s) – solid
(aq) – aqueous
TYPES OF SOLUTES
1. electrolytes
- when dissolved, solution conducts electricity
- H2O without ions is a poor conductor.
- are ionic compounds or acids (usually)
a) strong electrolytes
- Ions completely dissociate from formula unit or lattice.
NaCl (aq) = Na+ (aq) + Cl- (aq)
- Water surrounds ions to dissolve lattice.
b) weak electrolytes
- Ions partially dissociate from formula unit.
HC2H3O2 (aq) + H+ (aq) + C2H3O2- (aq)
a lot
a little
a little
- Equilibrium (balance) exists between dissociated ions and undissociated
compounds. More in 1190.
HC2H3O2 (l) + H2O (l)
2.) nonelectrolytes
- when dissolved, solution is still electrical insulator
- usually molecular compounds
C12H22O11 (s) + H2O (l)  C12H22O11 (aq)
Subtle point: Relatively insoluble compounds can be strong electrolytes.
ACIDS, BASES AND SALTS
Water (H2O) is made of two ions
H+ (aq) – hydrogen ion
OH- (aq) – hydroxide ion
Acids – substances that increase H+ (aq) conc.
Strong acids
- strong electrolytes, i. e., completely dissociate
HCl (aq) = H+ (aq) + Cl- (aq)
- memorize list
HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4
Weak acids
- weak electrolytes
HF (aq) ⇋ H+ (aq) + F- (aq)
HF (aq)  HF (aq) + H+ (aq) + F- (aq)
a lot
a little
a little
H3PO4 (aq) ⇋ H+ (aq) + H2PO4- (aq)
Bases – substances that increase OH- concentration
Strong bases
- strong electrolytes
- all alkali metal hydroxides
LiOH, NaOH, KOH, RbOH, CsOH
- some alkaline earth metal hydroxides
Ca(OH)2, Sr(OH)2, Ba(OH)2
Weak bases
- weak electrolytes
- usually increase OH- (aq) conc. “indirectly” by decreasing H+ (aq) conc.
NH3 (aq) + H2O (l) ⇋ NH4+ (aq) + OH- (aq)
C5H5N (aq) + H2O (l) ⇋ C5H5NH+ (aq) + OH- (aq)
Neutralization reaction
Acid + Base  Water + Salt
Salt
- ionic compound
- cations and anions remaining after water is made
HNO3 (aq) + KOH (aq)  H2O (l) + KNO3 (aq)
IONIC EQUATIONS
- molecular equations where all ions in aqueous solution are written in dissociated form
NET IONIC EQUATIONS
- ionic equations where spectator ions (ions unchanged in reaction) are not included
Example:
Molecular Equation
2 KBr (aq) + Pb(NO3)2 (aq)  2 KNO3 (aq) + PbBr2 (s)
Ionic Equation
2 K+ (aq) + 2 Br- (aq) + Pb2+ (aq) + 2 NO3- (aq)  2 K+ (aq) + 2 NO3- (aq) + PbBr2 (s)
Net Ionic Equation
Pb2+ (aq) + 2 Br- (aq)  PbBr2 (s)
Note removing spectator ions clarifies important part of reaction.
Example:
Molecular Equation
HClO4 (aq) + NaOH (aq)  H2O (l) + NaClO4 (aq)
Ionic Equation
H+ (aq) + ClO4- (aq) + Na+ (aq) + OH- (aq)  H2O (l) + Na+ (aq) + ClO4- (aq)
Net Ionic Equation
H+ (aq) + OH- (aq)  H2O (l)
Example:
Molecular Equation
Fe (s) + CuSO4 (aq)  FeSO4 (aq) + Cu (s)
Ionic Equation
Fe (s) + Cu2+ (aq) + SO42- (aq)  Fe2+ (aq) + SO42- (aq) + Cu (s)
Net Ionic Equation
Fe (s) + Cu2+ (aq)  Fe2+ (aq) + Cu (s)
METATHESIS REACTIONS
- Cations and anions in two ionic compounds switch places to form precipitate or
nonionic compound.
- Also known as double replacement reactions.
- Compulsion of ions to combine to form other substances is called chemical driving
force.
- Three driving forces in metathesis reactions
1.) formation of a precipitate (solid)
2.) formation of a nonelectrolyte or weak electrolyte
3.) formation of gas
CO2, SO2, H2S, H2
1. Formation of precipitate
**To determine if precipitate is formed, solubility rules must be memorized.**
Solubility Rules
Generally Soluble Compounds
Anion
Exceptions
NO3-, C2H3O2None
ClO3-, ClO4None
- Cl , Br , I
Ag+, Cu+, Pb2+, Hg22+
2SO4
Pb2+, Hg22+, Ca2+, Sr2+, Ba2+
Generally Insoluble Compounds
Anion
Exceptions
S2All alkali metals, NH4+, and Ca2+, Sr2+, Ba2+
CO32-, PO43All alkali metals, NH4+
CrO42-, C2O42All alkali metals, NH4+
2OH , O
All alkali metals and Ca2+, Sr2+, Ba2+
** Note all alkali metal and ammonium compounds are soluble.**
Using solubility rules to predict if reaction occurs
- Reaction happens if precipitate is formed.
- Reaction does not happen if precipitate is not formed.
Example: Does the following reaction occur and what are its products?
Na2S (aq) + Fe(NO3)2 (aq) 
To answer question, switch ions to form potential products.
Na2S (aq) + Fe(NO3)2 (aq) 
NaNO3 ( ) + FeS ( )
Is NaNO3 or FeS insoluble?
According solubility rules, FeS is insoluble.
Therefore reaction occurs.
Balanced molecular equation is
Na2S (aq) + Fe(NO3)2 (aq)  2 NaNO3 (aq) + FeS (s)
Ionic equation is
2 Na+ (aq) + S2- (aq) + Fe2+ (aq) + 2 NO3- (aq)  2 Na+ (aq) + 2 NO3- (aq) + FeS (s)
Net ionic equation is
Fe2+ (aq) + S2- (aq)  FeS (s)
Example: Does the following reaction occur and if so, what are its products?
Cesium chloride (aq) + Calcium nitrate (aq) 
CsCl (aq) + Ca(NO3)2 (aq)  CsNO3 ( ) + CaCl2 ( )
Is CsNO3 or CaCl2 insoluble?
According solubility rules, both are soluble.
No driving force is present; therefore, no reaction occurs.
Balanced molecular equation is
2 CsCl (aq) + Ca(NO3)2 (aq)  2 CsNO3 (aq) + CaCl2 (aq)
Ionic equation is
2 Cs+ (aq) + 2 Cl- (aq) + Ca2+ (aq) + 2 NO3- (aq)  2 Cs+ (aq) + 2 NO3- (aq) + Ca2+ (aq) + 2 Cl- (aq)
Net ionic equation is

Zip, zero, nothing, nada!!!
- All the ions are spectator ions, no reaction occurs.
Example: Does the following reaction occur and what are its products?
Ammonium sulfate (aq) + Barium acetate (aq) 
(NH4)2SO4 (aq) + Ba(C2H3O2)2 (aq)  NH4C2H3O2 ( ) + BaSO4 ( )
Is NH4C2H3O2 or BaSO4 insoluble?
According solubility rules, all sulfates are soluble, except Ca2+, Sr2+ and Ba2+.
Therefore reaction occurs.
Balanced molecular equation is
(NH4)2SO4 (aq) + Ba(C2H3O2)2 (aq)  2 NH4C2H3O2 (aq) + BaSO4 (s)
Ionic equation is
2 NH4+ (aq) + SO42- (aq) + Ba2+ (aq) + 2 C2H3O2- (aq)  2 NH4+ (aq) + 2 C2H3O2- (aq) + BaSO4 (s)
Net ionic equation is
Ba (aq) + SO42- (aq)  BaSO4 (s)
2+
2. Formation of nonelectrolyte (weak electrolyte)
Example
Ba(OH)2 (aq) + 2 HI (aq)  2 H2O (l) + BaI2 (aq)
Net ionic equation
OH- (aq) + H+ (aq)  H2O (l)
- Formation of nonelectrolyte (water) is the driving force of the neutralization
reaction.
Example
Hg2(C2H3O2)2 (aq) + 2 HNO3 (aq)  2 HC2H3O2 (aq) + Hg2(NO3)2 (aq)
Hg2
2+
Ionic equation
(aq) + 2 C2H3O2- (aq) + 2 H+ (aq) + 2 NO3- (aq)  Hg22+ (aq) + 2 NO3- (aq) + 2 HC2H3O2 (aq)
Net ionic equation
C2H3O2- (aq) + H+ (aq)  HC2H3O2 (aq)
- Formation of weak electrolyte (acetic acid) is the driving force of the
reaction.
3. Formation of gas
- Common gases formed in metathesis reactions are CO2 (via H2CO3), SO2 (via
H2SO3), H2S, H2.
- Carbonic acid is unstable at high concentrations.
H2CO3 (aq)  H2O (l) + CO2 (g)
- Sulfurous acid is unstable at high concentrations.
H2SO3 (aq)  H2O (l) + SO2 (g)
- Addition of acid to all carbonates causes formation of CO2 via decomposition of
carbonic acid.
Example
2 HClO4 (aq) + MgCO3 (s)  H2CO3 (aq) +Mg(ClO4)2 (aq)
or considering decomposition of carbonic acid
2 HClO4 (aq) + MgCO3 (s)  H2O (l) + CO2 (g) +Mg(ClO4)2 (aq)
Example
2 HNO3 (aq) + CaS (aq)  H2S (g) + Ca(NO3)2 (aq)
Net ionic equation
2 H+ (aq) + S2- (aq)  H2S (g)
- reaction of sulfide with acid results in formation of hydrogen sulfide gas.
SINGLE REPLACEMENT REACTIONS
- also known as oxidation-reduction reactions
- also known as redox reactions
Definitions
oxidation – process of substance losing ereduction – process of substance gaining eMnemonic: OIL RIG
Oxidation
Is
Loss
Reduction
Is
Gain
*Oxidation and reduction must occur together.*
- e- lost by substance must equal e- gained by other substance.
Example
2 Na (s) + 2 HCl (aq)  2 NaCl (aq) + H2 (g)
Net ionic equation
2 Na (s) + 2 H+ (aq)  2 Na+ (aq) + H2 (g)
- Na is losing electrons, it is being oxidized
- H+ is gaining electrons, it is being reduced
Example
Fe (s) + CuSO4 (aq)  FeSO4 (aq) + Cu (s)
Net ionic equation
Fe (s) + Cu2+ (aq)  Fe2+ (aq) + Cu (s)
- Fe is oxidized
- Cu2+ is reduced
Predicting Spontaneous Redox Reactions
- A redox (single replacement) reaction occurs when a metal ion in a compound is
replaced with another metal ion.
- Some metal atoms are more willing to give away electrons than other metal atoms.
- The willingness of a metal atom to give away its electrons is called its activity.
- In a reaction between metal ion (in aqueous phase) and metal, if metal is more
active than metal ion, reaction will proceed.
- We need to have an activity series to assess relative activity of metals.
Abbreviated Activity Series
Na
Al
Zn
Cr
Fe
Ni
Sn
Pb
H
Cu
Ag
Au
Increasing activity
(willingness to be ionized)
Example
SnSO4 (aq) + Ni (s)  NiSO4 (aq) + Sn (s)
- Ni replaces Sn2+
- Ni is more active than Sn
Does the reverse reaction occur?
NiSO4 (aq) + Sn (s)  SnSO4 (aq) + Ni (s) NO!!
- Sn is not more active than Ni
Example: Does the reaction, Pb(NO3)2 (aq) + Cu(s)  Pb (s) + Cu(NO3)2 (aq) occur?
Look at reactant metals: Pb2+ and Cu.
- Is copper more active than lead?
- No, therefore reaction does not occur.
Example: Does the reaction
Fe2O3 (s) + 3 Zn (s)  2 Fe (s) + 3 ZnO (s) occur?
- reactant metals are Fe2+ and Zn
Zn is more active than Fe, i.e., Zn is more willing to be an ion than Fe; therefore, yes,
the reaction does occur.
Because zinc is more active than iron, it is used to protect iron from rusting (galvanizing).
Example: Does the reaction Al2(SO4)3 (aq) + 2 Cr (s)  2 Al (s) + Cr2(SO4)3 (aq) occur?
- reactant metals are Al3+ and Cr
Al is more active than Cr; therefore, no, the reaction does not occur.
REACTION TYPE REVIEW
1) Combination
recognize
- formation reactions: predict products, know reactants
- hydration reactions: predict products
2) Combustion predict products, know reactants
3) Decomposition
recognize
- decomposition of carbonates: predict products
- decomposition of sulfites: predict products
- decomposition of chlorates, perchlorates: predict products
4) Single replacement reactions: predict products
5) Metathesis (double replacement) reactions: recognize
- neutralization: predict products
- precipitation: predict products
- gas formation: predict products, H2, H2S, CO2, SO2
- recognize chemical driving forces
- H2CO3 (aq)  H2O (l) + CO2(g)
- H2SO3 (aq)  H2O (l) + SO2(g)
6) Addition: recognize
7) Substitution: recognize
8) Elimination: recognize
SOLUTIONS
Definitions
solvent – bulk material used to dissolve substance
solute – material dissolved in solvent
Examples
1) salt water
solvent – water
solute – salt
2) dish water
solvent – water
solute – dish soap
3) engine coolant
solvent – ethylene glycol
solute – water
Concentration
- amount of solute per amount of solvent
- molarity
moles of solute mol
Definition: Molarity M  

liters of solution
L
- molarity can be thought of a conversion factor between volume and # of moles.
- we’ll consider other concentration units in 1190
Molarity calculations
Example: What is the molarity of a solution of NaCl that has 1.73 molNaCl dissolved
in 2.14 L of solution?
1.73 mol
mol
 0.808
 0.808 M
2.14 L
L
Example: What is the molarity of a KBr solution when 114.8 g of KBr is dissolved in
water to make 538.1 mL of solution?
How many moles of solute are to be dissolved?
114.8 g
1 mol
 0.9647 mol
119.00 g
0.9647 mol 1000 mL
mol

 1793
.
 1793
.
M
5381
. mL
L
L
- note the conversion of mL to L
Example: How many grams of gold are in 3.75 L of 0.024 M solution of Au(NO3)3?
We can convert between volume of solution and moles of solute using molarity as a
conversion factor.
0.024 mol
3.75 L 
 0.090 mol Au ( NO3 ) 3
L
0.090 molAu( NO3 )3
1 molAu
196.9665g Au

 18g Au
1 molAu( NO3 )3
molAu
DILUTIONS
- making a solution of lower concentration from a solution of higher concentration
Question: If a 5.0 M solution of NaCl is available, how does one make 250 mL of
2.0 M NaCl?
1) To make solution, we need to have the correct number of moles of NaCl.
Md 
mold
 mold  M d  Vd
Vd
mold  2.0
mol
 0.250 L  0.50 mol
L
2) Now consider how to get 0.50 molNaCl from a 5.0 M solution.
Mc 
Vc 
molc
molc
 Vc 
Vc
Mc
0.50 mol
 0.10 L
5.0 mol L
3) Therefore, to make 250 mL of 2.0 M NaCl, one needs 100 mL of 5.0 M NaCl.
Then add water to make 250 mL solution.
Reconsider step 2. Where did # of mold come from?
**Number of moles of solute is the same before and after dilution.**
i. e., mold = molc
Knowing mol = M  V yields Md  Vd = Mc  Vc
When generalized for any concentration unit, the result becomes the dilution equation
cd  Vd = cc  Vc
c – concentration
Example: Find the volume of 6.02 M HCl to make 500 mL of 0.125 M HCl.
cd  Vd  cc  Vc
 Vc 
cd  Vd 0.125 M  500 mL

 10.4 mL
cc
6.02 M
Lab Safety Aside: Never add water to acid, always add acid to water.
SOLUTION STOICHIOMETRY
- Previously, we dealt with comparing components of a chemical reaction by
converting mass of substances to moles.
- Now we will consider solutions where we will need to convert volumes to moles to
make comparisons.
Scheme:
Mass of
reactant
(g)
Mass of
product
(g)
M
Molar
mass
M
Molar
mass
Moles of
reactant
(mol)
Balanced
equation
Moles of
product
(mol)
M
Molarity
M
Molarity
Volume of
reactant
(L)
Volume of
product
(L)
GRAVIMETRIC ANALYSIS
Measure the concentration of solution by measuring the mass of a precipitate formed.
- Use grams of precipitate to find moles of solute.
- Divide by volume of solution to find concentration.
Example: 25.00 mL of Pb(NO3)2 solution with an unknown concentration reacts with
excess aqueous Rb3AsO4. After filtering and drying, 0.0814 g of precipitate is
found. What is the concentration of the lead(II) nitrate solution?
M(Pb3(AsO4)2) = 692.2 g/mol
3 Pb(NO3)2 (aq) + 2 Rb3AsO4 (aq)  Pb3(AsO4)2 (s) + 6 RbNO3 (aq)
0.0814g Pb3 (AsO4 )2 
1molPb3 (AsO4 )2

3molPb(NO3 )2
692.2g Pb3 (AsO4 )2 1molPb3 (AsO4 )2

0.0141 molPb(NO3 )2
1

0.02500 L
L
VOLUMETRIC ANALYSIS
Measure the concentration of solution by measuring the volume of solution need to find
stoichiometric equivalence between reactants using a titration.
- Use volume of titrant to find moles of analyte.
- Divide moles of analyte by volume of analyte to find concentration.
- Acid/base reactions are commonly used but not always.
Example: What is the concentration of a HBr solution when 50.00 mL is titrated with
41.88 mL of 0.1762 M KOH?
KOH (aq) + HBr (aq)  H2O (l) + KBr (aq)
0.04188 L 
0.1762 mol KOH 1molHBr
0.1476 mol HBr
1



 0.1476 M HBr
L
1mol KOH 0.05000 L
1L
LIMITING REAGENT PROBLEMS
Recall the following about limiting reagent problems.
- To find limiting reactant, calculate number of moles of product formed from each
number of moles of reactant.
- Limiting reactant will yield lowest number of moles produced.
Example: How many grams of Zn(OH)2 are produced when 350 mL of 0.152 M of
ZnSO4 is mixed with 250 mL of 0.275 M of LiOH?
First write balanced equation
ZnSO4 (aq) + 2 LiOH (aq)  Zn(OH)2 (s) + Li2SO4 (aq)
Now calculate possible amount of product that each reactant can produce.
0.350 L 
0.152 molZnSO4 1 molZn(OH)2

 0.0532 mol Zn(OH)2
L
1molZnSO4
0.275 molLiOH 1 molZn(OH)2

 0.0344 mol Zn(OH)2
L
2 molLiOH
Therefore, LiOH is the limiting reagent.
0.250 L 
0.0344 mol Zn(OH)2 
99.39 g Zn(OH)2
molZn(OH)2
 3.42g Zn(OH)2
Example: When 732 L of 1.81 M of Ag2SO4 is mixed with 1148 L of 2.07 M of KBr,
a) kilograms of AgBr (s) formed
b) concentration of all metal ions remaining
a) Ag2SO4 (aq) + 2 KBr (aq)  2 AgBr (s) + K2SO4 (aq)
Ag 2SO4 : 732 L 
1.81 molAg2SO4
KBr : 1148 L 
L

2 molAgBr
1molAg2SO4
 2650 mol AgBr
2.07 molKBr 1 molAgBr

 2376 molAgBr
L
1molKBr
KBr is the limiting reactant.
KBr : 2376 mol AgBr 
187.772 g AgBr
molAgBr
 446000 g AgBr  446 kg AgBr
In photographic film, AgBr decomposes on exposure to light which darkens the negative.
2 AgBr(s) + h  2 Ag(s) + Br2 (l)
b) Find concentration of metal ions remaining.
The metal ions we might have in solution are Ag+ and K+.
i) Consider the concentration of K+ first.
- K+ is a spectator ion.
- It hasn’t participated in any chemical change.
Number of moles of K+ ion is same as number of moles of KBr initially.
Volume of solution is 732 L + 1148 L = 1880 L.
cK 
2376 molK
1880 L
 1.26 M
Note that although KBr is limiting reagent, none of the K+ is used. This means that
we could be more precise by saying that Br- is the limiting reagent.
ii) Consider the concentration of Ag+.
Once reaction is complete, most of the Ag+ ion is now part of AgBr solid. But since
Ag+ ion is not limiting reagent some of it remains.
How much remains?
First of all how much Ag2SO4 went into AgBr solid.
2376 molKBr 
1 molAg2SO4
2 molKBr
 1188 molAg2SO4
Therefore subtract Ag2SO4 used from Ag2SO4 total.
total
- used
remaining
1325 mol
-1188 mol
137 mol
We need to be careful about dissociation of Ag2SO4.
137 molAg2SO4 
2 molAg
1molAg2SO4
 274 molAg
Thus concentration of Ag+ is
c Ag  
274 mol Ag
1880 L
 0.15 M