Parts of solutions
 Solution – homogeneous mixture
 Solute – part that dissolves
 Solvent – causes the dissolving
 Soluble – can be dissolved
 Miscible – liquids dissolve in each other

Saturation of Solutions

Aqueous Solutions

Dissolved in water

Water is a polar molecule

The oxygen atoms have a partial negative charge

The hydrogen atoms have a partial positive charge

The angle is 105 o

Hydration

The process of breaking apart ions of a salt.
 The “+” end of water attracts the anion
 The “-” end of water attracts the cation

Solubility

The ability to dissolve in a given amount of water

Usually g/100mL

Varies greatly

Depends upon ion attraction

Will dissolve nonionic substances if they have polar bonds

Electrolytes

Electrical current through a substance

Ions that are dissolved can move

Solutions are classified three ways

Types of Solutions

Strong electrolytes –Completely ionized when dissolved in water 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 electricity

Types of solutions continued

Acids – form H + ion when dissolved

Strong acids fall apart completely
H
2
SO
4
HNO
3
HCl HBr HI HClO
4
 Weak acids – do not dissociate completely
 Bases – form OH when dissolved
 Strong bases – KOH NaOH

Dissociation

Acids
HCl  H + (aq) + Cl (aq)
HNO
3
 H + (aq) + NO
3
(aq)
H
2
SO
4
 H + (aq) + HSO
4
(aq)


Strong bases
NaOH  Na + (aq) + OH (aq)
KOH  K + (aq) + OH (aq)

Nonelectrolytes

Dissolve in water but do not produce any ions

Example is ethanol (C
2
H
5
OH) molecules disperse in the water but doesn’t conduct electricity

Composition of Solutions

Concentration
1. Molarity (M) – moles of solute per volume of solution in liters
2. M = moles of solute liters of solution

Preparation of Molar Solutions
Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.
1.56 g HCl
1
1 mole HCl
36.5 g HCl
= 0.0427 mole HCl
26.8 mL
1
1L
1000 mL
= .0268 L
M = 0.0427 mole HCl
.0268 L
= 1.60 M HCl

Concentrations of Ions
Give the concentration of each type of ion in 0.50 M
Co(NO
3
)
2
Co(NO
3
)
2
(s) Co +2
(aq)
+ 2NO
3
-
(aq)
Co +2 1 x 0.50 M = 0.50 M Co +2
NO
3
2 x 0.50 M = 1.0 M NO
3
-

Calculate the number of moles of Cl -
1.75 L of 1.0 x 10 -3 M ZnCl
2
.
ions in
ZnCl
2
 Zn +2
(aq)
+ 2Cl -
(aq)
2 x 1.0 x 10 -3 = 2.0 x 10 -3 M Cl -
1.75 L 2.0 x 10 -3 mole Cl -
L
= 3.5 x 10 -3 mole Cl -

Standard solution – a solution where the concentration is accurately known
How much solid K
2
Cr
2
O
7 must by weighted out to make a solution? A chemist needs 1.00 L of an aqueous 0.200 M K
2
Cr
2
O
7 solution.

Dilution
1. Water is added to achieve a particular M
2. Moles of solute after = moles of solute before
3.
M
1
V
1
=M
2
V
2
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H
2
SO
4 solution?

Types of Solution Reactions
 Precipitation Reactions
1. A solid forms from two solutions
2. Precipitate – the insoluble solid
KNO
3
(aq) + BaCl
2
(aq) 

Three Types of Equations Used to Describe Reactions in a Solution

The formula equation gives the overall reaction.
 The complete ionic equation represents as ions all the reactants and products that are strong electrolytes.

The net ionic equation includes only those ions that undergo a change.


Write the formula equation, complete ionic equation, and the net ionic equation for
KCl (aq) + AgNO
3
(aq) 

Stoichiometry of Precipitation Reactions

Calculate the mass of solid NaCl that must be added to
1.50 L of a 0.100 M AgNO
3
Ag + ions in the form of AgCl.
solution to precipitate all the


When aqueous solutions of Na
2
SO
4 and Pb(NO
3
)
2 are mixed, PbSO
4
Pb(NO
3
)
2 formed when 1.25 L of 0.0500 M and 2.00 L of 0.0250 M Na
2
SO
4 are mixed.

Acid-Base Reactions

Acid is a proton donor (H + )

Base is a proton acceptor usually OH
— accepts a H +

H +
(aq)
+ OH
−
(aq)



Cations are surrounded and bound by water molecules protons are also solvated by water molecules
Two ways to show this
1.
H + (aq)
2.
H
3
O + (aq) – hydronium ion

Types of acid donors

Monoprotic
HCl, HNO
3 donates ____ H +

Diprotic
H
2
SO
4 donates ____ H +

Triprotic
H
3
PO
4 donates ____ H +

Neutralization Reactions

What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH?
HCl
(aq)
+ NaOH
(aq)
 NaCl
(aq)
+ H
2
O
(l)
H +(aq) + OH
−
(aq)
 H
2
O
(l)
.0250 L 0.350 mole OH − =
L NaOH

Acid-Base Titrations
 Volumetric analysis
Process of determining the amount of a substance by titration

Titration
Process of delivering solutions to one another

Equivalence point (stoichiometric point)
Point at which the titration has occurred

Endpoint
Point at which the indicator changes color

What volume of 0.812 M HCl, in milliliters, is required to titrate 1.33 g of NaOH to the equivalence point?

Oxidation-Reduction Reactions
 Electrons are transferred
 Spontaneous redox rxns can transfer energy
Electrons (electricity)
Heat

 L
E
O
 G
E

Redox Reactions Examples:
0 0 +1
2Na + Cl
2
 2NaCl
Each sodium atom oses one lectron
--1 oxidation


Each chlorine atom ains one lectron:
0
Cl + e __  Cl
-1 reduction


1. The oxidation number of any uncombined element is zero

2. The oxidation number of a monatomic ion equals its charge
 3. The oxidation number of oxygen in compounds is -
2

4. The oxidation number of hydrogen in compounds is +1


5. The sum of the oxidation numbers in the formula of a compound is 0.

6. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its charge ex. NO
3
—
SO
4
2
—

Reducing Agents and Oxidizing Agents

The substance reduced is the oxidizing agent

The substance oxidized is the reducing agent
0 +1
Na  Na + e --
Sodium is oxidized – it is the reducing agent
0 -- 1
Cl + e - Cl
Chlorine is reduced – it is the oxidizing agent

Oxidation-Number Changes in reactions

Can you identify what is being oxidized and what is being reduced?
2AgNO
3
(aq)
+ Cu
(s)
 Cu(NO
3
)
2
(aq)
+ 2Ag
(s)
The oxidation number of Ag decreases from +1 to 0
(reduction), copper’s oxidation number increase from 0 to +2 (oxidation)

Balancing Redox Equations

Look at the reaction between solid copper and silver ions in aqueous solution:
Cu (s) + Ag + (aq)  Ag (s) + Cu +2 (aq)
1 e -gained
Cu + Ag +  Ag + Cu +2
2 e
lost


Ultimately must have equal number of electrons gained and lost
Cu (s) +