Water: The Universal Solvent
• One of the most valuable
properties of water is its
ability to dissolve.
• An individual water
molecule has a bent
shape with a H-O-H bond
angle of approximately
105 degrees.
• Water is polar thus
having positive &
negative partial charges
on its ends.
Ionic Compounds in Water
• The positive ends of a water molecule are
attracted to negative cations and the
negative ends are attracted to positive
cations in an ionic compound – this is
called hydration.
• The ions
become hydrated
& move around
independently.
Covalent Compounds in Water
• Water also dissolves many nonionic
substances such as ethanol (C2H5OH).
• The reason for this is that ethanol is also
polar.
• Polar dissolves polar – “like dissolves like”.
• This is the reason water will not dissolve
oil.
Solutions as Electrolytes
• Strong conductor = strong electrolyte
• Weak conductor = weak electrolyte
• No conductor - nonelectrolyte
Solution Concentration
• Solution can be expressed in a variety of ways
but the most common method is molarity:
M = moles solute
volume (L) solution
• Example: calculate the molarity of a solution
prepped by dissolving 1.56 g of gaseous
hydrochloric acid in enough water to make 26.8
mL of solution.
– 1.60 M HCl
Concentration of Ions in Solution
•
•
(a)
(a)
When an ionic salt dissolves in water ions are
form in solution. The moles of ions formed
must be considered in concentration.
Example: what is the concentration each ion in
(a) 0.50M cobalt II nitrate (b) 2.0M iron III
perchlorate.
Co(NO3)2 (s) + H2O  Co2+ (aq) + 2NO3- (aq)
[Co2+] = 1 x 0.50M = 0.50M
[NO3-] = 2 x 0.50M = 1.0M
Fe(ClO4)3 (s) + H2O  Fe3+ (aq) + 3ClO4- (aq)
[Fe3+] = 2.0M
[ClO4-] = 6.0M
Concentration of Ions in Solution
• Example: Calculate the number of moles
of chloride ions in 1.75 L of 1.0 x 10-3M
zinc chloride.
(step 1) ZnCl2 (s)  Zn2+(aq) + 2Cl-(aq)
(step 2) [Cl-] = 2 x 1.0 x 10-3 = 2.0 x 10-3M
(step 3) 1.75 L x 2.0 x 10-3 mole Cl- =
1L
3.5 x 10-3 moles of chloride ions
Example: Concentration & Volume
• Typical blood serum is about 0.14M sodium
chloride. What volume of blood contains 1.0 mg
of sodium chloride?
– 0.12 mL of blood
• To analyze the alcohol content of a certain wine,
a chemist needs 50.00 mL of an aqueous
0.200M potassium dichromate solution. How
much solid potassium dichromate must be
weighed out to make this solution?
– 2.94 g K2Cr2O7
Dilution Formula
• This formula allows a chemist to prepare a
diluted solution from a concentrated one.
M1V1 = M2V2
or
McVc = MdVd
• Example: what volume of 16 M sulfuric
acid must be used to prepare 1.5 L of a
0.10 M sulfuric acid solution?
– 9.4 mL of H2SO4 must be diluted with 1.5 L
of water.
Types of Chemical Reactions
Precipitation Reactions
• When 2 aqueous solutions are mixed an
insoluble precipitate sometimes forms – also
known as double replacement or metathesis
reactions.
• It is important to remember that some ions are
the key players and some are just spectators:
– The formula equation gives the overall reaction.
– The complete ionic equation represents all ions
involved in the reaction.
– The net ionic equation includes only those solution
components undergoing a change, spectator ions are
not included.
The reaction of Pb(NO3)2 & NaI.
Write the formula equation, ionic & net ionic equations:
Stoichiometry of Precipitation Reactions
• Calculate the mass of solid sodium
chloride that must be added to 1.50 L of a
0.100 M silver nitrate solution to
precipitate all the silver ions in the form of
silver chloride.
• First find the # of moles Na necessary for
the # of moles of silver ions already
present, then convert to grams.
– 8.77 g NaCl
Example: Determine Mass of Product Formed
• When aqueous solutions of sodium sulfate
and lead II nitrate are mixed a precipitate
forms. Give the net ionic equation for the
reactions and calculate the mass of this
precipitate when 1.25 L of 0.0500 M lead II
nitrate and 2.00 L of 0.0250 M sodium
sulfate.
15.2 g PbSO4
Acid-Base Reactions
• An acid is a substance that produces H+
ions when dissolved in water.
• A base is a substance that produces OH−
ions when dissolved in water.
Acids and bases as Electrolytes
A, Strong acids and bases are strong electrolytes, as indicated by
the brightly lit bulb. B, Weak acids and bases are weak electrolytes.
Neutralization Reactions
Strong Acid—Strong Base:
Because both ionize completely, the H+
ions and OH- ions react with each other to
form water molecules.
Basic: HNO3 (aq) + NaOH (aq) 
Ionic: H+ (aq) + NO3- (aq) + Na+ (aq) + OH- (aq) 
Net Ionic:
H+ (aq) +
OH- (aq)

H2O (l)
Weak Acid – Strong Base:
A two-step process occurs:
• 1. The ionization of the weak acid
HB (aq)  H+ (aq) +
B- (aq)
• 2. The neutralization of the H+ ion by the
OH- from the strong base
H+ (aq) + OH- (aq)  H2O (l)
Net Ionic: HX
+
OH-

X-
+
H2O
Strong Acid – Weak Base:
This is also a two-step process:
•
1.The first step occurs when the weak
base reacts with water to produce OHions.
NH3 + H2O  NH4+ + OH-
•
2. In the second step the H+ ions from
the strong acid neutralize the OH- ions to
form water.
H+ + OH-  H2O
• Net Ionic: H+
+
X

XH+
Example: Neutralization Reactions
• 1)What volume of a 0.100 M HCl is needed to
neutralize 25.0 mL of 0.350 M NaOH?
8.75 x 10-2 L
• 2)In a certain experiment, 28.0 mL of 0.250 M
nitric acid and 53.0 mL of 0.320 M potassium
hydroxide are mixed. Calculate the moles of
water formed in the resulting reaction. What is
the [H+] and [OH-] after the reaction goes to
completion?
0.024 moles H2O, [H+] = 0, [OH-] = 0.123 M
Acid-Base Titration
• This is a volumetric analysis technique for
determining the amount (usually concentration)
of a substance.
• This involves the delivery (from a buret) of a
measured volume of a solution of known
concentration (the titrant) into a solution
containing the substance being analyzed (the
analyte).
• The neutralization point is known as the
equivalence point. This point is marked by an
indicator.
• The point when the indicator changes color is
known as the end point.
• The goal is to choose an indicator which has a
similar endpoint as the equivalence point your
reaction.
Titration Example #1
You perform an acid-base titration to standardize an HCl solution by
placing 50.00 mL of HCl in a flask with a few drops of indicator
solution. You put 0.1524 M NaOH into the buret, and the initial
reading is 0.55 mL. At the end point, the buret reading is 33.87 mL.
What is the concentration of the HCl solution?
Titration Example #2
In a titration, it is found that 25.0 mL of 0.500 M NaOH is
required to react with
• (a) a 15.0-mL sample of HCl. What is the molarity of
HCl?
0.833 M
• (b) a 15.0-mL sample of a weak acid, H2A. What is the
molarity of H2A, assuming the reaction to be
H2A(aq) + 2OH-(aq)  2H2O + A2-(aq)?
0.417 M
• (c) an aspirin tablet weighing 2.50 g. What is the
percentage of acetylsalicylic acid, HC9H7O4, in the
aspirin tablet? The reaction is
HC9H7O4 (s) + OH- (aq)  H2O + C9H7O4 - (aq)
90.0%
Oxidation-Reduction (Redox)
• Oxidation means the losing of electrons
(an increase in the oxidation #) and
reduction means the gaining of electrons
(a decrease in the oxidation #). The 2
occur together, they are opposite sides of
the same coin.
• A good way to remember LEO the lion
goes GER – losing electrons
oxidation…..gaining electrons reduction
OR OIL RIG – oxidation is losing,
reduction is gaining.
Oxidizing & Reducing Agents
• An oxidizing agent is the species that
accepts the electrons i.e. the H+ ion
above.
– Non-metals tend to be oxidizing agents.
• A reducing agent is the species that
donates the electrons i.e. the Zn above.
– Metals tend to be reducing agents.
Oxidation Numbers
• The first step to balancing any redox is assigning
oxidation numbers to reactants and products in
the equation – please reference pg. 89 in text:
– The oxidation # of an element in its elemental state is
0.
– The oxidation # of an element in a monatomic ion is
equal to the charge of that ion.
– Certain elements have the same oxidation in all or
almost all their compounds.
– The sum of the oxidation numbers in a neutral
species is 0; in a polyatomic ion, it is equal to the
charge of that ion.
Assigning Oxidation #’s Practice
• What is the oxidation number of
phosphorus
– in sodium phosphate?
• P = +5
– In the dihydrogen phosphate ion?
• P = +5
Balancing Redox Reactions
•
•
Assign oxidation numbers (Rules – Pg. 89)
Identify the oxidation and reduction reactions.
Split in 2 half reactions.
Balance the element being oxidized and reduced.
Balance the elements (that is being reduced or oxidized) oxidation
# by adding electrons. Oxidation adds to the right, reduction adds
to the left.
Balance the oxygens by adding water molecules.
Balance the hydrogens by adding H+ ions.
If the electrons on both sides are not the same you must find the
least common multiple between the 2 electrons. Multiply each
reaction to get the same number of electrons on both sides.
•
•
•
•
•
–
•
•
It is important to check that all atoms and charges balance at
this point. THE MISTAKES ARE LIKELY TO HAPPEN HERE!!!
Combine the reactions and simplify if necessary.
If in basic solution add OH- ions to both sides to produce water
molecules on the side with H+ ions. Simplify the water molecules
if necessary.
Zn (s) + 2HCl (aq)  ZnCl2 (aq) + H2 (g)
• Oxidation:
• Reduction:
Zn  Zn+2 + 2e2H+ + 2e-  H2
Net ionic equation:
Zn (s) + 2H+ (aq)  H2 (g) + Zn+2 (aq)
Balance the following redox…
• (a) Fe2+(aq)+ MnO4- (aq) Fe3+(aq) Mn2+(aq)
(acidic solution)
• (b) Cl2(g) Cr(OH)3(s)  Cl- (aq) CrO42- (aq)
(basic solution)