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)
MC #1
• When 70. milliliter of 3.0-molar Na2CO3 is
added to 30. milliliters of 1.0-molar
NaHCO3 the resulting concentration of
Na+ is
(A) 2.0 M
(B) 2.4 M
(C) 4.0 M
(D) 4.5 M
(E) 7.0 M
MC #2
• A student wishes to prepare 2.00 liters of 0.100-molar
KIO3 (molecular weight 214 g/mol). The proper
procedure is to weigh out
(A) 42.8 grams of KIO3 and add 2.00 kilograms of H2O
(B) 42.8 grams of KIO3 and add H2O until the final
homogeneous solution has a volume of 2.00 liters
(C) 21.4 grams of KIO3 and add H2O until the final
homogeneous solution has a volume of 2.00 liters
(D) 42.8 grams of KIO3 and add 2.00 liters of H2O
(E) 21.4 grams fo KIO3 and add 2.00 liters of H2O
MC #3
• A 20.0-milliliter sample of 0.200-molar K2CO3
solution is added to 30.0 milliliters of 0.400molar Ba(NO3)2 solution. Barium carbonate
precipitates. The concentration of barium ion,
Ba2+, in solution after reaction is
(A) 0.150 M
(B) 0.160 M
(C) 0.200 M
(D) 0.240 M
(E) 0.267 M
MC #4
• The weight of H2SO4 (molecular weight
98.1 g/mol) in 50.0 milliliters of a 6.00molar solution is
(A) 3.10 grams
(B) 12.0 grams
(C) 29.4 grams
(D) 294 grams
(E) 300. grams
MC #5
• Given that a solution is 5 percent sucrose by
mass, what additional information is necessary
to calculate the molarity of the solution?
I. The density of water
II. The density of the solution
III. The molar mass of sucrose
(A) I only
(B) II only
(C) III only
(D) I and III
(E) II and III
MC #6
• A yellow precipitate forms when 0.5 M
NaI(aq) is added to a 0.5 M solution of
which of the following ions?
A) Pb2+ (aq)
B) Zn2+ (aq)
C) CrO42¯ (aq)
D) SO42¯ (aq)
E) OH¯ (aq)
MC #7
• When 100 mL of 1.0 M Na3PO4 is mixed with 100
mL of 1.0 M AgNO3, a yellow precipitate forms
and [Ag+] becomes negligibly small. Which of
the following is a correct listing of the ions
remaining in solution in order of increasing
concentration?
A) [PO43¯] < [NO3¯] < [Na+]
B) [PO43¯] < [Na+] < [NO3¯]
C) [NO3¯] < [PO43¯] < [Na+]
D) [Na+] < [NO3¯] < [PO43¯]
E) [Na+] < [PO43¯] < [NO3¯]
MC #8
• The volume of distilled water that should
be added to 10.0 mL of 6.00 M HCl(aq) in
order to prepare a 0.500 M HCl(aq) solution
is approximately
A) 50.0 mL
B) 60.0 mL
C) 100. mL
D) 110. mL
E) 120. mL
MC #9
• The net ionic equation for the reaction between
silver carbonate and hydrochloric acid is
(A) Ag2CO3 + 2H+ + 2 Cl¯ ---> 2 AgCl + H2O + CO2
(B) 2Ag+ + CO32¯ + 2 H+ + 2 Cl¯ ---> 2 AgCl + H2O + CO2
(C) CO32¯ + 2 H+ ---> H2O + CO2
(D) Ag+ + Cl¯ ---> AgCl
(E) Ag2CO3 + 2H+ ---> 2Ag+ + H2CO3
MC #10
5 Fe2+ + MnO4¯ + 8 H+ <===> 5 Fe3+ + Mn2+ + 4H2O
• In a titration experiment based on the equation above,
25.0 milliliters of an acidified Fe2+ solution requires 14.0
milliliters of standard 0.050-molar MnO4¯ solution to
reach the equivalence point. The concentration of Fe2+ in
the original solution is
(A) 0.0010 M
(B) 0.0056 M
(C) 0.028 M
(D) 0.090 M
(E) 0.14 M
FRQ #1
• A 1.2516 gram sample of a mixture of CaCO3 and Na2SO4 was analyzed
by dissolving the sample and completely precipitating the Ca2+ as
CaC2O4. The CaC2O4 was dissolved in sulfuric acid and the resulting
H2C2O4 was titrated with a standard KMnO4 solution.
(a) Write the balanced equation for the titration reaction, shown unbalanced
below:
MnO4- + H2C2O4 + H+  Mn2+ + CO2 + H2O
(i) Indicate which substance is the oxidizing agent and which
substance is the reducing agent.
(b) The titration of the H2C2O4 obtained required 35.62 milliliters of 0.1092
molar MnO4- solution. Calculate the number of moles of H2C2O4 that
reacted with the MnO4(c) Calculate the number of moles of CaCO3 in the original sample.
(d) Calculate the percentage by weight of CaCO3 in the original sample.
FRQ #2
• Permanganate ion, MnO4-, oxidizes sulfite ions
to sulfate ion. The manganese product depends
upon the pH of the reaction mixture. The mole
ratio of oxidizing to reducing agent is two to five
at pH 1 (acidic), and is two to one at pH 13
(basic). For each of these cases, write a
balanced equation for the reaction, and indicate
the oxidation state of the manganese in the
product containing manganese.
FRQ #3
• A 0.150 g sample of solid lead(II) nitrate is added to 125 mL of
0.100 M sodium iodide solution. Assume no change in volume
of the solution. The chemical reaction that takes place is
represented by the following equation:
Pb(NO3)2(s) + 2 NaI(aq)  PbI2(s) + 2NaNO3(aq)
(a) List an appropriate observation that provides evidence of a
chemical reaction between the two compounds.
(b) Calculate the number of moles of each reactant.
(c) Identify the limiting reactant. Show calculations to support
your identification.
(d) Calculate the molar concentration of NO3–(aq) in the mixture
after the reaction is complete.
FRQ #4
• Answer the following questions about acetylsalicylic acid,
the active ingredient in aspirin.
(a) The amount of acetylsalicylic acid in a single aspirin
tablet is 325 mg, yet the tablet has a mass of 2.00 g.
Calculate the mass percent of acetylsalicylic acid in the
tablet.
(b) A student dissolved 1.625 g of pure acetylsalicylic
acid in distilled water and titrated the resulting solution to
the equivalence point using 88.43 mL of 0.102 M
NaOH(aq). Assuming that acetylsalicylic acid has only
one ionizable hydrogen, calculate the molar mass of the
acid.
FRQ #5 (part I)
5 Fe2+(aq) + MnO4–(aq) + 8 H+(aq)  5 Fe3+(aq) + Mn2+(aq) + 4H2O(l)
• The mass percent of iron in a soluble iron(II) compound is measured
using a titration based on the balanced equation above.
(a) What is the oxidation number of manganese in the
permanganate ion, MnO4–(aq)?
(b) Identify the reducing agent in the reaction represented above.
Explain your reasoning.
• The mass of a sample of the iron(II) compound is carefully
measured before the sample is dissolved in distilled water. The
resulting solution is acidified with H2SO4(aq). The solution is then
titrated with MnO4–(aq) until the end point is reached.
(c) Describe the color change that occurs in the flask when the
end point of the titration has been reached. Explain why the color of
the solution changes at the end point.
FRQ #5 (part II)
(d)
Let the variables g, M, and V be defined as follows:
g = the mass, in grams, of the sample of the iron(II) compound
M = the molarity of the MnO4–(aq) used as the titrant
V = the volume, in liters, of MnO4–(aq) added to reach the end
point
• In terms of these variables, the number of moles of MnO4–(aq) added
to reach the end point of the titration is expressed as M x V. Using
the variables defined above, the molar mass of iron (55.85 g mol-1),
and the coefficients in the balanced chemical equation, write the
expression for each of the following quantities
(i) The number of moles of iron in the sample
(ii) The mass of iron in the sample, in grams
(iii) The mass percent of iron in the compound
(e) What effect will adding too much titrant have on the
experimentally determined value of the mass percent of iron in the
compound? Justify your answer.
Equations #1
(a) A sample of solid iron(III) oxide is reduced completely with solid carbon.
(i)
Balanced equation:
(ii) What is the oxidation number of carbon before the reaction, and what
is the oxidation number of carbon after the reaction is complete
(b) Equal volumes of equimolar solutions of ammonia and hydrochloric acid are
combined.
(i)
Balanced equation:
(ii) Indicate whether the resulting solution is acidic, basic, or neutral.
Explain.
(c) Solid mercury(II) oxide decomposes as it is heated in an open test tube in a
fume hood.
(i)
Balanced equation:
(ii) After the reaction is complete, is the mass of the material in the test
tube greater than, less than, or equal to the mass of the original sample?
Explain.
Equations #2
(a) A small piece of sodium is placed in a beaker of distilled water.
(i) Balanced equation:
(ii) The reaction is exothermic, and sometimes small flames are
observed as the sodium reacts with the water. Identify the product of
the reaction that burns to produce the flames
(b) Hydrogen chloride gas is oxidized by oxygen gas.
(i) Balanced equation:
(ii) If three moles of hydrogen chloride gas and three moles of
oxygen gas react as completely as possible, which reactant, if any,
is present in excess? Justify your answer.
(c) Solid potassium oxide is added to water.
(i) Balanced equation:
(ii) If a few drops of phenolphthalein are added to the resulting
solution, what would be observed? Explain.