Download Chapter 04

Chemistry
Second Edition
Julia Burdge
Lecture PowerPoints
Jason A. Kautz
University of Nebraska-Lincoln
4
Reactions in Aqueous
Solutions
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
1
4
Reactions in Aqueous Solutions
4.1 General Properties of Aqueous Solutions
Electrolytes and Nonelectrolytes
Strong Electrolytes and Weak Electrolytes
4.2 Precipitation Reactions
Solubility Guidelines for Ionic Compounds in Water
Molecular Equations
Ionic Equations
Net Ionic Equations
4.3 Acid-Base Reactions
Strong Acids and Bases
Brønsted Acids and Bases
Acid-Base Neutralization
4.4 Oxidation-Reduction Reactions
Oxidation Numbers
Oxidation of Metals in Aqueous Solutions
Balancing Simple Redox Equations
Other Types of Redox Reactions
4
Reactions in Aqueous Solutions
4.5 Concentration of Solutions
Molarity
Dilution
Solution Stoichiometry
4.6 Aqueous Reactions and Chemical Analysis
Gravimetric Analysis
Acid-Base Titrations
4.1
General Properties of Aqueous Solutions
A solution is a homogenous mixture of two or more substances.
The substance present in the largest amount (moles) is referred to as the
solvent.
The other substances present are called the solutes.
A substance that dissolves in a particular solvent is said to be soluble
in that solvent.
General Properties of Aqueous Solutions
An electrolyte is a substance that
dissolves in water to yield a solution that
conducts electricity.
NaCl(s)
H2O
Na+(aq) + Cl–(aq)
An electrolyte undergoes dissociation
and breaks apart into its constituent
ions.
General Properties of Aqueous Solutions
A non-electrolyte is a substance that
dissolves in water to yield a solution that
does not conduct electricity.
C12H22O11(s)
H2O
C12H22O11(aq)
The sucrose molecules remain intact
upon dissolving.
General Properties of Aqueous Solutions
An electrolyte that dissociates completely
is known as a strong electrolyte.
Water soluble ionic compounds
NaCl(s)
H2O
Na+(aq) + Cl–(aq)
H2O
H+(aq) + Cl–(aq)
Strong Acids
HCl(g)
Strong Bases
NaOH(s)
H2O
Na+(aq) + OH–(aq)
Aqueous Solutions
General Properties of Aqueous Solutions
A weak electrolyte is a compound that
produces ions upon dissolving but exists in
solution predominantly as molecules that are not
ionized.
Weak Acids
HC2H3O2(l)
–
H+(aq) + C2H3O2 (aq)
Weak Bases
NH3(g) + H2O(l)
+
NH4(aq) + OH–(aq)
General Properties of Aqueous Solutions
The double arrow,
directions.
, denotes a reaction that occurs in both
NH3(g) + H2O(l)
+
NH4(aq) + OH–(aq)
When both the forward and reverse reactions occur at the same rate, the
reaction is in a state of dynamic equilibrium.
General Properties of Aqueous Solutions
Classify each of the following as a nonelectrolyte, a weak electrolyte, or a
strong electrolyte.
Citric acid, H3C6H5O7
Potassium phosphate, K3PO4
Glucose, C6H12O6
Solution:
H3C6H5O7: Citric acid is a weak acid and therefore a weak electrolyte.
K3PO4:
Potassium phosphate is a soluble salt that contains both
cations and anions and is therefore a strong electrolyte.
C6H12O6:
Glucose contains no ions and is therefore a molecular
compound and a nonelectrolyte.
4.2
Precipitation Reactions
An insoluble product that separates from a solution is called a precipitate.
2NaI(aq) + Pb(NO3)2(aq)
PbI2(s) + 2NaNO3(aq)
Precipitation Reactions
A chemical reaction in which a precipitate forms is called a precipitation
reaction.
Precipitation Reactions
Water is a good solvent for ionic compounds because it is a polar molecule.
The polarity of water results from electron distributions within the molecule.
The oxygen atom has an attraction for the hydrogen atoms’ electrons and is
therefore partially negative compared to hydrogen.
The oxygen atom is partially negative

+

–
+

The hydrogen atoms are partially positive
Precipitation Reactions
Hydration occurs when water molecules remove the individual ions from
an ionic solid surround them so the substances dissolves.
Precipitation Reactions
Solubility is defined as the maximum amount of solute that will dissolve in
a given quantity of solvent at a specific temperature.
Precipitation Reactions
Classify the following as soluble or insoluble in an aqueous solution.
Ba(NO3)2
AgI
Mg(OH)2
Precipitation Reactions
Classify the following as soluble or insoluble in an aqueous solution:
Ba(NO3)2
AgI
Mg(OH)2
Solution:
Ba(NO3)2:
Soluble; compounds containing nitrate are soluble with
no exceptions.
AgI:
Insoluble; although many compounds containing the
iodide are soluble; when I‒ is paired with Ag+, the
compound is insoluble.
Mg(OH)2:
Insoluble; most metal hydroxides are insoluble;
exceptions include group 1A hydroxides and Ba(OH)2.
Precipitation Reactions
Identify the precipitate in the following reaction:
Pb(NO3)2 (aq) + 2NaI(aq) 2NaNO3(?) + PbI2(?)
Copyright McGraw-Hill 2009
Precipitation Reactions
Identify the precipitate in the following reaction:
Pb(NO3)2 (aq) + 2NaI(aq) 2NaNO3(?) + PbI2(?)
Solution:
NaNO3:
Soluble; all nitrates are soluble with no exceptions; all salts that
contain group 1A cations are soluble with no exceptions.
PbI2:
Insoluble (precipitates out of solution); although many
compounds containing the iodide are soluble; when I‒ is paired
with Ag+, the compound is insoluble.
Pb(NO3)2 (aq) + 2NaI(aq) 2NaNO3(aq) + PbI2(s)
Precipitation Reactions
In a molecular equation compounds are represented by chemical
formulas as though the exist in solution as molecules or formula units
Na2SO4(aq) + Ba(OH)2(aq)
2NaOH(aq) + BaSO4(s)
Precipitation Reactions
In the reaction between aqueous Na2SO4and Ba(OH)2
Na2SO4(aq) + Ba(OH)2(aq)
2NaOH(aq) + BaSO4(S)
the aqueous species are represented as follows:
Na2SO4(aq) → 2Na+(aq) + SO42–(aq)
Ba(OH)2(aq) → Ba2+(aq) + 2OH–(aq)
NaOH(aq) → Na+(aq) + OH–(aq)
In an ionic equation compounds that exist completely or predominately as
ions in solution are represented as those ions.
2Na+(aq) + SO42– (aq) + Ba2+(aq) + 2OH–(aq)
2Na+(aq) + 2OH–(aq) + BaSO4(s)
Precipitation Reactions
An equation that includes only the species that are actually involved in the
reaction is called a net ionic equation.
Ions that appear on both sides of the equation are called spectator ions.
2Na+(aq) + SO42– (aq) + Ba2+(aq) +2OH–(aq)
2Na+(aq) + 2OH–(aq) + BaSO4(s)
Spectator ions do not participate in the reaction.
Ba2+(aq) + SO42– (aq)
BaSO4(s)
Precipitation Reactions
To determine the molecular, ionic and net ionic equations:
1) Write and balance the molecular equation, predicting the products by
assuming that the cations trade anions.
2) Write the ionic equation by separating strong electrolytes into their
constituent ions.
3) Write the net ionic equation by identifying and canceling spectator ions
on both sides of the equation.
4) If both the reactants and products are all strong electrolytes, there is no
net ionic equation; no reaction takes place.
Precipitation Reactions
Write the molecular, ionic, and net ionic equations for the combination of
Fe(NO3)2(aq) with Na2CO3(aq).
Solution:
Step 1: Write and balance the molecular equation, predicting the products
by assuming that the cations trade anions; checking the solubility of
each product.
Fe(NO3)2(aq) + Na2CO3(aq) → FeCO3(s) + 2NaNO3(aq)
Precipitation Reactions
Write the molecular, ionic, and net ionic equations for the combination of
Fe(NO3)2(aq) with Na2CO3(aq).
Solution:
Step 2: Write the ionic equation by separating strong electrolytes into their
constituent ions.
Fe2+(aq) + 2NO3– (aq) + 2Na+(aq) + CO32–(aq) → FeCO3(s) + 2Na+(aq) + 2NO3–(aq)
Precipitation Reactions
Write the molecular, ionic, and net ionic equations for the combination of
Fe(NO3)2(aq) with Na2CO3(aq).
Solution:
Step 3: Write the net ionic equation by identifying and canceling spectator
ions on both sides of the equation.
Fe2+(aq) + 2NO3– (aq) + 2Na+(aq) + CO32–(aq) → FeCO3(s) + 2Na+(aq) + 2NO3–(aq)
2–
Fe2+(aq) + CO3 (aq) → FeCO3(s)
4.3
Acid-Base Reactions
Acids can be either strong or weak.
A strong acid is a strong electrolyte.
Acid-Base Reactions
A weak acid is a weak electrolyte; it does not dissociate completely.
Acetic acid, HC2H3O2, is an example.
HC2H3O2(l)
–
H+(aq) + C2H3O2 (aq)
acidic proton
Most acids are weak acids.
Acid-Base Reactions
Strong bases are strong electrolytes.
Strong bases are the hydroxides of Group 1A and heavy Group 2A
Sodium hydroxide, NaOH, is an example.
NaOH(s)
H2O
Na+(aq) + OH–(aq)
Acid-Base Reactions
An Arrhenius acid is one that ionizes in water to produce H+ ions.
HCl(g)
H2O
H+(aq) + Cl–(aq)
An Arrhenius base is one that dissociates in water to produce OH– ions.
NaOH(s)
H2O
Na+(aq) + OH–(aq)
Acid-Base Reactions
A Brønsted acid is a proton donor.
A Brønsted base is a proton acceptor.
In these definitions, a proton refers to a hydrogen atom that has lost its
electron—also known as a hydrogen ion (H+).
NH3(g) + H2O(l)
NH3 is a Brønsted base:
accepts a proton
to
+
become NH4
H2O is a Brønsted acid:
donates a proton to
become OH–
+
NH4(aq) + OH–(aq)
Acid-Base Reactions
Brønsted acids donate protons to water to form the hydronium ion (H3O+).
hydrogen ion (H+)
proton
All refer to the same aqueous species
hydronium ion (H3O+)
HF(aq) + H2O(l)
H2O
H3O+(aq) + F–(aq)
Acid-Base Reactions
A monoprotic acid has one proton to donate.
Hydrochloric acid is an example:
HCl(g)
H+(aq) + Cl–(aq)
one equivalent of solvated
hydrogen ion
Acid-Base Reactions
A polyprotic acid has more than one acidic hydrogen atom.
Sulfuric acid, H2SO4, is an example of a diprotic acid; there are two acidic
hydrogen atoms.
Polyprotic acids lose protons in a stepwise fashion:
Step 1:
H2SO4(aq)
–
H+(aq) + HSO4 (aq)
In H2SO4, the first ionization is strong.
Step 2:
–
HSO4(aq)
2–
H+(aq) + SO4 (aq)
In H2SO4, the second ionization
occurs only to a very small extent.
Acid-Base Reactions
Bases that produce only one mole of hydroxide per mole of compound are
called monobasic.
Sodium hydroxide is an example:
NaOH(s)
H2O
Na+(aq) + OH–(aq)
one equivalent of hydroxide
Acid-Base Reactions
Some strong bases produce more than one hydroxide per mole of
compound.
Barium hydroxide is an example of a dibasic base.
Ba(OH)2(s)
H2O
Ba2+(aq) + 2OH–(aq)
two equivalents of hydroxide
Acid-Base Reactions
A neutralization reaction is a reaction between an acid and a base.
Generally, a neutralization reaction produces water and a salt.
HCl(aq)
acid
+
NaOH(aq)
→
base
H2O(l)
water
The net ionic equation of many acid-base reactions is:
H+(aq) + OH–(aq)
H2O(l)
+
NaCl(aq)
salt
4.4
Oxidation-Reduction Reactions
An oxidation-reduction (or redox) reaction is a chemical reaction in which
electrons are transferred from one reactant to another.
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
Zn metal loses 2 electrons and is
oxidized to Zn2+
Zn2+ is called the reducing agent
Zn(s)
+
Cu2+(aq)
→
Zn2+(aq)
+
Cu2+ gains 2 electrons and is
reduced to Cu metal
Cu is called the oxidizing agent
Cu(s)
Oxidation-Reduction Reactions
Zn(s)
+
Cu2+(aq)
→
Zn2+(aq)
+
Cu(s)
Oxidation-Reduction Reactions
A redox reaction is the sum of an oxidation half-reaction and a reduction
half-reaction.
Oxidation (lose 2e–)
Zn(s)
+
Cu2+(aq)
→
Zn2+(aq)
+
Cu(s)
Reduction (gain 2e–)
Oxidation half-reaction:
Zn(s)
Reduction half-reaction:
Cu2+(aq) + 2e–
Overall redox reaction:
Cu2+(aq) + Zn(s)
Zn2+(aq) + 2e–
Cu(s)
Zn2+(aq) + Cu(s)
Oxidation-Reduction Reactions
The oxidation number is the charge an atom would have if electrons were
transferred completely.
H2(g)
+
F2(g)
→
2HF(g)
Oxidation number:
0
0
+1 –1
Total contribution to
charge:
0
0
+1 –1
N2(g)
+
3H2(g)
→
2NH3(g)
Oxidation number:
0
0
–3 +1
Total contribution to
charge:
0
0
–3 +3
The oxidation number is sometimes called the oxidation state.
Oxidation-Reduction Reactions
To assign oxidation numbers:
1) The oxidation number of an element, in its elemental form, is zero.
2) The oxidation numbers in any chemical species must sum to the overall
charge on the species.
•
must sum to zero for any molecule
•
must sum to the charge on any polyatomic ion
•
the oxidation number of a monoatomic ion is equal to the
charge on the ion
Oxidation-Reduction Reactions
To assign oxidation numbers:
3)Know the elements that nearly always have the same oxidation number.
Oxidation-Reduction Reactions
Assign the oxidation numbers to the elements in the compound KMnO4.
Step 1: Start with the oxidation numbers you know:
K
Mn O4
Oxidation number:
+1
+7
–2
Total contribution to
charge:
+1
+7
–8
Step 2: The numbers in the boxes (total contribution to charge) must sum
to zero (KMnO4 is a neutral compound).
Oxidation-Reduction Reactions
Assign the oxidation numbers to the elements in the compound H2SO4.
Solution:
Step 1: Start with the oxidation numbers you know:
H2
S
O4
Oxidation number:
+1
+6
–2
Total contribution to
charge:
+2
+6
–8
Step 2: The numbers in the boxes (total contribution to charge) must sum
to zero (the chemical species is neutral).
Oxidation-Reduction Reactions
–
Assign the oxidation numbers to the elements in the ion ClO3.
Solution:
Step 1: Start with the oxidation numbers you know:
–
Cl
O3
Oxidation number:
+5
–2
Total contribution to
charge:
+5
–6
Step 2: The numbers in the boxes (total contribution to charge) must sum
to negative one (the chemical species is a –1 anion).
Oxidation-Reduction Reactions
In a displacement reaction, an atom or an ion in a compound is replaced by
an atom of another element.
Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)
0
+2 –1
+2 –1
0
0
+2 –2
+2 –2
0
Zinc displaces, or replaces copper in the dissolved salt.
Zn is oxidized to Zn2+
Cu2+ is reduced to Cu
When a metal is oxidized by an aqueous solution, it becomes an
aqueous ion.
Oxidation-Reduction Reactions
Metals listed at the top are
called active metals.
Metals listed at the bottom are
called noble metals.
An element in the series will be
oxidized by the ions of any
element that appears below it in
the table.
Increasing ease of oxidation
The activity series is a list of metals (and hydrogen) arranged from top to
bottom in order of decreasing ease of oxidation.
Element
Oxidation Half-Reaction
Zinc
Zn → Zn2+ + 2e–
Iron
Fe → Fe2+ + 2e–
Nickel
Ni → Ni2+ + 2e–
Hydrogen
H2 → 2H+ + 2e–
Copper
Cu → Cu2+ + 2e–
Silver
Ag → Ag+ + e–
Gold
Au → Au3+ + 3e–
Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)
Cu(s) + ZnCl2(aq) → no reaction
Oxidation-Reduction Reactions
Which of the following reactions will occur?
Sn(s) + CuBr2(aq)
Ag(s) + NaCl(aq)
?
?
?
Increasing ease of
oxidation
Co(s) + BaI2(aq)
Element
Oxidation Half-Reaction
Sodium
Na → Na+ + 1e–
Barium
Ba → Ba2+ + 2e–
Cobalt
Co → Co2+ + 2e–
Tin
Sn → Sn2+ + 2e–
Hydrogen
H2 → 2H+ + 2e–
Copper
Cu → Cu2+ + 2e–
Silver
Ag → Ag+ + e–
Solution:
Co(s) + BaI2(aq)
Sn(s) + CuBr2(aq)
Ag(s) + NaCl(aq)
No reaction. Cobalt is below barium.
Cu(s) + Sn2+(aq)
No reaction. Silver is below sodium.
Oxidation-Reduction Reactions
Redox reactions must have both mass balance and charge balance.
Cr(s) + Ni2+(aq) → Cr3+(aq) + Ni(s)
Oxidation half-reaction:
Cr(s)
Reduction half-reaction:
Ni2+(aq) + 2e–
Cr3+(aq) + 3e–
Ni(s)
Before adding half-reactions, the electrons must balance.
Oxidation-Reduction Reactions
Prior to adding the two half-reactions, balance the electrons.
Step 1: Multiply the oxidation half-reaction by 2
Oxidation half-reaction:
2 Cr(s)
Cr3+(aq) + 3e–
Step 3: Multiply the reduction half-reaction by 3
Reduction half-reaction:
3 Ni2+(aq) + 2e–
Oxidation half-reaction:
2Cr(s)
Reduction half-reaction:
3Ni2+(aq) + 6e–
3Ni2+(aq) + 2Cr(s)
Ni(s)
2Cr3+(aq) + 6e–
2Ni(s)
2Ni(s) + 2Cr3+(aq)
Oxidation-Reduction Reactions
Give the overall balanced equation for the following reactions.
Cd(s) + Ag+(aq)
Solution:
Cu(s) + H+(aq)
Cd(s) + 2Ag+(aq)
?
?
Increasing ease of
oxidation
Cu(s) + H+(aq)
Element
Oxidation Half-Reaction
Barium
Ba → Ba2+ + 2e–
Cadmium
Cd → Cd2+ + 2e–
Tin
Sn → Sn2+ + 2e–
Hydrogen
H2 → 2H+ + 2e–
Copper
Cu → Cu2+ + 2e–
Silver
Ag → Ag+ + e–
No reaction. Copper is below hydrogen.
2Ag(s) + Cd2+(aq)
Oxidation-Reduction Reactions
Combination reactions can involve oxidation and reduction.
N2(g)
+
3H2(g)
→
2NH3(g)
0
0
–3 +1
0
0
–3 +3
Hydrogen is oxidized from 0 to +1
Nitrogen is reduced from 0 to –3
Oxidation-Reduction Reactions
Decomposition can also be a redox reaction.
NaH(s)
→
2Na(s)
+
3H2(g)
+1 –1
0
0
+1 –1
0
0
Na+ is reduced to Na
H– is oxidized to H2
Oxidation-Reduction Reactions
Disproportionation reactions occur when one element undergoes both
oxidation and reduction.
oxidation
reduction
2H2O2(aq)
→
2H2O(l)
+
O2(g)
+1 –1
+1 –2
0
+2 –2
+2 –2
0
Oxygen in H2O2 (and other peroxides) has an oxidation number of –1.
Oxidation-Reduction Reactions
Combustion is also a redox process.
CH4(g)
+
2O2(g)
→
CO2(g)
+
2H2O(l)
–4 +1
0
+4 –2
+1 –2
–4 +4
0
+4 –4
+2 –2
4.5
Concentration of Solutions
Molarity (M), or molar concentration, is defined as the number of moles
of solute per liter of solution.
molarity =
moles solute
liters solution
Other common
rearrangements:
L=
mol
M
mol = M  L
Concentration of Solutions
Determine the molarity of a 5.00 L solution that contains 235 g of sucrose,
C12H22O11.
Solution:
Step 1: Determine the molar mass of sucrose. (342.30 g/mol)
Step 2: Determine the moles of sucrose. (0.68653 moles sucrose)
Step 3: Determine the molarity of the solution:
moles solute
molarity =
liters solution
molarity =
0.68653 mol sucrose
= 0.137 M
5.00 L
Concentration of Solutions
How many millilitres of 3.50 M NaOH can be prepared from 75.00 grams of
solid NaOH?
Solution:
Step 1: Convert grams to moles. (1.87509 moles NaOH)
Step 2: Use the molarity equation to find liters; convert to mL:
molarity =
liters of solution =
moles solute
liters solution
1.87509 mol NaOH
= 0.53574 L = 536 mL
3.50 M
Concentration of Solutions
Dilution is the process of preparing a less concentrated solution from a
more concentrated one.
moles of solute before dilution = moles of solute after dilution
Concentration of Solutions
In an experiment, a student needs 250.0 mL of a 0.100 M CuCl2 solution. A
stock solution of 2.00 M CuCl2 is available.
How much of the stock solution is needed?
Solution:
Use the relationship that moles of solute before dilution =
moles of solute after dilution.
Mc x Lc = Md x Ld
(2.00 M CuCl2)(Lc) = (0.100 M CuCl2)(0.2500 L)
Lc = 0.0125 L or 12.5 mL
To make the solution:
1) Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask.
2) Carefully dilute to the calibration mark.
Concentration of Solutions
What volume of 6.0 M H2SO4 is needed to prepare 500.0 mL of a solution
that is 0.25 M H2SO4?
Solution:
Use the relationship that moles of solute before dilution =
moles of solute after dilution.
Mc x Lc = Md x Ld
(6.0 M H2SO4)(Lc) = (0.25 M H2SO4)(0.5000 L)
Lc = 0.021 L or 21 mL
Concentration of Solutions
What are the concentrations of ions in a solution that is 0.750 M in barium
nitrate? [Ba(NO3)2]
Solution:
Step 1: Determine if barium nitrate is a strong electrolyte using solubility
rules and, if so, write the equation for the dissociation.
Ba(NO3)2(s)
H2O
Ba2+(aq) + 2NO3‒(aq)
Step 2: Determine the concentration of Ba2+(aq) ions using the
stoichiometric ratio.
0.750 mol Ba(NO3 )2
1 mol Ba2+
1 mol Ba2+
[Ba ] = [Ba(NO3 )2 ] 
=

= 0.750 M Ba2+
1 mol Ba(NO3 )2
L
1 mol Ba(NO3 )2
2+
Square brackets around a chemical species indicates concentration.
Concentration of Solutions
What are the concentrations of ions in a solution that is 0.750 M in barium
nitrate? [Ba(NO3)2]
Ba(NO3)2(s)
H2O
Ba2+(aq) + 2NO‒3(aq)
Solution:
Step 3: Determine the concentration of NO3–(aq) ions using the
stoichiometric ratio.
2 mol NO3
0.750 mol Ba(NO3 )2
2 mol NO3
[NO ] = [Ba(NO3 )2 ] 
=

= 1.50 M NO3
1 mol Ba(NO3 )2
L
1 mol Ba(NO3 )2

3
4.6
Aqueous Reactions and Chemical Analysis
Gravimetric analysis is an analytical technique based on the
measurement of mass.
Gravimetric analysis is highly accurate.
Applicable only to reactions that go to completion or have nearly 100 %
yield.
Aqueous Reactions and Chemical Analysis
One common type of gravimetric analysis involves the isolation of a
precipitate.
Typical steps involve:
1) Mass an unknown solid.
2) Dissolve the unknown in water.
3) React the unknown with an excess amount of a substance that is
known to form a precipitate.
4) Filter, dry and weigh the precipitate.
5) Use the formula mass and the mass of the precipitate to find the %
mass of the unknown ion.
Aqueous Reactions and Chemical Analysis
A 0.825 g sample of an ionic compound containing chloride ions and an
unknown metal is dissolved in water and treated with excess silver nitrate.
If 1.725 g of AgCl precipitate forms, what is the percent by mass of Cl in
the original sample?
Solution:
Step 1: Find the percent by mass chlorine in AgCl.
% Cl =
35.45 g Cl
 100 = 24.7 %
143.35 g AgCl
Step 2: Multiply the percent of Cl by the mass of the precipitate to obtain
the mass of Cl in the sample.
0.247 x 1.725 g AgCl = 0.427 g Cl in the sample
Aqueous Reactions and Chemical Analysis
A 0.825 g sample of an ionic compound containing chloride ions and an
unknown metal is dissolved in water and treated with excess silver nitrate.
If 1.725 g of AgCl precipitate forms, what is the percent by mass of Cl in the
original sample?
Solution:
Step 3: Divide the mass of Cl in sample by the total mass of sample;
multiply by 100 to determine percent.
% Cl in unknown =
0.427 g Cl
 100 = 51.7% Cl
0.825 g sample
Aqueous Reactions and Chemical Analysis
Quantitative studies of acid-base neutralization reactions are most
conveniently carried out using a technique known as a titration.
A titration is a volumetric technique
that uses burets.
The point in the titration where the
acid has been neutralized is called
the equivalence point.
Aqueous Reactions and Chemical Analysis
The equivalence point is usually signalled by a color change.
The color change is brought about
by the use of an indicator.
Indicators have distinctly different
colors in acidic and basic media.
The indicator is chosen so that the
color change, or endpoint, is very
close to the equivalence point.
Phenolphthalein is a common
indicator.
Aqueous Reactions and Chemical Analysis
Typical steps of a titration include:
1) Prepare an unknown solution
2) Add an appropriate indicator to visualize the endpoint.
3) Carefully add a standard solution with a buret until the endpoint is
reached.
4) Using solution stoichiometry, calculate the molarity of the
unknown solution.
Aqueous Reactions and Chemical Analysis
Sodium hydroxide solutions are commonly used in titrations.
NaOH solutions must be standardized as the concentrations change over
time. (NaOH reacts with CO2 that slowly dissolves into the solution
forming carbonic acid.)
The acid potassium hydrogen phthalate (KHP) is frequently used to
standardize NaOH solutions.
acidic proton of KHP;
KHP is a monoprotic acid
Aqueous Reactions and Chemical Analysis
A sample of 1.00 g of KHP was used to standardize a sodium hydroxide
solution. The standardization required 15.00 mL of NaOH to reach the
endpoint.
What is the concentration of the NaOH solution?
Solution:
Step 1: Use the molar mass of KHP to determine the moles of KHP.
1.00 g KHP
moles of KHP =
= 0.0048971 moles KHP
204.2 g/mol
Aqueous Reactions and Chemical Analysis
A sample of 1.00 g of KHP was used to standardize a sodium hydroxide
solution. The standardization required 15.00 mL of NaOH to reach the
endpoint.
What is the concentration of the NaOH solution?
Solution:
Step 2: Using the net ionic equation, convert to moles of NaOH
KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O(l)
1 mol NaOH
moles of NaOH = 0.0048971 mol KHP 
= 0.0048971 mol NaOH
1 mol KHP
Aqueous Reactions and Chemical Analysis
A sample of 1.00 g of KHP was used to standardize a sodium hydroxide
solution. The standardization required 15.00 mL of NaOH to reach the
endpoint.
What is the concentration of the NaOH solution?
Solution:
Step 3: Use the molarity equation to calculate molarity of the sodium
hydroxide solution.
molarity of NaOH =
mol NaOH
0.0048971 moles NaOH
=
= 0.326 M NaOH
liters of solution
0.01500 L solution
Aqueous Reactions and Chemical Analysis
How many milliliters of a 1.42 M H2SO4 solution are needed to neutralize
95.5 mL of a 0.336 M KOH solution?
Solution:
Step 1: Write and balance the chemical equations that corresponds to the
neutralization reaction:
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
Step 2: Use the molarity and volume given to determine the moles of
KOH.
moles of KOH = 0.336 M KOH x 0.0955 L = 0.032088 moles KOH
Aqueous Reactions and Chemical Analysis
How many milliliters of a 1.42 M H2SO4 solution are needed to neutralize
95.5 mL of a 0.336 M KOH solution?
Solution:
Step 3: Using the moles of KOH and the stoichiometric ratios from the
balanced equation, convert to moles of H2SO4.
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
0.032088 mol KOH 
1 mol H2SO4
= 0.016044 mol H2SO4
2 mol KOH
Aqueous Reactions and Chemical Analysis
How many millilitres of a 1.42 M H2SO4 solution are needed to neutralize
95.5 mL of a 0.336 M KOH solution?
Solution:
Step 4: Using the moles of H2SO4 and the concentration given, calculate
the millilitres of solution.
0.016044 mol H2SO4
= 0.0113 L or 11.3 mL
1.42 M solution
4
Chapter Summary: Key Points
Properties of Aqueous Solutions
Electrolytes and Nonelectrolytes
Strong Electrolytes and Weak
Electrolytes
Precipitation Reactions
Solubility of Ionic Compounds in
Water
Molecular Equations
Ionic Equations
Net Ionic Equations
Acid-Base Reactions
Strong Acids and Bases
Brønsted Acids and Bases
Acid-Base Neutralization
Oxidation-Reduction Reactions
Oxidation Numbers
Oxidation of Metals in Aqueous
Solutions
Balancing Simple Redox Equations
Other Types of Redox Reactions
Concentration of Solutions
Molarity
Dilution
Solution Stoichiometry
Aqueous Reactions and Chemical
Analysis
Gravimetric Analysis
Acid-Base Titrations