Download Chapter 4 Student Notes

Chemistry 20H
Chemistry: The Central Science
Chapter 4 Aqueous Reactions and Solution Stoichiometry
Objectives
1.
Discuss the general properties of aqueous solutions. (4.1)
2.
Be able to predict whether a precipitate will form when two solutions are mixed (double replacement,
exchange, or metathesis reactions). (4.2)
3.
Write complete ionic equations and net ionic equations for precipitation reactions. (4.2)
4.
Discuss the nature of acids and bases. (4.3)
5.
Distinguish between strong and weak acids and bases. (4.3)
6.
Be able to list several strong acids and bases. (4.3)
7.
Be able to predict whether a compound is a strong, weak, or non-electrolyte. (4.3)
8.
Be able to identify neutralization reactions and predict the products of such reactions. (4.3)
9.
Be able to predict which neutralization reactions will form gases. (4.3)
10.
Discuss the process of oxidation and reduction. (4.4)
11.
Be able to determine the oxidation number of each atom in any element or compound. (4.4)
12.
Be able to identify displacement reactions (single replacement). (4.4)
13.
Use activity series to determine whether a displacement reaction will proceed. (4.4)
14.
Be able to calculate the concentration of solutions and incorporate what we know about moles. (4.5)
15.
Be able to perform dilution calculations. (4.5)
16.
Be able to solve problems involving solution stoichiometry. (4.6)
17.
Use titration to collect data for solution stoichiometry. (4.6)
Vocabulary
See the Summary and Key Terms on pages 156 & 157 of the text.
2
4.1
General Properties of Aqueous Solutions
o
o
o
o
A solution is a homogeneous mixture of two or more substances.
A solution is made when one substance (the solute) is dissolved in another (the solvent).
The solute is the substance that is present in the smallest amount.
Solutions in which water is the solvent are called aqueous solutions.
Electrolytic Properties
o
o
o
All aqueous solutions can be classified in terms of whether or not they conduct electricity.
If a substance forms ions in solution, then the substance is an electrolyte and the solution conducts
electricity. An example is NaCl.
If a substance does not form ions in solution, then the substance is a nonelectrolyte and the solution
does not conduct electricity. An example is sucrose.
Ionic Compounds in Water
o
o
When an ionic compound dissolves in water, the ions are said to dissociate.
o
o
o
o
o
o
This means that in solution, the solid no longer exists as a well-ordered arrangement of ions in
contact with one another.
Instead, each ion is surrounded by a shell of water molecules.
This tends to stabilize the ions in solution and prevent cations and anions from recombining.
The positive ions have the oxygen atoms of water pointing towards the ion; negative ions have
the hydrogen atoms of water pointing towards the ion.
The transport of ions through the solution causes electric current to flow through the solution.
Dissociation equations
o
We can write equations to illustrate the dissociation of ionic compounds in water:

o
HC2H3O2 (aq)  H1+(aq) + C2H3O21- (aq)
Like all chemical equations, dissociation equations must be balanced:

(NH4)3PO4 (aq)  3 NH4 (aq) + PO4 (aq)
Write dissociation equations for the ionic substances in questions 4.15 and 4.16
3
Molecular Compounds in Water
o
o
o
When a molecular compound (e.g. CH3OH ) dissolves in water, there are no ions formed.
Therefore, there is nothing in the solution to transport electric charge and the solution does not
conduct electricity.
There are some important exceptions:
o
o
For example, NH3(g) reacts with water to form NH4+(aq) and OH1–(aq).
For example, HCl(g) in water ionizes to form H1+(aq) and Cl1- (aq).
Strong and Weak Electrolytes
o
o
Compounds whose aqueous solutions conduct electricity well are called strong electrolytes.
These substances exist only as ions in solution.
o
o
Example: NaCl
 NaCl(aq)  Na1+(aq) + Cl1-(aq)
The single arrow indicates that the Na1+ and Cl1- ions have no tendency to recombine to form
NaCl molecules.
o
o
In general, soluble ionic compounds (metal + non-metal) are strong electrolytes.
A few molecular substances, like HCl are also strong electrolytes.
o
o
o
Compounds whose aqueous solutions conduct electricity poorly are called weak electrolytes
These substances exist as a mixture of ions and un-ionized molecules in solution.
The predominant form of the solute is the un-ionized molecule.
o
o
o
o
Example: acetic acid, HC2H3O2 (CH3COOH)
 HC2H3O2 (aq)  H1+(aq) + C2H3O21-(aq)
The double arrow means that the reaction is significant in both directions.
It indicates that there is a balance between the forward and reverse reactions.
This balance produces a state of chemical equilibrium.
o
Organic acids tend to be weak electrolytes
o
NOTE: There is no relationship between the strength (degree of dissociation) and concentration (how
much is dissolved).
4.2 Precipitation Reactions
o
o
Reactions that result in the formation of an insoluble product are known as precipitation reactions.
A precipitate is an insoluble solid formed by a reaction in solution.
Example: Pb(NO3)2 (aq) + 2 KI(aq)  PbI2 (s) + 2 KNO3 (aq)
o
4
Solubility Guidelines for Ionic Compounds
o
o
The solubility of a substance at a particular temperature is the amount of that substance that can be
dissolved in a given quantity of solvent at that temperature.
A substance with a solubility of less than 0.01 mol/L is regarded as being insoluble.
Experimental observations have led to empirical guidelines for predicting solubility.
o
Solubility guidelines for common ionic compounds in water:
o
o
o
o
o
o
o
o
Compounds containing alkali metal ions or ammonium ions are soluble.
Compounds containing NO31- or C2H3O21- are soluble.
 An exception is Ag C2H3O2
Compounds containing Cl1-, Br1- or I1- are soluble.
 Exceptions are the compounds of Ag1+, Hg22+, and Pb2+.
Compounds containing SO42- are soluble.
 Exceptions are the compounds of Sr2+, Ba2+, Hg22+, and Pb2+.
Compounds containing S2- are insoluble.
 Exceptions are the compounds of NH41+, the alkali metal cations, and Ca2+, Sr2+, and
Ba2+.
Compounds of CO32- or PO43- are insoluble.
 Exceptions are the compounds of NH41+ and the alkali metal cations.
Compounds of OH1- are insoluble.
 Exceptions are the compounds of NH41+, the alkali metal cations, and Ca2+, Sr2+, and
Ba2+.
Complete questions 4.19 and 4.20
Exchange (Metathesis, Double Replacement) Reactions
o
o
Exchange reactions, or metathesis reactions, involve swapping ions in solution:
o AX + BY  AY + BX.
Many precipitation and acid-base reactions exhibit this pattern.
Ionic Equations
o
o
o
o
o
Consider 2 KI(aq) + Pb(NO3)2 (aq)  PbI2 (s) + 2 KNO3 (aq)
Both KI(aq) + Pb(NO3)2 (aq) are colorless solutions. When mixed, they form a bright yellow precipitate of
PbI2 and a solution of KNO3.
The final product of the reaction contains solid PbI2, aqueous K1+, and aqueous NO31- ions.
Sometimes we want to highlight the reaction between ions.
The molecular equation lists all species in their molecular forms:
o
o
o
Pb(NO3)2 (aq) + 2 KI(aq)  PbI2 (s) + 2 KNO3 (aq)
The complete ionic equation lists all strong soluble electrolytes in the reaction as ions:
o
Pb2+(aq) + 2 NO31-(aq) + 2 K1+(aq) + 2 I1-(aq)

PbI2(s) + 2 K1+(aq) + 2 NO31-(aq)
5
o
o
o
Only strong electrolytes dissolved in aqueous solution are written in ionic form.
Weak electrolytes and nonelectrolytes are written in their molecular form.
The net ionic equation lists only those ions which are not common on both sides of the reaction:
o
o
Pb2+(aq) + 2 I1-(aq)  PbI2 (s)
Note that spectator ions, ions that are present in the solution but play no direct role in the reaction,
are omitted in the net ionic equation.
Write the balanced molecular equation, complete ionic equation and net ionic equation for the
following:
 FeS
a)
FeCl2 (aq) + K2S
b)
AlBr3 (aq) + NaOH
c)
(NH4)3PO4 (aq) + Ca(NO3)2 (aq)  NH4NO3 (aq) + Ca3(PO4)2 (s)
d)
Aqueous solutions of silver nitrate and sodium carbonate react to produce a precipitate of silver
carbonate and aqueous sodium nitrate.
e)
Aqueous solutions of barium chloride and potassium sulphate react to produce a precipitate of barium
sulphate and aqueous potassium chloride.
(aq)
(aq)
(s)
+ KCl
(aq)
 NaBr (aq) + Al(OH)3 (s)
Complete questions 4.21 to 4.24. In each reaction with a precipitate write the balanced molecular
equation, complete ionic equation and net ionic equation
4.3 Acid-Base Reactions
Acids
o
o
o
o
o
o
o
o
Acids are substances that are able to ionize in aqueous solution to form H 1+ (Arrhenius definition).
H1+ lacks an electron, so it is a bare proton.
Acids are often called proton donors in acid-base reactions.
An example is HC2H3O2 (acetic acid).
Since H1+ is a naked proton, we refer to acids as proton donors and bases as proton acceptors.
Common acids are HCl, HNO3, vinegar, and vitamin C.
Acids that ionize to form one H1+ ion are called monoprotic acids (HCl, HNO3).
Acids that ionize to form two H1+ ions are called diprotic acids (H2SO4).
Bases
o Bases are substances that are able to ionize in aqueous solution to form OH 1- (Arrhenius definition).
o Bases are substances that accept or react with the H1+ ions formed by acids (proton acceptor).
o Hydroxide ions, OH1-, react with the H1+ ions to form water:
o H1+(aq) + OH1-(aq)  H2O(l)
o Common bases are NH3 (ammonia), NaOH (sodium hydroxide), Draino, and milk of magnesia.
o Compounds that do not contain OH1- ions can also be bases:
o Proton transfer between NH3 (a weak base) and water (a weak acid) is an example of an acid–
base reaction.
o Since there is a mixture of NH3, H2O, NH41+, and OH1- in solution, we write
 NH3 (aq) + H2O(l)  NH41+(aq) + OH1-(aq)
6
Strong and Weak Acids and Bases
o
o
o
o
o
Strong acids and strong bases are strong electrolytes.
They are completely ionized in solution.
Strong bases include: Group 1A metal hydroxides, Ca(OH)2, Ba(OH)2, and Sr(OH)2.
Strong acids include: HCl, HBr, HI, HClO3, HClO4, H2SO4, and HNO3.
We write the ionization of HCl as:
o HCl  H1+ + Cl1-
o
o
o
o
Weak acids and weak bases are weak electrolytes.
Therefore, they are partially ionized in solution.
HF(aq) is a weak acid; most acids are weak acids.
We write the ionization of HF as:
1o HF  H1+
Identifying Strong and Weak Electrolytes
o
o
o
o
Compounds can be classified as strong electrolytes, weak electrolytes, or nonelectrolytes by looking at
their solubility.
Strong electrolytes:
o Ionic compounds are usually strong electrolytes.
o Molecular compounds that are strong acids are strong electrolytes.
Weak electrolytes:
o Weak acids and bases are weak electrolytes.
Nonelectrolytes:
o All other compounds.
Complete questions 4.35 to 4.38
Neutralization Reactions and Salts
o
A neutralization reaction occurs when an acid and a base react:
o
o
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
(acid) +
(base)  (water) + (salt)
o
o
o
o
In general an acid and a base react to form a salt.
A salt is any ionic compound whose cation comes from a base and anion from an acid.
The other product, H2O, is a common weak electrolyte.
A typical example of a neutralization reaction is the reaction between an acid and a metal hydroxide:
o Mg(OH)2 (milk of magnesia) is a suspension.
o As HCl is added, the magnesium hydroxide dissolves, and a clear solution containing Mg 2+ and
Cl1- ions is formed.
o Molecular equation:
 Mg(OH)2 (s) + 2 HCl(aq)  MgCl2 (aq) + 2 H2O(l)
o Complete ionic equation:
 Mg(OH)2 (s) + 2 H1+(aq) + Cl1-(aq)  Mg2+(aq) + 2 Cl1-(aq) + 2 H2O(l)
o Net ionic equation:
 Mg(OH)2 (s) + 2 H1+(aq)  Mg2+(aq) + 2 H2O(l)
o
Note that the magnesium hydroxide is an insoluble solid; it appears in the net ionic equation.
7
Complete questions 4.39 and 4.40
Acid-Base Reactions with Gas Formation
o
There are many bases besides OH1- that react with H1+ to form molecular compounds:
o
o
o
o
o
o
Reaction of sulfides with acid gives rise to H2S(g).
Sodium sulfide (Na2S) reacts with HCl to form H2S(g).
Molecular equation:
 Na2S(aq) + 2 HCl(aq)  H2S(g) + 2 NaCl(aq)
Complete ionic equation:
 2 Na1+(aq) + S2-(aq) + 2 H1+(aq) + 2 Cl1-(aq)  H2S(g) + 2 Na1+(aq) + 2 Cl1-(aq)
Net ionic equation:
 2 H1+(aq) + S2–(aq)  H2S(g)
Carbonates and hydrogen carbonates (or bicarbonates) will form CO 2 (g) when treated with an acid:
o
o
o
Sodium bicarbonate (NaHCO3; baking soda) reacts with HCl to form bubbles of CO2 (g).
Molecular equation:
 NaHCO3 (s) + HCl(aq) 
(aq) + H2CO3 (aq)  H2O(l) + CO2 (g) + NaCl(aq)
Net ionic equation:
 H1+(aq) + HCO31-(aq)  H2O(l) + CO2 (g)
Complete question 4.43
4.4 Oxidation-Reduction Reactions
Oxidation and Reduction
o
o
Oxidation-reduction, or redox, reactions involve the transfer of electrons between reactants.
When a substances loses electrons, it undergoes oxidation:
o Ca(s) + 2 H1+(aq)  Ca2+(aq) + H2 (g)
o The neutral Ca has lost two electrons to 2 H1+ to become Ca2+.
o We say Ca has been oxidized to Ca2+.
o
When a substance gains electrons, it undergoes reduction:
o 2 Ca(s) + O2 (g)  2 CaO(s).
o In this reaction the neutral O2 has gained electrons from the Ca to become O2– in CaO.
o We say O2 has been reduced to O2–.
o
In all redox reactions, one species is reduced at the same time as another is oxidized.
8
Oxidation Numbers
o
o
o
Electrons are not explicitly shown in chemical equations.
Oxidation numbers (or oxidation states) help up keep track of electrons during chemical reactions.
Oxidation numbers are assigned to atoms using specific rules:
1. For an atom in its elemental form, the oxidation number is always zero.
2. For any monatomic ion, the oxidation number equals the charge on the ion.
 Nonmetals usually have negative oxidation numbers.
3. The oxidation number of oxygen is usually –2.
 The major exception is in peroxides (containing the O22– ion).
 The oxidation number of hydrogen is +1 when bonded to nonmetals and –1 when
bonded to metals.
 The oxidation number of fluorine is –1 in all compounds. The other halogens have an
oxidation number of –1 in most binary compounds.
4. The sum of the oxidation numbers of all atoms in a neutral compound is zero.
5. The sum of the oxidation numbers in a polyatomic ion equals the charge of the ion.
o
The oxidation of an element is evidenced by its increase in oxidation number; reduction is accompanied
by a decrease in oxidation number.
Example 1:
o
o
o
H2SO4
hydrogen generally has an O.N. of 1+
oxygen generally has an O.N. of 2sulfur is the problem, we get the answer mathematically:
2(H) + 1(S) + 4(O) = 0
(the substance has no total charge)
2(1+) + S + 4(2-) = 0
(2+) + S + (8-) = 0
S = 0 - (2+) - (8-)
S = 6+
Example 2:
o
Cr2O72-
The oxidation number of S is 6+
(the dichromate ion)
oxygen is usually 22(Cr) + 7(O) = 22Cr + 7(2-) = 22Cr + (14-) = 22Cr = 2- - (14-)
2Cr = 12+
Cr = 6+
The oxidation number of Cr is 6+
9
Complete questions 4.49 and 4.50
Recognizing Oxidation and Reduction in Chemical Reaction Equations
Consider the following balanced chemical equation:
3 Cu2+ + 2 Al  3 Cu + 2 Al3+
Let’s apply what we know about oxidation numbers to this equation:
O.N.
3 Cu2+
+ 2 Al 
2+
0
3 Cu
+
2 Al3+
0
3+
In this reaction the copper goes from 2+ on the reactant side to 0 on the product side; it’s charge got lower, so
it is reduced (it gained electrons). The aluminum goes from 0 to 3+; it’s charge got greater, so it is oxidized (it
lost electrons):
reduction (gained 2 electrons each)
┌───────────┐
3 Cu2+ + 2 Al  3 Cu + 2 Al3+
O.N.
2+
0
0
3+
└─────────┘
oxidation (lost 3 electrons each)
notice there are 3 coppers in the balanced equation; the total electrons gained is 3(2 e 1-) = 6 electrons.
there are 2 aluminums in the equation; the total electrons lost is 2(3 e 1-) = 6 electrons.
A redox equation is balanced if:
it is balanced for atoms on each side.
the total electrons lost and gained are equal.
A bit of terminology; the substance that is oxidized can be said to have caused the other substance to be
reduced. It is the reducing agent. The substance that is reduced can be said to have caused the other
substance to be oxidized. It is the oxidizing agent.
Our equation now looks like this:
oxidizing agent
reduction (gained 2 electrons each)
┌───────────┐
3 Cu2+ + 2 Al  3 Cu + 2 Al3+
O.N.
2+
0
0
3+
└─────────┘
oxidation (lost 3 electrons each)
reducing agent
10
Complete questions 4.51 and 4.52
Oxidation of Metals by Acids and Salts
o
o
The reaction of a metal with either an acid or a metal salt is called a displacement reaction (also
called single replacement).
The general pattern is:
o A + BX 
o
Example: It is common for metals to produce hydrogen gas when they react with acids. Consider the
reaction between Mg and HCl:
o Mg(s) + 2 HCl(aq)  MgCl2 (aq) + H2 (g)
o In the process the metal is oxidized and the H1+ is reduced.
o
Example: It is possible for metals to be oxidized in the presence of a salt:
o Fe(s) + Ni(NO3)2 (aq)  Fe(NO3)2 (aq) + Ni(s)
o The net ionic equation shows the redox chemistry well:
o Fe(s) + Ni2+(aq)  Fe2+(aq) + Ni(s)
o In this reaction iron has been oxidized to Fe2+, while the Ni2+ has been reduced to Ni.
o
Always keep in mind that whenever one substance is oxidized, some other substance must be reduced.
Complete questions 4.53 and 4.54
11
The Activity Series
o
o
o
Allows us to predict whether a specific redox reaction will occur.
We can list metals in order of decreasing ease of oxidation.
This list is an activity series:
o
o
o
o
The metals at the top of the activity series are called active metals.
The metals at the bottom of the activity series are called noble metals.
A metal in the activity series can only be oxidized by a metal ion below it.
If we place Cu into a solution of Ag1+ ions, then Cu2+ ions can be formed because Cu is above Ag in the
activity series:
o Cu(s) + 2 AgNO3 (aq)  Cu(NO3)2 (aq) + 2 Ag(s)
o or
o Cu(s) + 2 Ag1+(aq)  Cu2+(aq) + 2 Ag(s)
Complete questions 4.56 to 4.58
12
4.5 Concentrations of Solutions
o
The term concentration is used to indicate the amount of solute dissolved in a given quantity of solvent
or solution.
Molar Concentration (Molarity)
o
o
o
o
Solutions can be prepared with different concentrations by adding different amounts of solute to
solvent.
The moles of solute per liter of solution is the molar concentration (molarity) of the solution.
The units of molar concentration is moles per litre (mol/L, mol·L-1)
The text uses the unit M, meaning molarity, to stand in for mol/L.
Expressing the Concentration of an Electrolyte
o
when a substance dissolves it is called a solvation reaction:
o
o
C12H22O11 (s)  C12H22O11 (aq)
sugar is a non-electrolyte:
soluble ionic compounds undergo dissociation in solution; they break up into their constituent ions:
o
NaCl(s)  Na1+(aq) + Cl1-(aq)
(this represents both solvation and dissociation)
o
to write a dissociation equation for any ionic compound you must find out the identity of the cation and
anion, then write the equation, paying attention to the stoichiometry:
o
Al2(SO4)3 (s) contains the Al3+ and SO42- ions. When it dissociates you get 2 Al3+ and 3 SO42- ions:

Al2(SO4)3 (s)  2 Al3+(aq) + 3 SO42-(aq)
o
this means that the concentration of the ions may be different than the calculated concentration of
the substance in solution:
o
If the aluminum sulfate is a 0.200 mol/L solution, then

Al2(SO4)3 (s)

0.200 mol/L

2 Al3+(aq)
+
3 SO42-(aq)
2(0.200 mol/L)
3(0.200 mol/L)
 = 0.400 mol/L
= 0.600 mol/L
Write the equation for the dissociation of the following ionic substances. Calculate the
concentration of each ion in solution:
a)
0.35 mol/L NaOH
b)
1.12 mol/L (NH4)2CO3
c)
0.056 mol/L V3(PO4)5
13
Interconverting Molarity, Moles, and Volume
o
The definition of molarity contains three quantities: molarity, moles of solute, and liters of solution.
o If we know any two of these, we can calculate the third.
o Dimensional analysis is very helpful in these calculations.
Examples:
1)
Calculate the molar concentration when 0.256 mol of CuSO4 is dissolved in 1.1 L of solution.
molar concentration
c
2)
=
moles of solute
volume of solution
=
n
v
=
0.256 mol
1.1 L
=
0.23 mol/L (2 s.d.)
Calculate the molar concentration when 28.2 g of KNO 3 is dissolved in 750. mL of solution.
-
given 28.2 g ; need to convert mass to moles
molar mass:
KNO3
1xK
1xN
3xO
moles of substance
=
0.279 mol
volume must be reported in L:
750. mL x
-
1 x 39.10 g/mol
1 x 14.01 g/mol
3 x 16.00 g/mol
101.11 g/mol
28.2 g
101.11 g/mol
=
-
=
=
=
1L
1000 mL
= 0.750 L
molar concentration:
c
=
n
V
=
=
0.279 mol
0.750 L
0.372 mol/L
14
3)
Calculate the volume of solution made with 0.455 mol of sodium phosphate in a 2.00 M solution.
c
=
n
V
V
=
n
c
=
0.455 mol
2.00 mol/L
=
4)
0.228 L
How many moles are present in 4.50 L of a 0.025 mol/L solution of magnesium nitrate?
c
=
n
V
n
=
cV
=
(0.025 mol/L)(4.50 L)
=
0.11 mol
Complete questions 4.61 to 4.69, odd
Dilution
o
o
o
o
many substances, especially acids, are received in a concentrated form (hydrochloric acid is 12.4
mol/L). This is called the stock solution.
to use the chemicals in the lab they are usually diluted to a concentration much less than they are
received.
the problem is to decide how much of the stock solution you need to make the solution you want.
For instance; you have a 12.4 mol/L stock solution of HCl and you want to make 2.00 L of 0.100 mol/L
solution:
Since you know the volume and concentration of the dilute solution, you can calculate the
number of moles in the solution:
n
= cv
= (0.100 mol/L)(2.00 L)
= 0.200 mol
15
-
This represents the number of moles of HCl you need from the stock solution. Now you
have the number of moles and the concentration, so you can calculate the volume of
concentrated HCl you need to add to 2.00 L of water to make the dilute solution:
v
=
n
c
=
0.200 mol
12.4 mol/L
= 0.0161 L (1000 mL/L)
= 16.1 mL of concentrated HCl is added to water to make 2.00 L of a
0.100 mol/L solution.
o
Fortunately, there is an easier way to do this. Since the moles taken from the stock solution (n s) is the
same number of moles that goes into the dilute solution (nd)
o
and
o
o
ns = nd
o
ns = csvs
o
nd = cdvd
so the formula for dilution is:
o
csvs = cdvd
Complete questions 4.71 to 4.74
4.6 Solution Stoichiometry and Chemical Analysis
o
o
We can apply what we know to stoichiometry problems.
Consider this problem:
o 150. mL of 0.500 mol/L lead (II) nitrate is added to a solution containing sufficient potassium
iodide. What mass of lead (II) iodide will precipitate ? Assume the lead (II) iodide is
completely insoluble.
o
o
treat this as a normal stoichiometry problem:
Step 1. Write and balance the chemical equation
Pb(NO3)2 (aq) + 2 KI(aq)  PbI2 (s) + 2 KNO3 (aq)
o
Step 2. Convert given information to moles
moles of Pb(NO3)2 = (150. mL)(1 L / 1000 mL)(0.500 mol/L) = 0.0750 mol Pb(NO3)2
o
Step 3. Use the mole ratio to determine the moles of PbI 2
the mole ratio is 1:1 so there are 0.0750 mol PbI2
16
o
Step 4. Convert moles of unknown to desired units
PbI2 = 207.2 g/mol + 2(126.90 g/mol) = 461.00 g/mol
mass of PbI2 = (0.0750 mol)(461.00 g/mol) = 34.6 g
o
Step 5. Make a concluding statement
The given amount of lead (II) nitrate will produce 34.6 g of lead (II) iodide.
o
These problems are often connected to a lab technique called titration.
Titrations
o
o
o
o
A common way to determine the concentration of a solution is via titration.
The level of acidity or basicity is very important, especially in products manufactured for human
use.
We determine the concentration of one substance by allowing it to undergo a specific chemical
reaction, of known stoichiometry, with another substance whose concentration is known (standard
solution).
When it involves acids and bases this method involves neutralizing a sample of an unknown acid solution
with a known concentration of base (or vice-versa). At the neutralization point the moles of acid equals the
moles of base. Since both acid and base are in solution, the following formula is used to solve for moles,
volume and concentration:
c =
o
n
V
which is altered to
n = cV
o
The moles of base added is thus equal to:
nb = cbVb
o
and the moles of acid originally in solution is equal to
na = caVa
o
Since at the point of neutralization the moles of acid and base are equal, the following formula is
arrived at:
caVa = cbVb
o
Which is identical to the formula used in the dilution calculations.
17
o
Example Calculations:
What is the concentration of a base if 50.0 mL of the base is neutralized by 16.1 mL of a
0.100 mol/L solution of acid ?
cb
vb
ca
va
=
=
=
=
?
50.0 mL
0.100 mol/L
16.1 mL
cava = cbvb
cb =
cava
vb
cb =
(0.100 mol/L)(16.1 mL)
(50.0 mL)
cb = 0.0322 mol/L
Calculate the concentration of HCl if 100. mL of an HCl solution is neutralized by 56.3 mL of a
0.500 mol/L solution of NaOH
ca
va
cb
vb
=
=
=
=
?
100. mL
0.500 mol/L
56.3 mL
ca =
ca =
cbvb
va
(0.500 mol/L)(56.3 mL)
(100. mL)
ca = 0.282 mol/L HCl
Indicators
o
How does one know when the neutralization point has been reached when titrating? There are two ways. By
far the most popular method today is measuring the pH electronically using pH electrodes and computerized
titrators.
o
An older method uses indicators. These are chemicals which are pH sensitive. They undergo a colour change
as the pH shifts from acid to basic conditions. The neutralization point is detected when the solution changes
colour.
o
Indicators are the method of choice in student labs. They are also used in the field, where instrumentation is
limited.
o
Indicators are used not only in acid-base reactions, but in a broad range of chemical reactions,
including redox reactions
18
Complete questions 4.77 to 4.86