Download Chapter 4 Student Presentation

Chapter 4
Aqueous Reactions and
Solution Stoichiometry
Chapter Objectives
• Metathesis reactions (double replacement)
– precipitation reactions
– acid-base reactions
– redox reactions
• Single Replacement reactions (displacement)
– activity series
• Molar concentration
• Solution Stoichiometry
– titration
– indicators
General Properties of Aqueous Solutions
• Homogeneous mixtures of
two or more pure
substances.
• The solvent in greatest
abundance.
• All other substances are
solutes.
• If water is the solvent the
solution is aqueous
Electrolytic Properties
• whether or not solutions conduct electricity.
• If ions form in solution, the substance is an
electrolyte and the solution conducts electricity.
e.g. NaCl.
• If no ions form, the substance is a nonelectrolyte.
e.g. sucrose.
• Ions form in solution through the process of
dissociation.
Dissociation
• Solid does not exist as a well-ordered crystal.
• Each ion surrounded by a shell of water
molecules.
– cations have oxygen end of water pointing in.
– anions have hydrogen end of water pointing in.
• Ions move to cause electric current to flow
through solution.
• Electrolytes - Soluble ionic
compounds
• Non-electrolytes Molecular compounds
Electrolytes
• A strong electrolyte
dissociates completely
when dissolved in
water.
– ionic compounds,
many acids & bases.
• A weak electrolyte only
dissociates partially
when dissolved in
water.
– weak acids and
bases.
Solvation Equations
• A substance dissolves in a solvation reaction:
– sugar is a non-electrolyte:
C12H22O11
(s)
→ C12H22O11
(aq)
• soluble ionic compounds undergo dissociation in
solution; break up into constituent ions:
NaCl
(s)
→ Na1+(aq) + Cl1-(aq)
(this represents both solvation and dissociation)
• to write a dissociation equation for any ionic
compound identify the cation and anion, then
write the balanced equation:
• Al2(SO4)3
(s)
contains Al3+ and SO42- ions.
• When it dissociates you get 2 Al3+ and 3 SO42ions:
Al2(SO4)3
(s)
→ 2 Al3+(aq) + 3 SO42-(aq)
• (make sure the equation is balanced)
• Write dissociation equations for the
ionic substances in questions 4.15 and
4.16
Precipitation Reactions
• When a mixture of soluble
ions forms a compound
that is insoluble, a
precipitate is formed.
• Pb(NO3)2 (aq) + 2 KI(aq)  PbI2 (s) + 2 KNO3 (aq)
Solubility Guidelines for Ionic Compounds
• Solubility is the amount of a substance that
can be dissolved in a given quantity of solvent
at that temperature.
• Measured in g/100g H2O, or mol/L
• Solubility < 0.01 mol/L is insoluble.
• Experimental observations have led to
empirical guidelines for predicting solubility.
Solubility of Selected Ions (soluble > 0.1 mol/L)
Anion
any
any
nitrate ion (NO31-)
chlorate ion (ClO31-)
acetate ion
(CH3COO1-)
chloride (Cl1-),
bromide (Br1-), iodide
(I1-)
Sulfate (SO42-)
+
Cation
+ group 1 metal ion (Li1+, Na1+, K1+, Rb1+,
Cs1+)
+ ammonium ion (NH41+)
+ any
= Solubility
is soluble
+
+
+
+
is
is
is
is
+
+
Sulfide (S2-)
+
Hydroxide (OH1-)
+
+
Phosphate (PO43-)
Carbonate (CO32-)
Sulfite (SO32-)
Oxide (O2-)
+
Ag1+
any other cation
Ag1+, Pb2+, Hg22+, or Cu1+
Ag1+, Pb2+, Hg22+, Hg2+ or Cu1+
any other cation
Ca2+, Sr2+, Ba2+, Ra2+, Ag1+, Pb2+, Hg22+
Hg2+
any other cation
group 1 & 2 elements, NH41+
any other cation
group 1 elements & NH41+, Ca2+, Sr2+,
Ba2+
any other cation
group 1 elements & NH41+
any other cation
is soluble
is soluble
not soluble
soluble
not soluble
not soluble
soluble
is not soluble
is soluble
is
is
is
is
soluble
not soluble
soluble
not soluble
is soluble
is not soluble
Predicting Solubility
• Na2S
– soluble, because Na is an alkali metal (group 1)
• NH4Cl
– soluble, because the NH41+ ion is soluble with any anion
• CuNO3
– soluble, because NO31- is soluble with any cation
• AgCH3COO
– insoluble
• Mg(CH3COO)2
– soluble
• PbCl4
– soluble; in this compound the Pb is 4+
• PbCl2
– insoluble; in this compound the Pb is 2+
Predicting Solubility
•
•
•
•
•
•
•
•
•
•
KCl
NaNO3
AgCl
Al2(SO4)3
FeS
Ca3(PO4)2
BaSO4
Pb(ClO3)2
(NH4)2S
PbCl2
• Complete questions 4.19 and 4.20
Metathesis (Exchange) Reactions
• Metathesis comes from a Greek word
that means “to transpose”
AgNO3
+ KCl
 AgCl (s) + KNO3 (aq)
Metathesis (Exchange) Reactions
• Metathesis comes from a Greek word
that means “to transpose”
• It appears the ions in the reactant
compounds exchange, or transpose,
ions
AgNO3
+ KCl
 AgCl (s) + KNO3 (aq)
Metathesis (Exchange) Reactions
• Metathesis comes from a Greek word
that means “to transpose”
• It appears the ions in the reactant
compounds exchange, or transpose,
ions
AgNO3
+ KCl
 AgCl + KNO3
Metathesis (Exchange) Reactions
• Metathesis comes from a Greek word
that means “to transpose”
• It appears the ions in the reactant
compounds exchange, or transpose,
ions
• always indicate the phase of the
substance, where known
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
• most are reactions of ionic compounds
Chemical Equations of
Metathesis Reactions
Molecular Equation
The molecular equation lists the reactants
and products in their molecular form.
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Complete Ionic Equation
• All strong electrolytes are dissociated into
their ions.
• This more accurately reflects the species that
are found in the reaction mixture.
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq) 
AgCl (s) + K+ (aq) + NO3- (aq)
Net Ionic Equation
• To form the net ionic equation, cross out
anything that does not change from the left
side of the equation to the right.
Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) 
AgCl (s) + K+(aq) + NO3-(aq)
Net Ionic Equation
• To form the net ionic equation, cross out
anything that does not change from the left
side of the equation to the right.
• The only things left in the equation are those
things that change (i.e., react) during the
course of the reaction.
Ag+(aq) + Cl-(aq)  AgCl (s)
Net Ionic Equation
• To form the net ionic equation, cross out
anything that does not change from the left
side of the equation to the right.
• The only things left in the equation are those
things that change (i.e., react) during the
course of the reaction.
• Those things that didn’t change (and were
deleted from the net ionic equation) are called
spectator ions.
Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) 
AgCl (s) + K+(aq) + NO3-(aq)
Writing Net Ionic Equations
1. Write a balanced molecular equation.
2. Dissociate all strong electrolytes. Write
the complete ionic equation.
3. Cross out anything that remains
unchanged from the left side to the
right side of the equation.
4. Write the net ionic equation with the
species that remain.
Write the balanced molecular equation,
complete ionic equation and net ionic equation
for the following:
a) FeCl2 (aq) + K2S (aq) 
b) AlBr3 (aq) + NaOH (aq) 
c) (NH4)3PO4 (aq) + Ca(NO3)2 (aq) 
d) Aqueous solutions of silver nitrate and
sodium carbonate react.
e) Aqueous solutions of barium chloride and
potassium sulfate react.
• Complete questions 4.22 and 4.24. In
each reaction with a precipitate write
the balanced molecular equation,
complete ionic equation and net ionic
equation
• Hand in on:
Acids:
• 2 definitions:
»Substances that increase
the concentration of H1+
when dissolved in water
(Arrhenius).
»Proton donors (Brønsted–
Lowry).
»If an acid can donate
more than one proton it is
diprotic or triprotic.
Acids
Strong acids dissociate completely in water
There are only seven strong acids:
•
•
•
•
•
•
•
Hydrochloric (HCl)
Hydrobromic (HBr)
Hydroiodic (HI)
Nitric (HNO3)
Sulfuric (H2SO4)
Chloric (HClO3)
Perchloric (HClO4)
Bases:
• Substances that
increase the
concentration of
OH1- when dissolved
in water
(Arrhenius).
• Proton acceptors
(Brønsted–Lowry).
Bases
Strong bases
dissociate
completely in water.
The strong bases are
the soluble salts of
hydroxide ion:
• Alkali metals (LiOH,
NaOH, etc)
• Ca(OH)2, Sr(OH)2,
Ba(OH)2
Weak Acids and 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:
– HF  H1+ + F1-
Identifying Strong & Weak Electrolytes
• Compounds can be classified as by looking at
their solubility:
• Strong electrolytes:
– Ionic compounds.
– Strong acids are strong electrolytes.
• Weak electrolytes:
– Weak acids and bases.
• Nonelectrolytes:
– All other compounds.
• Complete questions 4.35 to 4.38
Acid-Base Reactions
In an acid-base
reaction, the acid
donates a proton
(H+) to the base.
Neutralization Reactions
When solutions of an acid and a base are
combined, the products are a salt and water.
Molecular Equation:
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
acid
base
salt
water
a salt is any non-acidic or basic ionic compound
Neutralization Reactions
When a strong acid reacts with a strong base, the
complete ionic equation is…
HCl
(aq)
+ NaOH (aq)  NaCl
(aq)
+ H2O (l)
H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq) 
Na+ (aq) + Cl- (aq) + H2O (l)
Neutralization Reactions
When a strong acid reacts with a strong base, the
net ionic equation is…
HCl
(aq)
+ NaOH (aq)  NaCl
(aq)
+ H2O (l)
H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq) 
Na+ (aq) + Cl- (aq) + H2O (l)
H+ (aq) + OH- (aq)  H2O (l)
Mg(OH)2 (s) + HCl (aq)
• Molecular Equation:
• Mg(OH)2 (s) + 2 HCl (aq) 
MgCl2 (aq) + 2 H2O (l)
• Complete Ionic Equation:
• Mg(OH)2 (s) + 2 H1+(aq) + 2 Cl1-(aq) 
Mg2+(aq) + 2 Cl1-(aq) + 2 H2O (l)
• Net Ionic Equation:
• Mg(OH)2 (s) + 2 H1+(aq) 
Mg2+(aq) + 2 H2O (l)
• Complete questions 4.39 & 4.40
Gas-Forming Reactions
• Reaction of sulfides with acid gives rise to
H2S(g).
• FeS (s) + 2 HCl (aq)  FeCl2 (aq) + H2S (g)
• Reaction of sulfites with acid gives rise to
SO2 (g).
• SrSO3 (s) + 2 HI (aq) 
SrI2 (aq) + SO2 (g) + H2O (l)
Gas Forming Reactions
• Acid with carbonate or hydrogen carbonate
forms CO2:
– CaCO3 (s) + HCl (aq) 
CaCl2 (aq) + CO2 (g) + H2O (l)
– NaHCO3 (aq) + HBr (aq) 
NaBr (aq) + CO2 (g) + H2O (l)
• Bases with ammonium compounds gives
ammonia gas:
– NH4Cl (aq) + NaOH (aq) 
NH3 (g)+ H2O (l) + NaCl (aq)
• Complete question 4.43
Oxidation-Reduction Reactions
• Involve the transfer of electrons
• An oxidation occurs when an atom or ion
loses electrons. (LEO)
• A reduction occurs when an atom or ion gains
electrons. (GER)
• In all redox reactions, one species is reduced
at the same time as another is oxidized.
Oxidation Numbers
• used to determine if an oxidationreduction reaction has occurred
• we assign a charge to each element in a
neutral compound or charged entity.
• these charges, real or virtual, are called
oxidation numbers
Rules of Oxidation Numbers
• Elements have an oxidation number of 0.
• The oxidation number of a monatomic ion is the
same as its charge.
• Oxygen has an oxidation number of 2- (except
peroxides, O22-)
• Hydrogen has an oxidation number of 1+ (except
hydrides, H1-)
• The sum of the oxidation numbers in a neutral
compound is 0.
• The sum of the oxidation numbers in a polyatomic
ion is the charge on the ion.
Calculation of Oxidation Numbers
• H2O
- H is 1+, O is 2• H2O2
– H is 1+, O is 1• AlH3
– Al is 3+, H is 1• CaCl2
– Ca is 2+, Cl is 1-
• PbS2
– Pb is 4+, S is 2• Mn3N7
– Mn is 7+, N is 3• H2SO4
– H is 1+, S is 6+, O
is 2• Cr2O72– Cr is 6+, O is 2-
• complete questions 4.49 and 4.50
Recognizing Oxidation and Reduction
• 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  3 Cu + 2 Al3+
2+
0
0
3+
• copper goes from 2+ to 0;
– 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)
• The total number of electrons lost and gained is the
same; 6
• 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 is the
reducing agent.
– The substance that is reduced is the
oxidizing agent.
Our equation now looks like this:
•
•
•
•
•
• O.N.
•
•
•
oxidizing agent
reduction (gained 2 electrons each)
┌───────────┐
3 Cu2+ + 2 Al  3 Cu + 2 Al3+
2+
0
0
3+
└────────────┘
oxidation (lost 3 electrons each)
reducing agent
Examples:
1. CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
2. 2 PbS(s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g)
• Complete questions 4.51 and 4.52
Displacement Reactions
• Also called single replacement reactions.
• In displacement reactions, one metal
replaces another metal in an ionic
compound or acids.
• The general pattern is:
– A + BX  AX + B
– where
• A is a neutral metal
• B is a metal cation
• X is any anion
Example:
• metals produce hydrogen gas with acids.
• Consider Mg and HCl:
– Mg(s) + 2 HCl(aq)  MgCl2 (aq) + H2 (g)
• the metal is oxidized
• H1+ is reduced.
Example:
• metals can be oxidized in the presence of a
salt:
– Fe(s) + Ni(NO3)2 (aq)  Fe(NO3)2 (aq) + Ni(s)
• The net ionic equation shows the redox
chemistry well:
– Fe(s) + Ni2+(aq)  Fe2+(aq) + Ni(s)
• iron has been oxidized to Fe2+
• Ni2+ has been reduced to Ni.
• Complete questions 4.53 and 4.54
Displacement Reactions
In this reaction,
silver ions oxidize
copper metal.
Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)
Displacement Reactions
The reverse reaction,
however, does not
occur.
Cu2+ (aq) + 2 Ag (s) 
x Cu (s) + 2 Ag+ (aq)
Why?
Activity Series
• is a list of metals in order of decreasing
ease of oxidation.
• allows us to predict whether a given
displacement reaction will occur.
Activity Series
• The metals at the top are active metals.
• The metals at the bottom are noble metals.
• A metal in the activity series can only be oxidized by a
metal ion below it.
• For example:
– If Cu + Ag1+ ions:
– Cu2+ ions are formed because Cu is above Ag in the
activity series:
– Cu(s) + 2 AgNO3 (aq)  Cu(NO3)2 (aq) + 2 Ag(s)
• or
– Cu(s) + 2 Ag1+(aq)  Cu2+(aq) + 2 Ag(s)
• Complete questions 4.56 to 4.58
Molar Concentration
• indicates the amount of solute
dissolved in a given quantity of
solvent or solution.
• is the most common measure of
concentration used in the lab.
• is the number of moles of solute, per
litre of solution (mol/L, mol∙L-1).
• is also called Molarity (M).
• for instance, 0.12 mol/L ≡ 0.12 M
Concentration of an Electrolyte
• the concentration of the ions may be
different than the calculated concentration
of the substance in solution:
• If the aluminum sulfate is a 0.200 mol/L
solution, then
Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)
• 0.200 mol/L 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:
– 0.35 mol/L NaOH
– 1.12 mol/L (NH4)2CO3
– 0.056 mol/L V3(PO4)5
Calculating Molarity
• Molar concentration (molarity)
calculations consist of 3 variables:
– concentration (mol/L)
– number of moles (mol)
– volume (L)
• If we know any two of these, we can
calculate the third.
• Dimensional analysis is very helpful in
these calculations.
Examples of Molarity
• Calculate the molar concentration when 0.256 mol
of CuSO4 is dissolved in 1.1 L of solution.
• molar concentration =
•
•
c
=
=
=
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 KNO3 is dissolved in 750. mL of solution.
• given 28.2 g ; need to convert mass to moles
• molar mass: KNO3
•
1xK=
1 x 39.10 g/mol
•
1xN=
1 x 14.01 g/mol
•
3 x O=
3 x 16.00 g/mol
•
101.11 g/mol
•
•
•
moles of substance
=
=
28.2 g
101.11 g/mol
0.279 mol
•
•
•
•
•
•
•
•
•
•
•
volume must be reported in L:
750. mL x
1L
1000 mL
= 0.750 L
molar concentration:
c
=
n
V
=
= 0.372 mol/L
0.279 mol
0.750 L
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
0.228 L
How many moles are present in 4.50 L of a 0.025
mol/L solution of magnesium nitrate?
•c =
•n =
n
V
cV
=
(0.025 mol/L)(4.50 L)
=
0.11 mol
• Complete questions 4.61 to 4.69
Other Measures of Concentration
• based on weight to weight (gram to gram)
concentration
• Parts per hundred (percent, %)
– 1 g of a solute per 100 grams of solution.
– One part in 102.
– equal to 1 g of a solute in 100 g of solution,
or 100 mL of solution.
Parts per .....
• Parts per thousand
– one gram for every 1000 g of solution.
– equivalent to:
• one drop of ink in a cup of water
• one second per 17 minutes.
– "Parts per thousand" is often used to record
the salinity of seawater.
Parts per ...
• Parts per million (ppm)
– one gram of solute for every 1 000 000 g
of solution (1 mg/kg, 1 μg/g)
– equivalent to:
• one drop of ink in a 150 litre (40
gallon) drum of water
• 1 L of ink in Riversdale pool (assume
106 L)
• 1 s in 11.57 days
– Used to measure levels of metals,
nutrients and toxic substances in the
environment
Parts per ...
• Parts per billion (ppb)
– one gram of solute for every 109
grams of solution (1 ng/g, 1 μg/kg)
– equivalent to:
• 1 gram of ink in Riversdale pool.
• 1 second in 31.71 years
– is the level of activity of hormones in
living system.
– some carcinogens and toxins are
active at this level.
Parts per ...
 Parts per trillion
– "ppt”
– one gram of solute for every 1012 grams of
solution (1 pg/g, 1 ng/kg).
– equivalent to:
• 1 gram of ink in Blackstrap Lake.
• 1 second in 31 700 years
Parts per....
 Parts per quadrillion (ppq)
– one gram of solute for every 1015 grams
of solution (1 fg/g, 1 pg/kg).
– equivalent to:
• 1 gram of ink in Lake Ontario.
• 1 second in 31 million years
– Very few analytical techniques can
measure with this degree of accuracy
– it is still used in some mathematical
models of toxicology and epidemiology.
Dilution
Dilution
• The stock solution is a more concentrated
form of solution (hydrochloric acid is 12.4
mol/L).
• use the chemicals in the lab in diluted form.
• need to decide how much of the stock
solution needed to make the solution you
want.
Dilution
• Have: 12.4 mol/L stock solution HCl
• Want: 2.00 L of 0.100 mol/L solution.
• First: Calculate the number of moles in the
dilute solution:
•
•
•
n
= cv
= (0.100 mol/L)(2.00 L)
= 0.200 mol
Dilution
• Second: Calculate volume of stock solution
needed:
•
•
•
•
v
=
n
c
=
0.200 mol
12.4 mol/L
= 0.0161 L (1000 mL/L)
= 16.1 mL HCl
• 16.1 mL HCl stock solution is added to water
to make 2.00 L of a 0.100 mol/L solution.
Dilution
• There is an easier way to do this:
• Moles taken from the stock solution (ns)
equals the number of moles that goes into
the dilute solution (nd)
•
• and
•
•
ns = nd
ns = csvs
nd = cdvd
• so the formula for dilution is:
•
csvs = cdvd
Dilution
• What volume of concentrated sulfuric
acid (18 M) is needed to make 500. mL
of 2.00 M dilute solution?
• CsVs = CdVd
• (18 mol/L)Vd = (2.00 mol/L)(500. mL)
• Vd = 56 mL concentrated H2SO4 is
needed.
Dilution
• What is the concentration of dilute
solution if 25.0 mL of glacial acetic acid
(24.0 mol/L) is added to make 2.00 L of
dilute solution?
• CsVs = CdVd
• (24.0 mol/L)(25.0 mL) = (Cd)(2.00 L)(1000
mL/1 L)
• Cd = 0.300 mol/L acetic acid.
Dilution
• What volume of a 1.00 mol/L stock
solution of aluminum chloride is needed
to make 500. mL of 3.0 x 10-4 mol/L
solution?
• CsVs = CdVd
• (1.00 mol/L)Vd = (3.0 x 10-4
mol/L)(500. mL)
• Vd = 0.15 mL AlCl3 stock solution is
needed.
• Complete questions 4.71 to 4.74
Mixing a Solution
Mixing Solutions
• Consider these solutions:
•
•
•
•
•
•
•
CaCl2 (aq) →
0.25 mol/L
Mg3(PO4)2 (aq)
0.011 mol/L
Ca2+(aq) +
2 Cl1-(aq)
1(0.25 mol/L) 2(0.25 mol/L)
= 0.25 mol/L
= 0.50 mol/L
→
3 Mg2+(aq)
+
3(0.011 mol/L)
= 0.033 mol/L
2 PO43-(aq)
2(0.011 mol/L)
= 0.022 mol/L
Mixing Solutions
• take 300. mL of the CaCl2 and 200. mL
of the Mg3(PO4)2 and we mix them.
• Assuming that no precipitate is formed,
what will the concentration of each ion
in solution be after mixing ?
Mixing Solutions
• we use the dilution formula:
CsVs = CdVd
• Cs and Vs is the original concentration and
volume of each ion
– 0.25 mol/L and 300. mL for the Ca2+
• Vd is the total volume of the mixture
– 300. mL + 200. mL.
• We are looking for Cd, the ion concentration
after mixing:
Mixing Solutions
• Cd =
CsVs
Vd
• For the Ca2+ ion the calculation is as follows:
• [Ca2+] = (0.25mol/L)(300.mL)
(300. mL + 200. mL)
•
= 0.15 mol/L
Mixing Solutions
• [Cl1-] =
•
(0.50 mol/L)(300. mL)
(300. mL + 200. mL)
= 0.30 mol/L
• [Mg2+] =
•
(0.033 mol/L)(200. mL)
(300. mL + 200. mL)
= 0.013 mol/L
• [PO43-] = (0.022 mol/L)(200. mL)
(300. mL + 200. mL)
•
= 0.0088 mol/L
• Complete question 4.70
Using Molarities in Stoichiometric
Calculations
• Consider this problem:
– 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.
Step 1
• Write and balance the chemical equation:
• Pb(NO3)2 (aq) + 2 KI(aq)  PbI2 (s) + 2 KNO3 (aq)
Step 2
• List information:
• Pb(NO3)2 (aq) + 2 KI(aq)  PbI2 (s) + 2 KNO3 (aq)
150. mL
g?
0.500 mol/L
Step 3
• Convert given information to moles:
• Pb(NO3)2 (aq) + 2 KI(aq)  PbI2 (s) + 2 KNO3 (aq)
150. mL
g?
0.500 mol/L
• (0.500 mol/L) (150. mL)(1 L / 1000 mL)
= 0.075 mol
Step 4
• Use mole ratio to find moles of unknown:
• 0.0750 mol Pb(NO3)2 x
1 mol PbI2
1 mol Pb(NO3)2
= 0.0750 mol PbI2
Step 5
• Convert answer to desired units
• 0.0750 mol Pb(NO3)2 x
1 mol PbI2
1 mol Pb(NO3)2
= 0.0750 mol PbI2
• 0.0750 mol PbI2 x (207.20 + 2(126.90)) g/mol
= 34.6 g PbI2
• Aqueous cadmium chloride reacts with
sodium sulfide to produce bright-yellow
cadmium sulfide. What mass of cadmium
sulfide would be produced if 50.00 mL of a
3.91 M solution of sodium sulfide reacts with
sufficient cadmium chloride?
• Solutions of calcium nitrate and sodium
phosphate react. If 256 mL of a 0.356 M
solution of calcium nitrate reacts with
sufficient sodium phosphate, what mass of
precipitate would result?
Titration
• The analytical technique in which one can
calculate the concentration of a solute in a
solution.
• is most often done in connection with acid –
base reactions.
• one substance, of unknown concentration and
known volume is placed in a flask.
• the other substance of known concentration is
placed in a burette. This allows the volume to
be read easily.
Titration
• We are dealing with substances in solution.
• The moles of base is equal to:
nb = cbVb
• and the moles of acid in solution is equal to:
na = caVa
• Since at the point of neutralization the moles
of acid and base are equal, the following
formula is arrived at:
caVa = cbVb
caVa = cbVb
• What is the concentration of HCl if 50.00 mL
of HCl solution is neutralized by 16.15 mL of
0.100 M NaOH ?
– Ca = ?
– Va = 50.00 mL
– Cb = 0.100 mol/L
– Vb = 16.15 mL
Ca =
(0.100 mol/L)(16.15 mL)
50.00 mL
= 0.0323 M NaOH
caVa = cbVb
• What volume of 0.250 mol/L KOH will be
neutralized by 160. mL of 0.400 mol/L
HNO3 ?
– Ca = 0.400 mol/L
– Va = 160. mL
– Cb = 0.250 mol/L
– Vb = ?
Vb =
(0.400 mol/L)(160. mL)
0.250 mol/L
= 256 mL KOH
Indicators
• chemicals which are pH sensitive.
• undergo a colour change as the pH shifts from
acid to basic conditions.
• The neutralization point is detected when the
solution changes colour.
• Indicators are the method of choice in student
labs. They are also used in the field, where
instrumentation is limited.
• Indicators are used in acid-base reactions and
a broad range of chemical reactions, including
redox reactions
Titration
• Complete questions 4.77 to 4.86