Unit 4: Stoichiometry
The last one!!! 
1
Reaction Assumptions
• Reaction (rxn) assumptions:
– Rxns are spontaneous; when reactants are mixed
they will react to form products
– Rxns are fast
– Rxns are quantitative; they go to completion
– Rxns are stoichiometric; they will react in simple
whole number ratios of chemical amounts
expressed in the balanced chemical equation,
regardless of conditions
Stoichiometry
Stoichiometry is the process of using a chemical equation to
calculate the quantity of reactants and products involved in a
reaction. The balanced equation for a chemical reaction
describes the stoichiometry of the reaction.
• Balanced equations have to be used in chemical reactions in
order to have correct stoichiometric calculations
• Ex: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
1mol of C3H8 reacts with 5mol of O2 to produce
3mol CO2 and 4mol H2O
• Balanced equations allow us to set up the mole ratio between
the substance we have and the one we want.
3
Mole Ratio
The mole ratio is a conversion factor based on the
balanced equation that allows us to find amounts of
different substances by using the substance we have.
• The mole ratio is always the same between two
substances in a chemical equation no matter the
amounts involved.
• Ex: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
the mole ratio between C3H8 and CO2 is always going to
be 1:3; or the mole ratio between CO2 and O2 will always
be 3:5
4
Mole-to-Mole Calculations
Using Stoichiometry
They give you moles, they ask for moles.
Ex: What number of moles of O2 will be produced by the decomposition
of 5.8 mol of water?
Step 1: Write and balance the chemical equation
Step 2: Set up the mole ratio between what you have (water) and what
you need (oxygen)
Step 3: Use the mole ratio as a conversion factor to answer the
question
5
Mass-to-Mole Calculations
Using Stoichiometry
They give you mass, they ask for moles.
Ex: Aluminum metal reacts vigorously with solid iodine to produce solid
aluminum iodide. What mass of I2(s) should we weigh out to react exactly with
35.0g of aluminum?
Step 1: Write and balance the chemical equation
Step 2: Set up the mole ratio between what you have (aluminum) and what
you need (iodine)
Step 3: Use the Factor-Label Method (aka Chain Method or Unit-Factor
Method) to convert the mass to moles and with the mole ratio to answer the
question
7
Mass-to-Mole Calculations
Using Stoichiometry
• The procedure we just did:
2Al(s)
+
35.0 g Al
3I2(s)
495 g I2

1
3
Use molar mass
of Al (26.98 g/mol)
Use molar mass
of I2 (253.8 g/mol)
1
3
mol Al
 2AlI3(s)
Use mole ratio
2
3 𝑚𝑜𝑙 𝐼2
2 𝑚𝑜𝑙 𝐴𝑙
2
mol I2
9
Mass-to-Mass Calculations
(Gravimetric Stoichiometry)
• Stoichiometry of a chemical reaction is related to the
amount of reactants used and products formed
• Gravimetric stoichiometry relates specifically to masses
of reactants or products.
• We will be using the factor-label method in this course
– Referred to as ‘chained method’ in textbook
• It’s the standard in science, it’s efficient.
Mass-to-Mass Calculations
Using Stoichiometry
They give you mass, they ask for mass (gravimetric
stoichiometry)
Step 1: Write and balance the chemical reaction
Step 2: Convert masses to moles, find molar masses of
the substances you have and need.
Step 3: Use the mole ratio and factor-label method to
answer the question
11
Mass-to-Mass Calculations
Using Stoichiometry
Ex: Solid lithium hydroxide has been used in space vehicles to remove
exhaled carbon dioxide from the living environment. The products are
solid lithium carbonate and liquid water. What mass of gaseous carbon
dioxide can 875g of lithium hydroxide absorb?
12
Calculations Using Stoichiometry Summary
2Al(s)
+
3I2(s)
g Al
g I2
1
6
Use molar mass of
Al (26.98 g/mol)
Use molar mass of
I2 (253.8 g/mol)
2
5
mol Al
 2AlI3(s)
Use mole ratio
3
3 𝑚𝑜𝑙 𝐼2
2 𝑚𝑜𝑙 𝐴𝑙
4
mol I2
14
Practice!
• HW:
• Gravimetric Stoichiometry Booklet – for marks
• Read p.286-292, p. 278-283
• Use the factor-label method
• Is it:
– Mole-to-mole? (easy-peasy, one-step calculation)
– Mass-to-mole? (need molar mass for given
substance)
– Mass-to-mass? (need molar mass of both given
substance and wanted substance)
– (need mole ratios for all)
15
Limiting Reactants/Reagents
A chemical rxn is limited by the substance that is in the least
amount (according to the mole ratio)  limiting reagent
Analogy:
For every car, you need 4 tires. So the car to tire ratio is 1:4.
What if the auto manufacturer had 8 car bodies and ordered 48
tires?
16
Limiting Reactants/Reagents
• With 8 cars, how many tires do we need? Use factor-label method:
• Is the # of tires we need more or less than what we have (48 tires)?
If we have more tires than we need = excess tires
If we have less tires than we need = limited
There was excess of tires but a limited amount of car bodies.
The same idea applies to chemical reactions; one reagent is given in
excess quantity (excess reagent) and another is in limited quantity
(limiting reagent).
17
Limiting and Excess Reagents
•
Consider the following reaction outlined and illustrated
Cu(s) + 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
Net ionic eq’n:
•
The copper atoms have become copper(II) ions (∴ the sol’n is blue) and
the silver ions (in silver nitrate) have formed pure silver solid around the
copper coil (white precipitate).
• Since there was still copper left at the end of an extended
period of time, more copper was present than was needed.
• Copper was in excess and ∴ is called the excess reagent.
– This means that the reaction proceeded until all the silver ions
were used up in the reaction.
– Silver ions were the limiting reagent (the amount of copper(II)
ions you can make is limited by the amount of silver ions you
have.)
Limiting & Excess Reagents
Group Work!
You are making cookies and are missing a key ingredient – eggs. You have
plenty of the other ingredients, except that you only have 1.33 cups of butter
and no eggs.
The recipe calls for 2 cups of butter and 3 eggs (plus the other ingredients) to
make 6 dozen cookies. You telephone a friend and have him bring you some
eggs.
a)
How many eggs do you need if you are going to use only your 1.33 cups of
butter?
b)
If you use all the butter (and get enough eggs), how many cookies can you
make?
When your friend arrives, he has a surprise for you—to save time he has
broken the eggs in a bowl for you. You ask him how many he brought, and he
replies, “All of them, but I spilled some on the way over.”
You weigh the eggs and find that they weigh 62.1 g. Assuming that an
average egg weighs 34.21 g:
a)
How much butter is needed to react with all the eggs?
b)
How many cookies can you make?
c)
Which will you have left over, eggs or butter?
d)
How much is left over?
Limiting Reagents
• Limiting reagent
– The reactant that limits the amount of product that
can be formed
– The reaction will stop when all of the limiting
reactant is consumed
• Excess reagent
– The reactant that remains when the reaction stops.
– The excess reactant remains because there is
nothing left for it to react with (the limiting reactant
is already completely consumed)
22
Limiting Reagents
Ex: Nitrogen can be prepared by passing gaseous ammonia over solid
copper (II) oxide at high temperatures. The other products of the rxn
are solid copper and water vapor. How many grams of N2 are formed
when 18.1 g of NH3 is reacted with 90.4 g of CuO?
Step 1: Write and balance the chemical equation
Step 2: Convert masses of reactants to moles
23
Limiting Reactants
Ex: Nitrogen can be prepared by passing gaseous ammonia over solid copper
(II) oxide at high temperatures. The other products of the rxn are solid copper
and water vapor. How many grams of N2 are formed when 18.1 g of NH3 is
reacted with 90.4 g of CuO?
Step 3: Determine the limiting reagent and use that to find the product they’re
asking for (use Factor Label Method)
- Pick one of the reactants and calculate how many moles of the other
reactant is needed to react with it:
-
Compare the # moles needed to # moles we have:
If we have more moles than what we need  excess reagent
If we have less moles than what we need  limiting reagent
24
Limiting Reactants
Ex: Nitrogen can be prepared by passing gaseous ammonia over solid
copper (II) oxide at high temperatures. The other products of the rxn
are solid copper and water vapor. How many grams of N2 are formed
when 18.1 g of NH3 is reacted with 90.4 g of CuO?
Step 3: Determine which reacting is limiting and use that to find the
product they’re asking for (use Factor Label Method)
- ∴ ____ is the limiting reagent. Use this to find the product they’re
asking for:
25
Summary for Solving Stoichiometry Problems
with Limiting Reagents
• When they ask you for how much product is produced
and they give you amounts of reactants:
– Write and balance the chemical equation
– Convert known masses of reactants to moles
– Determine the limiting reagent (using the moles of one
reactant and the mole ratio between the reactants)
– Use the factor label method to determine the mass of the
product they’re asking for (using the limiting reagent, the
mole ratio between the product and the limiting reagent and
the molar mass of the product)
28
Practice!
• Practice HW: Limiting Reagents
Worksheet, p. 327 # 1, 4, 5
• Read: p. 284, 320-327
29
Percent Yield
• Theoretical yield (the amount of product you could have gotten in
the lab)
– The maximum amount of product produced using stoichiometry of the
chemical reaction and the given amount(s) of reactant(s)
– Theoretical yields are rarely produced due to the presence of side rxns
(other rxns that consume one or more of the reactants/products)
• Actual yield (the amount of product you actually got in the lab)
– Amount of the product actually obtained during the chemical reaction
• Percent yield (what you got over what you could have gotten – like
a test score)
– Comparison of actual yield to theoretical yield expressed as a
percentage
% 𝑦𝑖𝑒𝑙𝑑 =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
× 100%
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
30
Determining Percent Yield
Ex: Methanol can be produced by the reaction between carbon
monoxide and hydrogen. Suppose 6.85g of CO(g) is reacted with 0.860g
of H2(g). What is the percent yield of methanol if 3.57g of CH3OH(l) is
actually produced? What is the remaining amount of excess reagent?
(We need to find the theoretical yield of methanol and then compare it
to the actual yield.)
Step 1: Write and balance the chemical equation
Step 2: Convert the mass of both reactants to moles
31
Determining Percent Yield
Ex: Methanol can be produced by the reaction between carbon monoxide and
hydrogen. Suppose 6.85g of CO(g) is reacted with 0.860g of H2(g). What is the
percent yield of methanol if 3.57g of CH3OH(l) is actually produced? What is the
remaining amount of excess reagent?
(We need to find the theoretical yield of methanol and then compare it to the
actual yield.)
Step 3: Find the limiting reagent by using the mole ratio between H2 and CO
32
Determining Percent Yield
Ex: Methanol can be produced by the reaction between carbon monoxide and
hydrogen. Suppose 6.85g of CO(g) is reacted with 0.860g of H2(g). What is the percent
yield of methanol if 3.57g of CH3OH(l) is actually produced? What is the remaining
amount of excess reagent?
(We need to find the theoretical yield of methanol and then compare it to the actual
yield.)
Step 4: Use the limiting reagent to find the theoretical yield of methanol (using mole
ratios and molar mass of methanol)
33
Determining Percent Yield
Ex: Methanol can be produced by the reaction between carbon monoxide and
hydrogen. Suppose 6.85g of CO(g) is reacted with 0.860g of H2(g). What is the
percent yield of methanol if 3.57g of CH3OH(l) is actually produced? What is the
remaining amount of excess reagent?
(We need to find the theoretical yield of methanol and then compare it to the
actual yield.)
Step 5: Calculate the percent yield % 𝑦𝑖𝑒𝑙𝑑 = 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 × 100%
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
34
Determining Percent Yield
Ex: Methanol can be produced by the reaction between carbon
monoxide and hydrogen. Suppose 6.85g of CO(g) is reacted with 0.860g
of H2(g). What is the percent yield of methanol if 3.57g of CH3OH(l) is
actually produced? What is the remaining amount of excess reagent?
MAKE SURE YOU’VE ANSWERED WHAT THEY ASKED!
Step 6: Calculate the mass of excess reagent used (a moles-to-mass
stoichiometry question) using factor-label method
Step 7: Calculate the mass of excess reagent (subtraction)
35
Percent Yield
Ex: 2S(s) + 3O2(g)  2SO3(g), if 3.2g S reacts with 9.0g O2
Determine the:
a)
b)
c)
d)
Limiting reagent
Theoretical yield of SO3 in grams
% yield if 5.0g SO3 was formed
Remaining amount of excess reagent
S
8.0 g SO3
63%
4.2 g O2
40
Percent Error
• Usually there is difference between the two yields.
This is due to experimental errors, which can
include:
– limitations of the equipment, there is always
uncertainty in measurement. This has nothing to do
with the operator doing anything wrong/incorrect.
– the purity of the grade of chemical used
– washing of any precipitate involved (some dissolving
or lost through the filter paper)
– any qualitative judgments. These include such things
as color or color changes.
Percent Error
• The predicted value is the accepted value, frequently
calculated from stoichiometry or as otherwise provided.
• The experimental value is the value as calculated from the
evidence in an investigation (experiment)
% 𝑒𝑟𝑟𝑜𝑟 =
𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒−𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒
𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒
× 100%
• For most school labs, investigations usually involve a total
of all experimental errors in the range of 5 – 10 %.
• Percent yields below 90% are unacceptable and generally
suggest there has been some “human error” which is NOT
an acceptable category of experimental error
Percent Yield vs Percent Error
Ex 1: A solution is made by dissolving 9.8g of barium chloride and is
completely reacted with a second solution containing dissolved sodium
sulfate.
a)
b)
c)
Predict the mass of precipitate expected.
If 10.0g of precipitate is actually formed, calculate the percent yield.
Calculate the percent error.
11 g
91%
9.1%
Percent Yield vs Percent Error
Ex 2: When 30.0g carbon are heated with silicon dioxide, 28.2g carbon
monoxide are produced. What is the percent yield of this reaction?
What is the percent error?
SiO2(s) + 3C(s)  SiC(s) + 2CO(g) 60.5% yield, 39.5% error
Percent Yield vs Percent Error
Ex 3: Aluminium reacts with oxygen to produce aluminum oxide. The
reaction of 50.0g aluminum and sufficient oxygen has a 75.0% yield.
How many grams of aluminum oxide are produced?
70.9 g
Percent Yield vs Percent Error
Ex 4: Acetylene, C2H2, used in welders’ torches, burns according to the
following equation:
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(g)
a)
b)
c)
How many molecules of oxygen are needed to react with 22.0g acetylene?
1.27×1024 molecules O2
How many grams of carbon dioxide could be produced from the complete
reaction of the acetylene in part (a)?
74.4 g
If the reaction in part (a) produces 64.0g CO2, what is the percent yield for
the reaction? Percent error? 86.0% yield, 14.0% error
Practice!
• Practice HW: Percent Yield/Error
Worksheet; p. 327 #3, 8; p. 393 #8, 10
50
Gas Stoichiometry
• Chemical rxns with gases, same rules and
methods of gravimetric stoichiometry apply! (Yay!)
• The only difference is that we also use volumes of
gases.
• So we use the molar volume of gases at SATP
(24.8L/mol) and STP (22.4L/mol) to find moles
and/or volume. If they give you mass, then you
convert mass to moles and then use molar
volume to find volume.
• If the question does NOT indicate conditions that
are STP or SATP, then we have to use the ideal
gas law to find moles and/or volume.
Gas Stoichiometry
• If it is a volume-to-volume calculation and the
conditions are NOT changing, then the volume
ratio is the same as the mole ratio (Law of
Combining Volumes) and you just do a very
simple calculation using the volume ratio like
you would a mole ratio.
• If it is a volume-to-volume calculation and
conditions ARE changing, then you have to
convert volume to moles, use mole ratio and
then convert moles to volume (using factorlabel method). It’s like replacing “mass” with
“volume.”
Gas Stoichiometry
Ex 1: The first step in the industrial manufacture of
sulfuric acid is the complete combustion of sulfur,
S8(s). What mass of sulfur is required to produce
112L of sulfur dioxide at STP?
160g
Ex 2: Coal can undergo an incomplete combustion in
the absence of a plentiful supply of air to produce
deadly carbon monoxide gas. What volume of
carbon monoxide is produced at SATP by the
incomplete combustion of 150kg of coal (C(s))?
3.10×105 L
Gas Stoichiometry
Ex 3: The first recorded observation of hydrogen
gas was made by the famous alchemist Paracelsus
when he added iron to sulfuric acid. Calculate the
volume of hydrogen gas at STP produced by adding
10.0 g of iron to an excess of sulfuric acid to also
produce iron (III) sulfate.
6.02 L
Ex 4: Ammonia reacts with sulfuric acid to form the
important fertilizer, ammonium sulfate. What mass
of ammonium sulfate can be produced from 75.0 L
of ammonia at 10°C and 110 kPa?
200 g
Practice
• Practice HW: Gas Stoichiometry
Worksheet; p. 296 #2, 3
• Read: p. 294-298
55
Solution Stoichiometry
• The majority of stoichiometric work in research
and industry involves aqueous solutions.
Solutions are easy to transport and handle, and
solution reactions are relatively easy to control.
• The general method of gravimetric stoichiometry
remains the same in solution stoichiometry. (Yay!)
• The major difference is the concentration and
volume of solutions are used as conversion
factors to convert to and from the chemical
amount of substance.
Solution Stoichiometry
Ex 1: What is the amount/molar concentration of a
KOH(aq) solution if 12.8mL of this solution is required to
react with 25.0mL of 0.110 mol/L H2SO4(aq)? 0.430 mol/L
Ex 2: What volume of 0.125mol/L NaOH(aq) is required to
react completely with 15.0mL of 0.100 mol/L Al2(SO4)3(aq)?
72.0 mL
Solution Stoichiometry
Ex 3: In a chemical analysis, a 10.0mL sample of
H3PO4(aq) was reacted with 18.2mL of 0.259mol/L
NaOH(aq). Calculate the amount concentration of the
phosphoric acid.
0.157 mol/L
Ex 4: The concentration of magnesium ions (assume
magnesium chloride) in seawater was analyzed and found
to be 5.00×10–3mol/L. What volume of 0.200mol/L
sodium hydroxide solution would be needed in an
industrial process to precipitate all of the magnesium ions
from 1.00L of seawater?
50.0 mL
Solution Stoichiometry
Ex 5: A student has accidentally spilled 100.0mL of 3.0mol/L
sulfuric acid onto the lab bench. The lab instructor sprinkles
some baking soda (sodium hydrogen carbonate) on the spill to
neutralize the acid according to the following chemical
equation:
H2SO4(aq) + NaHCO3(s)  CO2(g) + H2O(l) + Na2SO4(aq)
a)
What mass of baking soda would the instructor need to
sprinkle on this spill to neutralize and clean it up?
50.g
b)
Calculate the volume of carbon dioxide that would be
produced at a temperature of 25°C and a pressure of 100kPa.
15.L
Practice!
• HW Practice: Solution Stoichiometry
Worksheet; p. 303 #1, 2, 5
• Read: p. 300-302
60
Erlenmeyer
Titration Steps
64
Titration Steps
65
Acid-Base Titration: Terminology
• Titration – is the process of carefully measuring
and controlling the addition of a solution (the
titrant) from a burette into a measured fixed
volume of another solution (sample) until the
reaction is judged to be complete.
• Titrant – the solution in the burette during a
titration
• Sample/Analyte/Titrand – the solution of unknown
concentration that is being titrated in the
Erlenmeyer Flask
Titration Calculations
• The equivalence point will occur when the
chemical amount of acidic solution is equal to
the chemical amount of the basic solution.
• This is detected by the endpoint – a sudden
colour change or pH change what you see in
the lab
Ex 1: Titration of 10.00mL of CH3COOH(aq) with 0.202mol/L of NaOH(aq)
Titrant: ________________ Sample: _______________
Equation:
Trial
Final burette reading (mL)
1
14.8
2
26.9
3
39.8
4
13.6
Initial burette reading (mL)
0.7
13.9
26.9
0.5
blue
green
green
green
Volume of NaOH used (mL)
Colour of the endpoint
1. What is the molar concentration of the acetic acid used?
(Answer: 0.263mol/L)
69
Ex 2: Titration of 25.0mL of Na2CO3(aq) with 0.352mol/L HI(aq)
Titrant: __________________ Sample: __________________
Equation:
Trial
Final burette reading (mL)
Initial burette reading (mL)
Volume of HI used (mL)
1
16.5
0.6
2
31.8
16.5
3
47.0
31.8
4
16.4
1.2
2. What is the molar concentration of the sodium carbonate
solution?
70
Ex 3: Titration of 10.0mL of Mg(OH)2(aq) with 0.150mol/L H2CO3(aq)
Titrant: __________________
Sample: __________________
Equation:
Trial
Final burette reading (mL)
Initial burette reading (mL)
Volume of H2CO3 used (mL)
1
12.8
0.2
2
25.3
12.8
3
37.9
25.3
3. What is the molar concentration of the magnesium hydroxide
solution?
71
Ex 4: Titration of 10.00mL of NH3(aq) with 1.48mol/L HCl(aq)
Titrant: __________________
Sample: __________________
Equation:
Trial
Final burette reading (mL)
Initial burette reading (mL)
Volume of HCl used (mL)
1
29.1
15.0
2
43.0
29.1
3
14.4
0.4
4. What is the molar concentration of the ammonia solution?
72
pH Curves
Tell us 5 things:
• Type of rxn
• # of rxns
• Buffer regions
• Endpoint
• Equivalence point
80
Endpoint – (steep slopes) sudden observable Δ
(like colour or pH) has occurred
• The region of rapid change in pH (the steep slope in the
graphinflection) indicates that the buffer is used up. The
midpoint are
of the
inflection/change
is called
Endpoints
VISIBLE
colour Δ’s
or pHthe
Δ’s that you can
______ of the titration.
only
see in the LAB. Endpoints tell you that you’ve
• The endpoint indicates that neutralization has occurred in an acidreached the equivalence point.
base titration.
• Neutralization is the pH where the chemical amount of acidic
Wesolution
use endpoint
equivalence
point
and basic&solution
are equal.
(Thisinterchangeably,
does not mean equal
though
endpoint
does not
always
match
exactly
volumes!)
Neutralization
is NOT
always
at pHup
7, only
if it isto
a strong
acid mixed with
a strong base (or a weak acid with a weak base).
equivalence
point.
– All endpoints are
!
84
Equivalence Point – volume of titrant
added to reach neutralization or endpoint
• The pH of the solution at the equivalence point
for a strong monoprotic acid-strong monoprotic
base rxn will be 7.
• The pH of the solution at the equivalence point
for any other acid-base reaction must be
determined experimentally, by plotting a
titration pH curve.
pH Curves
Tell us 5 things:
• Type of rxn
• # of rxns
• Buffer regions
• Endpoint
• Equivalence point
87
Strong Base Titrated
with a Strong Acid
0.48mol/L NaOH(aq) is titrated with 0.50mol/L HCl(aq)
Since the initial pH is above 7,
this indicates that the:
Sample is a _____________
(_________), and the titrant is
an _____________ (_______).
Equivalence point is
_________________, and the
endpoint is
_________________.
89
Strong Acid Titrated
with a Strong Base
0.50mol/L HCl(aq) is titrated with 0.48mol/L NaOH(aq)
Since the initial pH is below 7,
this indicates that the:
Sample is an _____________
(_________), and the titrant is
a _____________ (________).
Equivalence point is
_________________, and the
endpoint is
_________________.
90
Important Point!
• Only for the rxn between a
strong monoprotic acid and a
strong monoprotic base, is the
pH 7 at the equivalence point.
91
Choosing Acid-Base Indicators for Titration
Practice
• For each titration curve on the following slide,
a) Determine the volume of titrant used to reach the
equivalence point
b) Determine the pH at the equivalence point (to the
nearest whole number)
c) Suggest a suitable indicator for a titration analysis
to be done using these reagents.
Practice!
• Practice HW: Titration Worksheet; p. 339
#1, 3, 6, 9
• Read: p. 328-339
97