Solutions Part II:
Molarity & Solution
Stoichiometry
Chapter 4 Sec 7-9
of Jespersen 7TH Ed
Dr. C. Yau
Spring 2015
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Molarity
Molarity (M) is a unit of concentration.
# mol solute
Molarity 
# L solution
Know this well!
Always remember that…M is a fraction in
disguise.
Note that it's "L of solution" not "L of water!"
When you see 3.00 M HCl you should
immediately be thinking…
3.00 mol HCl
3.00 M HCl 
1 L solution
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Molarity vs. Molar Mass
This concept and use of molarity is as
important as that of molar mass.
You learned molar mass gives you...
# grams
# moles
Now you must learn molarity gives you...
# mol solute
# L soln
Learn it NOW!...the sooner the better!
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Molarity
Molarity is NOT a quantity.
3 M HCl means there are 3 moles of HCl per
liter of solution.
It does NOT mean you have 3 moles, nor
does it mean you have one liter.
It’s like going to the store to buy apples and
the price is $2 per pound.
It doesn’t mean you have to pay $2, nor
does it mean you have to buy one pound.
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Molarity
There are 2 important things you should
remember about molarity.
# mol solute
1) As a fraction, Molarity 
# L solution
it is a conversion unit between…
moles solute
L soln
If you know how many moles of solute, you
can calculate the volume.
If you know the volume, you can calculate
the number of moles of solute.
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Molarity
2) The equation
# mol solute
Molarity 
# L solution
can be rearranged to a very useful format:
# moles = Molarity x Volume (in L)
KNOW THIS WELL!!
You will often need to know the # moles
of the solute in a solution. REMEMBER
… # moles = M x V Sometimes written as
n=MxV
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Example 4.10 p.185
To study the effect of dissolved salt on the
rusting of an iron sample, a student prepared
a solution of NaCl by dissolving 1.461 g of
NaCl in a total volume of 250.0 mL. What is
the molarity of this solution?
Example 4.11 p. 186
How many milliliters of 0.250 M NaCl solution
must be measured to obtain 0.100 mol of
NaCl?
Do Pract Exer 26 & 27 p.187
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Example 4.12 p.188
Strontium nitrate is used in fireworks to
produce brilliant red colors. Suppose we
need to prepare 250.0 mL of 0.100 M
Sr(NO3)2 solution. How many grams of
strontium nitrate are required?
Tips: 1) You see V and you see M and
immediately you should be thinking
MxV = # moles.
2) You have moles and question is
asking about grams, what do we need?
Practice Exercises #28, 29, 30 on p.189
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Preparing a Solution of a
Specified Molarity
Be sure you can describe exactly how you
would prepare a solution of a specified
molarity (such as in Example 5.12).
Answer was 5.29 g strontium nitrate to
prepare the 250.0 mL of 0.100 M soln.
Because molarity is based of L of solution
and not L of water, you cannot just add
250.0 mL of water to 5.29 g Sr(NO3)2
It requires a special apparatus called the
"volumetric flask."
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The volumetric flask, unlike the
graduated cylinder, has only one
graduation mark. The one shown
on the right is for 250 mL. It has
one mark on the neck.
When it is filled with a liquid up to
that mark, it contains 250.00 mL of
the liquid.
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Volumetric Flasks
They come in standard sizes:
5 mL, 10 mL, 25 mL, 50 mL, 100 mL
250 mL, 500 mL, 1 L, 2 L etc.
We don’t have
ones like 30 mL
or 75 mL, etc.
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Describe exactly how you would prepare 250 mL
of 0.100 M Sr(NO3)2.
You should know that Sr(NO3)2 is an ionic
compound and therefore it is a solid to begin
with. We would need to weigh out a certain
amount of the solid.
First you would calculate how much you need to
weigh out. Previously the answer we obtained
was 5.29 g.
Ans. I would weigh out 5.29 g of Sr(NO3)2,
transfer it quantitatively into a 250-mL
volumetric flask, dilute to the mark with
distilled water and shake it to mix thoroughly.
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Preparation of a Solution by Dilution
of a More Concentrated Solution.
Example 4.13 p.189
How can we prepare 100.0 mL of 0.0400 M
K2Cr2O7 from 0.200 M K2Cr2O7?
Tips: There is no mention of grams, so
do not try to use molar mass!
You should first recognize this is a
"dilution problem."
A "dilution problem" utilizes a special
equation.
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EQUATION FOR DILUTION PROBLEMS
Recognize that we are talking about 2
solutions: one dilute and one
concentrated.
Mdil x Vdil
=
Mconc x Vconc
 # mol solute 
 # mol solute 

  

 in dilute soln 
 in conc. soln 
We often just write… M1 V1 = M2 V2
V1 and V2 can be any unit of volume as
long as they are the same units.
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Example 4.13 p.189
How can we prepare 100.0 mL of 0.0400 M
K2Cr2O7 from 0.200 M K2Cr2O7?
Instead of Mdil Vdil we can just assign one solution
with M1 V1.
M1 = 0.0400 M
M2 = 0.200 M
V1 = 100.0 mL
V2 = ? mL
M1 V1 = M2 V2
First we solve for V2, to get…
so
M1 x V1
V2 
M2
0.0400M x 100.0 mL
V2 
 20.0 mL
0.200 M
Do Pract Exer 31 & 32 p.190
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Prep of 100.0 mL of 0.0400 M K2Cr2O7
from 0.200 M K2Cr2O7
Fig.5.25
(b)
(c)
(d)
p.191 (a)
(a) Measure 20.0 mL of 0.200 M K2Cr2O7 w/ pipet.
(b) Transfer to 100-mL volumetric flask.
(c) Dilute to the mark with distilled water.
(d) After mixing, we have 100.0 mL of 0.0400 M. 16
Example: (not from book)
What is the molar concentration of
sodium ions in 0.500 M sodium sulfate?
What is the molar concentration of IO3- in
0.240 M KIO3?
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Example: (not from book)
3.0 L of 2.0 M NaCl is mixed with 4.0 L of
4.0 M CaCl2. What is the molar
concentration of chloride ions in the
resulting solution?
HINT: What units would the answer
have?
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NOTE OF CAUTION
The equation of M1 V1 = M2 V2
can be used ONLY for dilution problems.
The relationship of # mol = # mol
works only because the # mol does not
change in the two solutions.
The # mol of solute transferred from the
concentrated solution must equal to the #
mol of solute that ends up in the dilute
solution.
This is true only because no rxn is involved!
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Solution Stoichiometry
You have had stoichiometry problems and
were working with grams and molar mass.
In "solution stoichiometry" we often just use
molarity instead of molar mass.
How do you know whether you need molar
mass? If the problem only involves only
volume and does not mention mass, you
would not need it.
"Stoichiometry" still means you need the
coefficients of the balanced equation.
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Example 4.14 p.191
One of the solids present in photographic
film is silver bromide, AgBr. Suppose we
wish to prepare AgBr by the following
precipitation reaction.
2AgNO3 + CaBr2
2 AgBr + Ca(NO3)2
How many milliliters of 0.125 M CaBr2
solution must be used to react with the
solute in 50.0 mL of 0.115 M AgNO3.
Tips: Remember M x V = # moles !
Do Pract Exer 33 & 34 p.192
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2AgNO3 + CaBr2
(aq)
(aq)
2 AgBr + Ca(NO3)2
(s)
(aq)
23.0 mL
of 0.125 M CaBr2 soln
must be added
to react with
50.0 mL of
0.115 M AgNO3
to form solid AgBr.
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Example 4.15 p.193
What are the molar concentration of the ions
in 0.20 M aluminum sulfate?
Example 4.16 p.194
A student found that the sulfate ion
concentration in a solution of Al2(SO4)3
was 0.90 M. What was the concentration
of Al2(SO4)3 in the solution?
Pract Exer 35 & 36 p. 194
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Example 4.17 p.194
How many milliliters of 0.100 M Ag+ solution
are needed to react completely with 25.0
mL of 0.400 M CaBr2 solution? The net
ionic equation for the reaction is
Ag+ (aq) + Br- (aq)
AgBr (s)
Do Pract Exer 37 & 38 p.195
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Example 4.18 p.196
A chemist was asked to analyze a soln of
chlordane, C10H6Cl8, dissolved in a
hydrocarbon solvent that was discovered
by construction workers during demolition
of an old work shed.
This insecticide was banned for sale in the
US in 1988 by the EPA b/c of its potential
for causing cancer.
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Example 4.18 p.196 (cont’d.)
Rxns were carried out on a 1.446 g sample
of the soln which converted all of the
chlorine to chloride ion dissolved in water.
This aq soln required 91.22 mL of 0.1400 M
AgNO3 to precipitate all of the chloride ion
as AgCl.
What was the percentage of chlordane in
the original soln? The ppt’n rxn was...
Ag+ (aq) + Cl- (aq)
AgCl (s)
Chlordane = C10H6Cl8
Do Pract Exer 39 & 40 p.197
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Titration
"Titration" is a lab procedure to determine the amount
of substance present in a sample (usually in
percent or molar concentration) by volumetric
analysis.
The apparatus used include a buret and sometimes a
pipet to measure out volume.
The "equivalence point" is the instant a stoichiometric
amount of solution has been delivered from the buret.
(The molar ratio shown by the coefficients of the
balanced equation has been reached.)
The term "titrant" refers to the carefully measured volume
delivered from a buret at the equivalence point.
The "end point" is a point close to the equivalence
point that is made visible by the use of an indicator
that changes color at or near the equivalence point. 27
The 50-mL Buret:
0.00The graduation goes from 0.00 mL
at the top down to 50.00 mL at
mL
the bottom.
An initial buret reading is recorded,
and after the endpoint, the final
buret reading is recorded.
50.00The difference between the two
mL
(final reading – initial reading)
gives the volume of solution that
has been delivered through the
stopcock at the bottom. This
volume of solution is known as
the titrant.
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The Volumetric Pipet, also known as
the Transfer Pipet
Similar to the volumetric flask, there is
only one graduation mark on the narrow
top section of the pipet.
Pipets come in standard sizes, such as
1.00 mL, 5.00 mL, 10.00 mL, 25.00 mL,
50.00 mL. They seldom go to larger
volumes because the solution would be
too heavy to draw up into the pipet.
The liquid is drawn up into the barrel with the use of
a bulb (or pipet pump) up to the calibration mark and
then allowed to drain into the desired container. 29
Example 4.19 p.198
A student prepares a soln of hydrochloric acid
that is approximately 0.1 M and wishes to
determine its precise concentration. A 25.00
mL portion of the HCl soln is transferred to a
flask, and after a few drops of indicator are
added, the HCl solution is titrated with 0.0775
M NaOH solution.
The titration requires exactly 37.46 mL of the
standard NaOH solution to reach the end
point. What is the molarity of the HCl
solution?
Do Pract Exer 41 & 42 p.199
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Solving Multi-Concept Problems (p.200)
Milk of magnesia is a suspension of Mg(OH)2
in water. It can be made by adding a base to
a soln containing Mg2+. Suppose that 40.0
mL of 0.200 M NaOH soln is added to 25.0
mL of 0.300 M MgCl2 soln. What mass of
Mg(OH)2 will be formed, and what will be the
concentrations of the ions in the soln after
the rxn is complete?
What is the strategy?
What is the first step in solving this problem?
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