The Chemistry of Acids
and Bases
There she blows!!!
Acids and Bases: Molar Solutions
At the conclusion of our time together,
you should be able to:
1. Define molarity
2. Make a solution with a given molarity
3. Determine the molarity of a given solution
Some Definitions
A solution is a
homogeneous mixture
of 2 or more
substances in a single
phase.
The larger constituent is
usually regarded as the
SOLVENT and the
others as SOLUTES.
Some Definitions
The units are moles per
Liter.
Specifically, it’s moles of
solute per Liter of
solution.
Abbreviated as mol/L or
M.
The word Molarity.
Some Definitions
Moles measure amount
of solute usually in
grams
Molarity measures moles
of solute per Liter of
solution.
What Do You Get From a Pampered
Cow?
Spoiled Milk.
Concentration of Solute
The amount of solute in a solution is given by its
concentration.
Molarity(M ) =
moles solute
liters of solution
Concentration of Solute
Molarity(M ) =
moles solute
liters of solution
Therefore: x both sides by Liters =
MV = moles = grams/molar mass
PROBLEM: Using grams/molar mass = MV.
Handout #1
x g NaCl
58.44 g NaCl
= 58.44 g NaCl
=
1Mx1L
PROBLEM: Using grams/molar mass = MV.
Handout #2
x g NaNO3
85.00 g NaNO3
= 42.5 g NaNO3
=
0.5 M x 1 L
Over-Worked Mouse
Acids and Bases: Molar Solutions
Let’s see if you can:
1. Define molarity
2. Make a solution with a given molarity
PROBLEM: Using grams/molar mass = MV.
Backside #1
x g H2SO4
98.09 g H2SO4
= 245 g H2SO4
=
2.50 M x 1 L
Hopefully, you’re not this lost!!!
Acids and Bases: Molar Solutions
At the conclusion of our time together,
you should be able to:
1. Define molarity
2. Make a solution with a given molarity
Interesting Vanity Plate…
PROBLEM: Using grams/molar mass = MV.
Backside #4
x g Ca(OH)2 =
74.10 g Ca(OH)2
= 0.741 g Ca(OH)2
0.100 M x 0.100 L
Preparing Solutions

Determine the mass of solute.

Place in the appropriate
volumetric flask.

Add deionized water and swirl
until solute is dissolved.

Add water to the mark on the
neck of the flask.

Stopper and mix thoroughly.
Support Bacteria!!

They're the only culture some people
have.
Stoichiometry: Molar Solutions
Let’s see if you can:
1. Define molarity
2. Make a solution with a given molarity
Concentration of Solute
Molarity(M ) =
moles solute
liters of solution
Therefore:
MV = moles = grams/molar mass
Acids and Bases: Molar Solutions
At the conclusion of our time together,
you should be able to:
1. Define molarity
2. Make a solution with a given molarity
3. Determine the molarity of a given solution
Concentration of Solute
Molarity(M ) =
moles solute
liters of solution
Therefore:
MV = moles = grams/molar mass
Now let’s use this formula to solve another
type of problem:
Kid’s Letters
to God:
PROBLEM: Mr. T accidentally dropped 15.14 g of
silver (I) nitrate into 100.0 mL of deionized water.
Rather than throw the solution away, help him by
determining its molar concentration so that he can
label it and still use it..
15.14 g AgNO3 x
1 mol AgNO3 =
169.88 g AgNO3
= 0.8912 M AgNO3
M x 0.100 L
PROBLEM: Molarity Problems #2
249 g KI x
1 mol KI
166.00 g KI
= 3 M KI
=
M x 0.5 L
How Would You Make this
3M Solution??

Measure 249 g of KI

Place in a 0.500 L volumetric
flask.

Add deionized water and swirl
until solute is dissolved.

Add water to the mark on the
neck of the flask.

Stopper and mix thoroughly.
Most Caring Person!!
An elderly gentleman had recently lost his wife.
Upon seeing the man cry, a little 4 year old
neighbor boy went into the old gentleman's yard,
climbed onto his lap, and just sat there.
When his mother asked him what he had said to
the neighbor, the little boy just said,
"Nothing, I just helped him cry."
Acids and Bases: Molar Solutions
Let’s see if you can:
1. Define molarity
2. Make a solution with a given molarity
3. Determine the molarity of a given solution
How We Doing??
Need Help????
Exit Quiz #1

1. Calculate the molarity if 0.75 mol of
NaCl in placed in 300.0 mL of water.

2. Calculate the number of moles and
grams of HCl if there is 12.2 mL of 2.45
M HCl solution.

1. Calculate the molarity if 0.75 mol of NaCl in
placed in 300.0 mL of water.
0.75 mol NaCl
= X M x 0.3000 L
= 2.5 M NaCl

2. Calculate the number of moles and grams of
HCl if there is 12.2 mL of 2.45 M HCl solution.
X mol HCl
= 2.45 M x 0.0122 L
= 0.0299 mol HCl
0.0299 mol HCl x
36.46 g HCl
1 mol HCl
= 1.09 g HCl
Entrance Quiz #2

1. Calculate the molarity if 0.50 mol of
KBr in placed in 750 mL of water.

2. Calculate the number of moles and
molarity of HCl if there is 12.2 mL with
2.45 g of HCl in solution.

1. Calculate the molarity if 0.50 mol of KBr in
placed in 750 mL of water.
0.50 mol KBr
= X M x 0.75 L
= 0.67 M KBr

2. Calculate the number of moles and molarity
of HCl if there is 12.2 mL with 2.45 g of HCl.
2.45 g HCl
x
1 mol HCl
= 0.0672 mol HCl
36.46 g HCl
0.0672 mol HCl
= X M x 0.0122 L
= 5.51 M HCl
Stoichiometry: Molar Solutions
At the conclusion of our time together,
you should be able to:
1.
2.
3.
4.
Define molarity
Determine the molarity of a given solution
Make a solution with a given molarity
Dilute a given solution to a new molarity
If the Amount (moles) of
Solute #1 = #2
And Using the Formula:
Molarity(M ) =
moles solute
liters of solution
M1V1 = moles and M2V2 = moles
Therefore if moles of solute are constant:
M1V1 = M2V2
The Scientific Method
begins with
Questions about the World Around
You.
Ever Wonder Why?...
there are flotation devices under plane seats
instead of parachutes?
PROBLEM #1: Using M1V1 = M2V2
0.150 M NaOH x
= 0.125 M NaOH
0.125 L
=
M x 0.150 L
PROBLEM #5: Using M1V1 = M2V2
2.40 M KCl x
0.500 L = 1.00 M KCl x
= 1.20 L
Therefore: 0.700 L needs to be added
XL
Stoichiometry: Molar Solutions
Let’s see if you can:
1.
2.
3.
4.
Define molarity
Determine the molarity of a given solution
Make a solution with a given molarity
Dilute a given solution to a new molarity
I knew I shouldn’t have done that!!
How much water do I need to add to 250 mL
of 3.0 M HCl to dilute it to 1.0 M HCl?
3.0 M HCl
x
0.250 L
=
= 0.75 L Total, therefore 0.50 L
1.0 M HCl x L
Name ________Class Period _____ Clicker Number
Name ________Class Period _____ Clicker Number
"Making Molar Solutions A1"
(10 points)
Make 50.00 mL of a 0.100M BaCl22H2O solution.

__________ grams mass of solute needed

__________ Instructor initials (one point)
What are the products of the reaction? Balance the
equation. Write the balanced molecular, complete ionic
and net ionic equations below. Place the precipitate on
the 3rd line of the first row of lines. (5 points)
________ + ________  ________ + ________
___ ____ + ___ ____  ______ + ___ ____

____ + ___
 _____
 What is this the best molar ratio based on the
stoichiometry? Should you have put all 50.00 mL of
each reactant together to form the most product?
Circle yes or no. (2 points)
____________ : ____________
Yes
No

Preparing Solutions

Determine the mass of solute.

Place in the appropriate
volumetric flask.

Add deionized water and swirl
until solute is dissolved.

Add water to the mark on the
neck of the flask.

Stopper and mix thoroughly.
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O
in enough water to make 250 mL of solution.
Calculate the Molarity.
Step 1: Calculate moles of
NiCl2•6H2O
1 mol
5.00 g •
= 0.0210 mol
237.7 g
Step 2: Calculate Molarity
0.0210 mol
= 0.0841 M
0.250 L
[NiCl2•6 H2O ] = 0.0841 M
USING MOLARITY
What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a 0.0500 M
solution?
moles = M•V
X g H2C2O4 x
1 mol H2C2O4
90.04 g H2C2O4
= 1.13 g H2C2O4
= 0.0500 x 0.250 L
Learning Check
How many grams of NaOH are required to
prepare 400. mL of 3.0 M NaOH solution?
moles = MV
mol = 3.0 M x 0.400 L
mol = 1.2
1.2 mol NaOH x 40.00 g NaOH
48 g NaOH
1 mol NaOH
PROBLEM: Mr. T needs to make 0.500 L of a
0.100 M solution of lead (II) nitrate for a future lab.
Please help him by calculating the amount of solute
needed and by outlining in detail the steps he
would need to take to make this molar solution.
X g Pb(NO3)2
x
1 mol Pb(NO3)2
331.22 g Pb(NO3)2
= 16.6 g Pb(NO3)2
= 0.100 M x 0.500 L
PROBLEM # 4: Grams to make 100. mL of
0.100 M calcium hydroxide.
X g Ca(OH)2
x
1 mol Ca(OH)2
74.10 g Ca(OH)2
= 0.741 g Ca(OH)2
= 0.100 M x 0.100 L
PROBLEM # 5: 4.00 moles of nitric acid in
1.50 L of solution = what M.
4.00 mol HNO3
= 2.67 M HNO3
= X M x 1.50 L