1. Concentration = 1 “fishar”
# of fish
volume (L)
1 fish
Concentration =
1 (L)
Concentration = 1 “fishar”
V = 1000 mL
V = 1000 mL
V = 5000 mL
n = 2 fish
n = 4 fish
n = 20 fish
Concentration = 2 “fishar”
[ ] = 4 “fishar”
[ ] = 4 “fishar”

2. Concentration = # of moles volume (L) V = 1000 mL V = 1000 mL
n = 8 moles
[ ] = 32 molar
V = 1000 mL
V = 1000 mL
V = 5000 mL
n = 2 moles
n = 4 moles
n = 20 moles
Concentration = 2 molar
[ ] = 4 molar
[ ] = 4 molar

3. Making Molar Solutions
…from liquids
(More accurately, from stock solutions)

4. Concentration…a measure of solute-to-solvent ratio
concentrated vs dilute
“lots of solute” “not much solute”
“watery”
Concentration of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution
Knowing the concentration of solutes is important in controlling the stoichiometry of reactant for reactions that occur in solution
A concentrated solution contains a large amount of solute in a given amount of solution. A 10 mol/L solution would be called concentrated.
A dilute solution contains a small amount of solute in a given amount of solution. A 0.01 mol/L solution would be called dilute.
Add water to dilute a solution; boil water off to concentrate it.

5. Making a Dilute Solution
remove
sample
moles of
solute
initial solution
same number of
moles of solute
in a larger volume
mix
Making a Dilute Solution
diluted solution
Timberlake, Chemistry 7th Edition, page 344

6. Concentration “The amount of solute in a solution”
A. mass % = mass of solute
mass of sol’n
B. parts per million (ppm)  also, ppb and ppt
– commonly used for minerals or
contaminants in water supplies
C. molarity (M) = moles of solute
L of sol’n
– used most often in this class
% by mass – medicated creams
% by volume – rubbing alcohol
mol L M
MOLARITY - Most common unit of concentration
Most useful for calculations involving the stoichiometry of reactions in solution
Molarity of a solution is the number of moles of solute present in exactly 1 L of solution:
moles of solute
molarity = liters of solution
Units of molarity — moles per liter of solution (mol/L),
abbreviated as M
Relationship among volume, molarity, and moles is expressed as
VL M Mol/L = L (mol) = moles
(L)
There are several different ways to quantitatively describe the concentration of a solution, which is the amount of solute in a given quantity of solution.
1. Molarity
– Useful way to describe solution concentrations for reactions that are carried out in solution or for titrations
– Molarity is the number of moles of solute divided by the olume of the solution
Molarity = moles of solute = mol/L
liter of solution
– Volume of a solution depends on its density, which is a function of temperature
2. Molality
– Concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent
– Molality = moles of solute
kilogram solvent
– Depends on the masses of the solute and solvent, which are independent of temperature
– Used in determining how colligative properties vary with solute concentrations
3. Mole fraction
– Used to describe gas concentrations and to determine the vapor pressures of mixtures of similar liquids
– Mole fraction () = moles of component
total moles in the solution
– Depends on only the masses of the solute and solvent and is temperature independent
4. Mass percentage (%)
– The ratio of the mass of the solute to the total mass of the solution
– Result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb)
mass percentage = mass of solute  100%
mass of solution
parts per million (ppm) = mass of solute  106
parts per billion (ppb) = mass of solute  109
– Parts per million (ppm) and parts per billion (ppb) are used to describe concentrations of highly dilute solutions, and these measurements correspond to milligrams (mg) and micrograms (g) of solute per kilogram of solution, respectively
– Mass percentage and parts per million or billion can express the concentrations of substances even if their molecular mass is unknown because these are simply different ways of expressing the ratios of the mass of a solute to the mass of the solution
M =
mol L
D. molality (m) = moles of solute
kg of solvent

7. ppm 1 inch in 16 miles
ppb 1 inch in 16,000 miles
ppt 1 sec = 32,000 years

8. Glassware

9. Glassware – Precision and Cost
beaker vs. volumetric flask
When filled to 1000 mL line, how much liquid is present?
beaker
5% of 1000 mL =
volumetric flask
50 mL
1000 mL mL
Range:
950 mL – 1050 mL
Range:
mL– mL
imprecise; cheap
precise; expensive

10. Markings on Glassware Beaker 500 mL + 5% Range = 500 mL + 25 mL
Graduated Cylinder
1000 mL mL
Range = mL + 5 mL
475 – 525 mL
Volumetric Flask
500 mL mL
Range = – mL
TC 20oC “to contain at a temperature of 20 oC” 22
TD “to deliver” T s
“time in seconds”

11. Measure to part of meniscus w/zero slope.~ ~
water in
grad. cyl.
mercury in
grad. cyl. ~ ~ ~ ~
Measure to part of meniscus w/zero slope.
Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

12. How to mix solid chemicals
Lets mix chemicals for the upcoming soap lab.
We will need 1000 mL of 3 M NaOH per class.
How much sodium hydroxide will I need, for five classes, for this lab?
M =
mol L
3 M =
? mol
1 L
? = 3 mol NaOH/class
x 5 classes
15 mol NaOH
How much will this weigh?
1 23g/mol + 16g/mol g/mol
MMNaOH = 40g/mol
To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution.
Solute occupies space in the solution so the volume of the solvent that is needed is less than the desired volume of solution.
To prepare a particular volume of a solution that contains a specified concentration of a solute, calculate the number of moles of solute in the desired volume of solution and then covert the number of moles of solute to the corresponding mass of solute needed.
40.0 g NaOH
X g NaOH = mol NaOH =
600 g NaOH
1 mol NaOH
FOR EACH CLASS:
To mix this, add 120 g NaOH into 1L volumetric flask with
~750 mL cold H2O.
Mix, allow to return to room temperature – bring volume to 1 L.

13. How to mix a Standard Solution
Wash bottle
Volume marker
(calibration mark)
Weighed
amount
of solute
Use a VOLUMETRIC FLASK to make a standard solution of known concentration
Step 1> add the weighed amount of solute in the volumetric flask
Step 2> add distilled water (about half of final volume)
Step 3> cap volumetric flask, and shake to dissolve solute completely
Step 4> add distilled water to volume marker (calibration mark)
The solution process may be exothermic (release heat). This may cause the liquid to show a larger volume than is real. Allow the solution to return to ambient (room) temperature and check volume again.
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480

14. How to mix a Standard Solution
An aqueous solution consists of at least two components, the solvent (water) and the solute (the stuff dissolved in the water). Usually one wants to keep track of the amount of the solute dissolved in the solution. We call this the concentrations. One could do by keeping track of the concentration by determining the mass of each component, but it is usually easier to measure liquids by volume instead of mass. To do this measure called molarity is commonly used. Molarity (M) is defined as the number of moles of solute (n) divided by the volume (V) of the solution in liters.
It is important to note that the molarity is defined as moles of solute per liter of solution, not moles of solute per liter of solvent. This is because when you add a substance, perhaps a salt, to some volume of water, the volume of the resulting solution will be different than the original volume in some unpredictable way. To get around this problem chemists commonly make up their solutions in volumetric flasks. These are flasks that have a long neck with an etched line indicating the volume. The solute (perhaps a salt) is added to the flask first and then water is added until the solution reaches the mark. The flasks have very good calibration so volumes are commonly known to at least four significant figures.

15. Process of Making a Standard Solution from Liquids
Solutions can be made using liquids or solids (or gases).
To make a 5% solution v/v (volume to volume)
This means to add 5 mL of solute in 95 mL of solvent. The total is 5 mL / 100 mL or 5%.
For the diagram add 25 mL of liquid solute and add water to bring volume to 500 mL (about 475 mL water).
SAFETY NOTE: Always add acid concentrate to water…never add water to concentrated acid.
If you’ve seen what happens when water or ice crystals hit hot oil…a similar phenomenon occurs when water is added to concentrated acid.
The addition of water to concentrated dissipates a large amount of heat. This heat rapidly boils the acid and causes it to spatter.
If however, you start with a large volume of water and slowly add acid, the same amount of heat is generated.
This time, the large volume of water is capable of absorbing the heat. The solution will not splatter.
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 483

16. How to mix a dilute solution from a concentrated stock solution
A solution of a desired concentration can be prepared by diluting a small volume of a more-concentrated solution, a stock solution, with additional solvent.
– Calculate the number of moles of solute desired in the final volume of the more-dilute solution and then calculate the volume of the stock solution that contains the amount of solute.
– Diluting a given quantity of stock solution with solvent does not change the number of moles of solute present.
– The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is
(Vs) (M s) = moles of solute = (Vd) (M d).
Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

17. Reading a pipette 4.48 - 4.50 mL 4.86 - 4.87 mL 5.00 mL
Identify each volume to two decimal places
(values tell you how much you have expelled) mL mL
5.00 mL

18. MConc.VConc. = MDiluteVDilute
Dilution of Solutions
Solution Guide
Formula
Weight
Specific Gravity
Molarity
Reagent
Percent
To Prepare 1
Liter of one molar
Solution
Acetic Acid Glacial (CH3COOH)
60.05
1.05
17.45
99.8%
57.3 mL
Ammonium Hydroxide (NH4OH)
35.05
0.90
14.53
56.6%
69.0 mL
Formic Acid (HCOOH)
46.03
1.20
23.6
90.5%
42.5 mL
Hydrochloric Acid (HCl)
36.46
1.19
12.1
37.2%
82.5 mL
Hydrofluoric Acid (HF)
20.0
1.18
28.9
49.0%
34.5 mL
Nitric Acid (HNO3)
63.01
1.42
15.9
70.0%
63.0 mL
Perchloric Acid 60% (HClO4)
100.47
1.54
9.1
60.0%
110 mL
Perchloric Acid 70% (HClO4)
1.67
11.7
70.5%
85.5 mL
Phosphoric Acid (H3PO4)
97.1
1.70
14.8
85.5%
67.5 mL
Potassium Hydroxide (KOH)
Sodium Hydroxide (NaOH)
40.0
19.4
45.0%
Sulfuric Acid (H2SO4)
98.08
1.84
18.0
50.5%
51.5 mL
This chart quickly shows you the amount of concentrated acid needed to make 1 liter of a 1 M solution. If you need a 5 M solution, add 5x the amount of acid in the same volume.
MConc.VConc. = MDiluteVDilute

19. **Safety Tip: When diluting, add acid or base to water.**
Acids (and sometimes bases) are
purchased in concentrated form (“concentrate”) and are easily
diluted to any desired concentration.
Dilutions of Solutions 
**Safety Tip: When diluting, add acid or base to water.**
C = concentrate
D = dilute
Dilution Equation:
Concentrated H3PO4 is 14.8 M. What volume of concentrate
is required to make L of M H3PO4?
VC = L = 845 mL

20. How would you mix the above solution?
1. Measure out L of concentrated H3PO4 .
2. In separate container, obtain ~20 L of cold H2O.
3. In fume hood, slowly pour [H3PO4] into cold H2O.
4. Add enough H2O until L of solution is obtained.
Be sure to wear your safety glasses!

21. You have 75 mL of conc. HF (28.9 M); you need 15.0 L of
0.100 M HF. Do you have enough to do the experiment?
MCVC = MDVD
28.9 M (0.075 L) = M (15.0 L)
Yes;
we’re OK.
mol HAVE
>
1.50 mol NEED

22. Dilution
Preparation of a desired solution by adding water to a concentrate.
Moles of solute remain the same.

23. Dilution
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3

24. Preparing Solutions How to prepare 500 mL of 1.54 M NaCl solution
mass 45.0 g of NaCl
add water until total volume is 500 mL
500 mL
volumetric
flask
500 mL
mark
45.0 g NaCl
solute

25. Preparing Solutions 500 mL of 1.54M NaCl
molality
molarity
1.54m NaCl in kg of water
500 mL of 1.54M NaCl
mass 45.0 g of NaCl
add kg of water
mass 45.0 g of NaCl
add water until total volume is 500 mL
500 mL water
500 mL
volumetric
flask
45.0 g NaCl
500 mL
mark