Agenda
• Day 66 – Concentration
• Lesson: PPT,
• Handouts: 1. Concentration& Dilution
Handout. 2. Concentration of Solutions
Worksheet
• Text: 1. P. 398-401 - Concentration ( %,
ppm)
• HW: 1. Worksheets, P. 400 # 1-4, P.402 # 310 %, ppm
Solution Concentration
Concentration = quantity of solute
quantity of solution (not solvent)
There are 3 basic ways to express concentration: 1)
percentages, 2) very low concentrations, and 3) molar
concentrations
% concentration can be in V/V, W/W, or W/V
• Like most %s, V/V and W/W need to have the same units
on top and bottom.
• W/V is sort of in the same units; V is mostly water and
water’s density is 1 g/mL or 1 kg/L
3 g H2O2/100 mL solution  3 g H2O2/100 g solution
Solution and Concentration
• Other ways of expressing concentration
–Molarity(M): moles solute / Liter solution
–Mass percent: (mass solute / mass of solution) * 100
–Molality* (m) - moles solute / Kg solvent
–Mole Fraction(A) - moles solute / total moles solution
* Note that molality is the only concentration unit in which
denominator contains only solvent information rather than
solution.
% Concentration
• % (w/w) =
mass solute
mass solution
x 100
[ 3% w/w = 3 g/100 g]
• % (w/v) =
mass solute
volume solution
x 100
[ 3% w/v = 3 g/100 mL]
volume solute
• % (v/v) =
volume solution
x 100
[ 3% v/v = 3 mL/100 mL]
Units of Concentrations
amount of solute per amount of solvent or solution
Percent (by mass) =
Molarity (M) =
g solute
g solution
x 100 =
g solute
g solute + g solvent
moles of solute
volume in liters of solution
moles = M x VL
x 100
Solution Concentration
Expressing concentrations in parts per million (ppm) requires
the unit on top to be 1,000,000 times smaller than the unit
on the bottom
E.g. 1 mg/kg or g/g
• Notice that any units expressed as a volume must be
referring to a water solution (1L = 1kg)- density of water
• For parts per billion (ppb), the top unit would have to be
1,000,000,000 times smaller
1 ppm = 1 g/106 mL
1g = 1000mg
= 1 g/ 1000 L
1000mg/1000L
m solute
= 1 mg/L
Cppm =
x 10 6
m solution
= 1 mg/kg
1mg = 1000 g
1000g/1000g
= 1 g/g
Molarity
n
C=
V
Molar concentration is the most commonly used in
chemistry.
amount of solute (in moles)
Molar concentration = ----------------------------------------volume of solution (in litres)
UNITS: ( mol/L) or (mol .L-1) or M
Concentration: Percentage Examples
1. What is the % W/W of copper in an alloy when 10 g of Cu
is mixed with 250 g of Zn?
10 g / 260 g = 3.8 % W/W
2. What is approximate % V/V if 30 mL of pure ethanol is
added to 250 mL of water?
30 mL / 280 mL = 11% V/V (in reality may be off)
3. What is the % W/W if 8.0 g copper is added to enough zinc
to produce 100 g of an alloy?
8.0 g / 100 g = 8% W/W
Concentration: Molarity Example
If 0.435 g of KMnO4 is dissolved in enough water to
give 250. mL of solution, what is the molarity of
KMnO4?
As is almost always the case, the first step is to convert the mass of
material to moles.
0.435 g KMnO4 x 1 mol KMnO4 = 0.00275 mol
KMnO4
158.0 g KMnO4 is known, this can be
Now that the number of moles of substance
combined with the volume of solution — which must be in liters —
to give the molarity. Because 250. mL is equivalent to 0.250 L .
Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M
0.250 L solution
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
Step 2: Calculate Molarity
[NiCl2•6 H2O ] = 0.0841 M
Concentration: Mixed Example
A solution of H2O2 is 3% (w/v).
a) Calculate the mass of H2O2 in 250.0 mL of solution.
b) Calculate the mass of H2O2in 1L of solution.
c) Calculate the number of moles of H2O2 in 1L of
solution.
d) State the molar concentration of the solution
e) Calculate the ppm of H2O2 .
Concentration: Mixed ExampleAnswers
a)
3g
mass = 250mL
= 7.50g
100ml
b)
3g
mass = 1000mL
= 30g
100ml
c)
1mol
n = 30g
= 0.88mol
34g
d)
mol
M = 0.88
L
3g
e)
6
Cppm =
x 10 = 30000 ppm
100mL
Making
Molar
Solutions
From
Solids
Preparation of Solutions
1.0 L of water
was used to
make 1.0 L of
solution.
Notice the
water left
over.
What are molar solutions?
A molar solution is one that expresses “concentration” in
moles per volume. Usually the units are in mol/L
mol/L can be abbreviated as M
Molar solutions are prepared using:
a balance to weigh moles (as grams)
a volumetric flask to measure litres
L refers to entire volume, not water!
Because the units are mol/L, we can use the equation
M = n/L
Alternatively, we can use the factor label method.
Calculations with molar solutions
Q: How many moles of NaCl are required to make 7.5 L of a
0.10 M solution?
M=n/L, n = 0.10 M x 7.5 L = 0.75 mol
# mol NaCl = 7.5 L x 0.10 mol NaCl = 0.75 mol
1L
But in the lab we weigh grams not moles, so …
Q: How many grams of NaCl are required to make 7.5 L of
a 0.10 M solution?
# g NaCl =
7.5 L x 0.10 mol NaCl x 58.44 g NaCl =43.83 g
1L
1 mol NaCl
More Practice Questions
1. How many grams of nitric acid are present in 1.0 L of a
1.0 M HNO3 solution?
63 g
2. Calculate the number of grams needed to produce 1.00
L of these solutions:
a) 1.00 M KNO3
101 g
b) 1.85 M H2SO4 181 g
c) 0.67 M KClO3 82 g
3. Calculate the # of grams needed to produce each:1
a) 0.20 L of 1.5 M KCl b) 0.160 L of 0.300 M HCl
a) 22 g b) 1.75 g
c) 0.20 L of 0.09 mol/L AgNO3
d) 250 mL of 3.1 mol/L BaCl2
c) 3 g d) 0.16 kg
4. Give the molarity of a solution containing 10 g of each
solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)2
5. Describe how 100 mL of a 0.10 mol/L a) 0.041 mol/L
NaOH solution would be made.
b) 0.054 mol/L
Practice making molar solutions
1. Calculate # of grams required to make 100 mL
of a 0.10 M solution of NaOH (see above).
2. Get volumetric flask, plastic bottle, 100 mL
beaker, eyedropper. Rinse all with tap water.
3. Fill a beaker with distilled water.
4. Pour 20 - 30 mL of H2O from beaker into flask.
5. Weigh NaOH. Add it to flask. Do step 5 quickly.
6. Mix (by swirling) until the NaOH is dissolved.
7. Add distilled H2O to just below the colored
line.
8. Add distilled H2O to the line using eyedropper.
9. Place solution in a bottle. Place label (tape) on
bottle (name, date, chemical, molarity). Place
bottle at front. Rinse & return equipment.
Concentration and Dilution
How can a solution be made less concentrated?
More solvent can be added.
What is this process called?
Dilution
This process is used extensively in chemistry... the
concentration decreases in dilution, BUT what happens
to the moles of the solute?
– Do they increase?
– Decrease?
– Stay the same?
Agenda
•
•
•
•
•
Day 67 – Dilutions
Lesson: PPT,
Handouts: 1. Concentration& DilutionHandout
Text: 1. P. 403-411HW: 1. Worksheets, P. 405 # 2-7, P. 410 # 110, P.416-421- Review questions
Dilution of Solutions
Dilution of solutions
• Since moles are constant, the new
concentration may be found using the
following formula:
n1=n2
Initial concentration
Final volume
C1V1 = C2V2
Initial volume
Final concentration
Dilution Example #1
A stock solution of 1.00M of NaCl is available. How many
milliliters are needed to make a 100.0 mL of 0.750M?
What we know: the molarity of the stock solution which is
1.00 M, and the two components of the diluted solution
which are C2= 0.750M and V2= 100 mL.
C1V1 = C2V2
Dilution Example #2
Concentrated HCl is 12M. What volume is needed to
Make 2L of a 1M solution?
What we know: the molarity of the stock solution
which is 12M, and the two values for the diluted
solution which are C2=1M and V2=2L.
C1V1 = C2V2
Dilution Example #3
Calculate the final concentration if 2L of 3M of
NaCl and 4L of 1.50M of NaCl are mixed. Assume there
is no volume contraction upon mixing.
ntotal
For this, you must use the equation:
Ctotal =
Vtotal
Making
Molar
Solutions
From
Liquids
(More accurately, from stock
solutions)
Making molar solutions from liquids
Not all compounds are in a solid form
Acids are purchased as liquids (“stock solutions”). Yet,
we still need a way to make molar solutions of these
compounds.
The Procedure is similar:
Use pipette to measure moles (via volume)
Use volumetric flask to measure volume
Now we use the equation C1V1 = C2V2
Reading a pipette
Identify each volume to two decimal places
(values tell you how much you have expelled)
4.48 - 4.50
4.86 - 4.87
5.00
More Practice
1. How many mL of a 14 M stock solution must be used to
make 250 mL of a 1.75 M solution?
2. You have 200 mL of 6.0 M HF. What concentration
results if this is diluted to a total volume of 1 L?
3. 100 mL of 6.0 M CuSO4 must be diluted to what final
volume so that the resulting solution is 1.5 M?
4. What concentration results from mixing 400 mL of 2.0
M HCl with 600 mL of 3.0 M HCl?
5. What is the concentration of NaCl when 3 L of 0.5 M
NaCl are mixed with 2 L of 0.2 M NaCl?
6. What is the concentration of NaCl when 3 L of 0.5 M
NaCl are mixed with 2 L of water?
7. Water is added to 4 L of 6 M antifreeze until it is 1.5 M.
What is the total volume of the new solution?
8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out,
what is the concentration of the remaining HF?
1.
2.
3.
Dilution problems (1-6)
M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL
V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)
V1 = 0.03125 L = 31.25 mL
M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 L
M2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L)
M2 = 1.2 M
M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ?
V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M)
V2 = 0.4 L or 400 mL
Dilution problems (4 - 6)
4. # mol = (2.0
mol/L)(0.4 L) + (3.0 mol/L)(0.6 L)
= 0.8 mol + 1.8 mol = 2.6 mol
# L = 0.4 L + 0.6 L
# mol/L = 2.6 mol / 1 L = 2.6 mol/L
5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L)
= 1.5 mol + 0.4 mol = 1.9 mol
# mol/L = 1.9 mol / 5 L = 0.38 mol/L
6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L)
= 1.5 mol + 0 mol = 1.5 mol
# mol/L = 1.5 mol / 5 L = 0.3 mol/L
Or, using M1V1 = M2V2,
M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L
Dilution problems (7, 8)
7.
M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ?
V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M)
V2 = 16 L
8. The concentration remains 0.2 M, both volume
and moles are removed when the solution is
poured out. Remember M is mol/L. Just like the
density of a copper penny does not change if it is
cut in half, the concentration of a solution does
not change if it is cut in half.
Practice making molar solutions
• Calculate # of mL of 1 M HCl required to make 100 mL
of a 0.1 M solution of HCl
• Get volumetric flask, pipette, plastic bottle, 100 mL
beaker, 50 mL beaker, eyedropper. Rinse all with tap
water. Dry 50 mL beaker
• Place about 20 mL of 1 M HCl in 50 mL beaker
• Rinse pipette, with small amount of acid
• Fill flask about 1/4 full with distilled water
• Add correct amount of acid with pipette. Mix.
• Add water to line (use eyedropper at the end)
• Place solution in plastic bottle
• Label bottle. Place at front of the room.
• Rinse and return all other equipment.