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 # 3- 10 %, ppm
Slide 3
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
waters density is 1 g/mL or 1 kg/L 3 g H 2 O 2 /100 mL solution 3 g
H 2 O 2 /100 g solution
Slide 4
Solution and Concentration Other ways of expressing
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.
Units of Concentrations amount of solute per amount of solvent
or solution Percent (by mass) = g solute g solution x 100 g solute
g solute + g solvent x 100 = Molarity (M) = moles of solute volume
in liters of solution moles = M x V L
Slide 7
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/10 6 mL = 1 g/ 1000 L = 1 mg/L = 1 mg/kg = 1 g/g 1g = 1000mg
1000mg/1000L 1g = 1000mg 1000mg/1000L 1mg = 1000 g 1000 g/1000g 1mg
= 1000 g 1000 g/1000g
Slide 8
Molarity 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
Slide 9
10 g / 260 g = 3.8 % W/W 30 mL / 280 mL = 11% V/V (in reality
may be off) 8.0 g / 100 g = 8% W/W 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? 2.What is approximate % V/V if 30 mL of
pure ethanol is added to 250 mL of water? 3.What is the % W/W if
8.0 g copper is added to enough zinc to produce 100 g of an
alloy?
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Concentration: Molarity Example If 0.435 g of KMnO 4 is
dissolved in enough water to give 250. mL of solution, what is the
molarity of KMnO 4 ? Now that the number of moles of substance is
known, this can be combined with the volume of solution which must
be in liters to give the molarity. Because 250. mL is equivalent to
0.250 L. As is almost always the case, the first step is to convert
the mass of material to moles. 0.435 g KMnO 4 x 1 mol KMnO 4 =
0.00275 mol KMnO 4 158.0 g KMnO 4 Molarity KMnO 4 = 0.00275 mol
KMnO 4 = 0.0110 M 0.250 L solution
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PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to
make 250 mL of solution. Calculate the Molarity. Step 1: Calculate
moles of NiCl 2 6H 2 O Step 2: Calculate Molarity NiCl 2 6 H 2 O
[NiCl 2 6 H 2 O ] = 0.0841 M
Slide 12
Concentration: Mixed Example A solution of H 2 O 2 is 3% (w/v).
a)Calculate the mass of H 2 O 2 in 250.0 mL of solution.
b)Calculate the mass of H 2 O 2 in 1L of solution. c)Calculate the
number of moles of H 2 O 2 in 1L of solution. d)State the molar
concentration of the solution e)Calculate the ppm of H 2 O 2.
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Concentration: Mixed ExampleAnswers a) b) c) d) e)
Slide 14
Making Molar Solutions From Solids
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Preparation of Solutions
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1.0 L of water was used to make 1.0 L of solution. Notice the
water left over.
Slide 17
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.
Slide 18
Calculations with molar solutions Q: How many moles of NaCl are
required to make 7.5 L of a 0.10 M solution? 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? M=n/L, n = 0.10 M x 7.5 L = 0.75
mol # mol NaCl = 7.5 Lx 0.10 mol NaCl 1 L = 0.75 mol # g NaCl = 7.5
Lx 0.10 mol NaCl 1 L =43.83 g x 58.44 g NaCl 1 mol NaCl
Slide 19
More Practice Questions 1.How many grams of nitric acid are
present in 1.0 L of a 1.0 M HNO 3 solution? 2.Calculate the number
of grams needed to produce 1.00 L of these solutions: a) 1.00 M KNO
3 b) 1.85 M H 2 SO 4 c) 0.67 M KClO 3 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 c) 0.20 L of 0.09 mol/L AgNO 3 d) 250 mL of 3.1 mol/L BaCl 2
4.Give the molarity of a solution containing 10 g of each solute in
2.5 L of solution: a)H 2 SO 4 b)Ca(OH) 2 5.Describe how 100 mL of a
0.10 mol/L NaOH solution would be made. 63 g 101 g 181 g82 g a) 22
gb) 1.75 g c) 3 g d) 0.16 kg a) 0.041 mol/L b) 0.054 mol/L
Slide 20
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 H 2 O 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 H 2 O to just below the colored line.
8.Add distilled H 2 O 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.
Slide 21
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?
Slide 22
Agenda Day 67 Dilutions Lesson: PPT, Handouts: 1.
Concentration& DilutionHandout Text: 1. P. 403-411- HW: 1.
Worksheets, P. 405 # 2-7, P. 410 # 1- 10, P.416-421- Review
questions
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Dilution of Solutions
Slide 24
Dilution of solutions Since moles are constant, the new
concentration may be found using the following formula: n 1 = n 2 C
1 V 1 = C 2 V 2 Initial volume Initial concentrationFinal volume
Final concentration
Slide 25
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 C
2 = 0.750M and V 2 = 100 mL. C 1 V 1 = C 2 V 2
Slide 26
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 C 2 =1M and V 2 =2L. C 1 V 1 = C 2 V 2
Slide 27
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. For this, you must use the
equation:
Slide 28
Making Molar Solutions From Liquids (More accurately, from
stock solutions)
Slide 29
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 C 1 V 1
= C 2 V 2
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Identify each volume to two decimal places (values tell you how
much you have expelled) 4.48 - 4.504.86 - 4.875.00 Reading a
pipette
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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 CuSO 4 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? More Practice
Slide 32
Dilution problems (1-6) 1. M 1 = 14 M, V 1 = ?, M 2 = 1.75 M, V
2 = 250 mL V 1 = M 2 V 2 / M 1 = (1.75 M)(0.250 L) / (14 M) V 1 =
0.03125 L = 31.25 mL 2. M 1 = 6 M, V 1 = 0.2 L, M 2 = ?, V 2 = 1 L
M 2 = M 1 V 1 / V 2 = (6 M)(0.2 L) / (1 L) M 2 = 1.2 M 3. M 1 = 6
M, V 1 = 100 mL, M 2 = 1.5 M, V 2 = ? V 2 = M 1 V 1 / M 2 = (6
M)(0.100 L) / (1.5 M) V 2 = 0.4 L or 400 mL
Slide 33
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
M 1 V 1 = M 2 V 2, M 1 = 0.5 M, V 1 = 3 L, M 2 = ?, V 2 = 5 L
Slide 34
Dilution problems (7, 8) 7. M 1 = 6 M, V 1 = 4 L, M 2 = 1.5 M,
V 2 = ? V 2 = M 1 V 1 / M 2 = (6 M)(4 L) / (1.5 M) V 2 = 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.
Slide 35
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.