8.6: concentration
 Amount concentration, (c), is the amount (in moles) of solute dissolved per litre of solution;
unit symbol mol/L
Equation:
amount concentration = amount of solute (in mol)
volume of solution (in L)
c=
n
v (L)
 A stock solution is a concentrated solution that is to be diluted to a lower concentration prior to
actual use.
Practice:
1. The food industry uses calcium chloride, CaCl2, to make processed foods, such as
pickles, taste saltier without adding sodium chloride. What is the amount
concentration of a 750 mL sample of a calcium chloride solution that contains 1.5
mol of calcium chloride? [ans: 2.0 mol/L]
Given:
n=1.5 mol
V=750 mL
Convert 750 mL to L
= 750 mL x 1L
1000 mL
= 0.750 L
C =?
C=n
V(L)
= 1.50 mol
0.750 L
= 2.0 mol/L
2. Phosphoric acid, H3PO4, is the active ingredient in many consumer products
designed to remove rust. What is the amount concentration of a solution that
contains 63 g of phosphoric acid in 250 mL?
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Preparing Solutions
 In the laboratory, standard solutions are prepared in volumetric flasks, a pear-shaped container
with an elongated neck.
 volumetric flask is designed to contain a certain volume of solution, at a specific temperature (at
20 °C).
 Solutions are prepared by first dissolving a known quantity of solute in about one quarter of the
volume of the solvent, such as water.
 Solvent is carefully added until the bottom of the solution’s meniscus lines up with the
calibration mark on the flask.
Sample Problem:
A solution of sodium oxalate, Na2C2O4, is used to analyze for the presence of calcium ions in water.
Water quality technicians typically use a 0.100 mol/L solution for this analysis.
What volume, in millilitres, can be prepared using 33.5 g of solid sodium oxalate?
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8.7: Preparing Dilutions
Dilution is the process of reducing the concentration of a solution by adding more
solvent.
Example of dilution is preparing orange juice from a can of frozen concentrate, by
diluting the concentrate with three cans of water.
The concentrate is diluted to four times its original volume (the original can + 3 cans of
water = 4 cans).
Diluting 1.0 L of a 12 mol/L hydrochloric acid solution to 4.0 L decreases its
concentration by a factor of 4. The final concentration is 3.0 mol/L.
a) The concentrated potassium
chromate solution in the small flask
b) The concentrated solution is added to
the larger volumetric flask.
d) The concentrated solution is diluted to a
C) The small flask is rinsed with distilled water and
the rinse water is transferred to the larger flask.
larger volume by adding water
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Other Volumetric Glassware
 A graduated cylinder is not precise enough for analytical work involving small volumes,
pipettes are used.
 A volumetric pipette is used to deliver small and fixed volume of solution. For example,
a 10 mL volumetric pipette delivers 10.00 +/- 0.02 mL of solution.
 This type of pipette can deliver a range of volumes, from 0.1 mL to 10.0 mL.
Dilution Calculations
 Solutions dilution can be prepared by simply applying some common sense
 Example, doubling the volume of the solution by adding water decreases its
concentration by a factor of 2, similarly, increasing the volume by ten times decreases
the concentration by a factor of 10.
 During dilution, the amount (in moles) of solute present does not change.
 The concentration changes because the volume of the solution increases
C=n
V increases (because water is
added)
The amount of solute in a solution is given by n = CV.
The Dilution Equation:
C1V1 =C2V2
Practice
1. What is the final concentration when 4.5 L of 3.0 mol/L car battery acid is diluted to
20.0 L? T/I [ans: 0.68 mol/L]
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8.8: Concentrations and consumer Products
1. Percentage Concentration
 A more general equation for the concentration, c, of a solution is:
C =quantity of solute
quantity of solution
 Percentage Volume/Volume (% V/V): The figure indicates that the liquid is 50 %
isopropyl
alcohol by volume.
Cv/v = 50 mL x 100 %
100 mL
In general,
cv/v = Vsolute x 100 %
Vsolution
2. Percentage Weight/Volume (% W/V):
 The concentrations of consumer products may also be given in terms of a percentage
weight/volume, % W/V or cW/V. A product label of hydrogen peroxide bottle may
indicate 6 % W/V.
 This means that 6 g of every 100 mL of the solution is the solute: hydrogen peroxide.
cw/v = 6 g x 100 %
100 mL
=6%
In general,
cw/v = msolute x 100 %
Vsolution
 The term “weight” on a product label actually refers to the mass of the solute.
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3. Percentage Weight/Weight (% W/W)
 The concentrations of some medications and concentrated acids are often expressed as
a percentage weight/weight, % W/W or cW/W.
 The product label, concentration is 5 % W/W benzoyl peroxide.
This means that about 5 g of every 100 g is benzoyl peroxide
cw/w = 5 g x 100 %
100 g
=5%
In general,
cw/w = msolute x 100 %
msolution
Extremely Low Concentrations
 Very dilute concentrations are often expressed in parts per million
(ppm, 1:106), parts per billion (ppb, 1:109), and even parts per trillion (ppt, 1:1012)
 Example, a bottle of water may contain sodium ions at a concentration of 36 ppm. This
means that there are 36 parts of sodium for every million parts of solution.
Masses and Volumes of Very Dilute Solutions
 A solute concentration of 1 ppm means that there is 1 g of solute in every 1 000 000 g or
1 000 000 mL of solution
c=
1 gsolute
1 000 000 gsolution
= 1 x 10-6 g solute/g solution
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cppm = msolute x 10
msolution
cppb = msolute
9
x 10
msolution
cppt =
msolute
12
x 10
msolution
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Sample Problem:
A 1.5 L sample of pool water contains 4.5 mg of chlorine. Determine the concentration of chlorine in
ppm.
Given: VH2O = 1.5 L; mCl = 4.5 mg
Required: chlorine concentration in parts per million, cppm
Analysis: cppm = msolute x 106
msolution
1.5 L of pool water has a mass of 1.5 kg, because density as water (1.0 g/mL) or 1Kg/L,
Since 1 kg = 106 mg, 1.5 kg = 1.5 x 106 mg
Solution: cppm =mCl x 106
mH2O
= 4.5 x 106 mg
1.5 x 106 mg = cppm = 3.0 ppm
Statement: The chlorine concentration of the pool water is 3.0 ppm.
Practice
1. In 2009, testing of certain brands of baby shampoos found unsafe concentrations of
Methanal. This chemical is thought to form from the breakdown of ingredients in the
shampoo. Exposure to 250 ppm of methanal may cause a rash.
What mass of methanal is present in 500 g of shampoo if its concentration is 250
Ppm? T/I [ans: 0.13 g]
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2. The maximum acceptable concentration of fluoride in tap water is 1.5 mg/L. T/I
(a) Express this concentration in ppm. [ans: 1.5 ppm]
(b) What volume of tap water contains 1.0 g of fluoride? [ans: 670 L]
3. A 250 mL sample of tap water is found to contain 12 ppb of an antibiotic. Determine
the mass of the antibiotic in the sample. T/I [ans: 3.0 x 10-6 g]
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