Chapter 6
Solutions
25 March 2014
Solubility of Gases: Henry’s Law
• Henry’s law – the number of moles of a gas dissolved
in a liquid at a given temperature is proportional to the
partial pressure of the gas above the liquid
• Gas solubility in a liquid is directly proportional to the
pressure of the gas in the atmosphere in contact with
the liquid
• Gases are most soluble at low temperatures
• Solubility decreases significantly at higher
temperatures
– Carbonated beverages – CO2 solubility less when warm
1
Concentration Based on Mass
• Concentration - amount of solute dissolved
in a given amount of solution
• Concentration of a solution has an effect on
– Physical properties
• Melting and boiling points
– Chemical properties
• Solution reactivity
Weight/Volume Percent
• Amount of solute = mass of solute in grams
• Amount of solution = volume in milliliters
amount of solute
concentration 
amount of solution
• Express concentration as a percentage by
multiplying ratio by 100% = weight/volume
percent or % (W/V)
%
W
grams of solute

 100%
V milliliter s of solution
2
Calculating Weight/Volume
Percent
Calculate the percent composition or % (W/V)
of 2.00 x 102 ml containing 20.0 g sodium
chloride
20.0 g NaCl, mass of solute
2.00 x 102 ml, total volume of solution
% (W/V) = 20.0 g NaCl / 2.00 x 102 ml x 100 %
= 10.0 % (W/V) sodium chloride
Calculate Weight of Solute
from Weight/Volume Percent
Calculate the number of grams of glucose in
7.50 x 102 ml of a 15.0% solution
%
W
grams of solute

 100%
V milliliters of solution
15.0 % (W/V) = Xg glucose/7.50 x 102 ml x 100 %
Xg glucose x 100 % = (15.0 % W/V)(7.50 x 102 ml)
Xg glucose = 113 g glucose
3
Weight/Weight Percent
%
W
grams solute

100%
W grams solutions
• Weight/weight percent is most useful for
solutions of 2 solids whose masses are easily
obtained
• Calculate % (W/W) of platinum in gold ring (an
alloy) with 14.00 g Au and 4.500 g Pt
[4.500 g Pt / (4.500 g Pt + 14.00 g Au)] x 100 %
= 4.500 g / 18.50 g x 100 % = 24.32 % Pt
Parts Per Thousand (ppt) and
Parts Per Million (ppm)
• As percentage (“per centum”) is the number of
parts of solute in 100 parts of solution, ppt and
ppm change the calculation only by orders of
magnitude
– ppt = (g solute / g solution) x 103
– ppm = (g solute / g solution) x 106
– ppt and ppm are most often used for very dilute
solutions (pollution in environmental studies, e.g.)
Note: ppt could in some literature mean parts per trillion
4
Solubility of Ca(OH)2
s = 0.173 g in 100 ml of water at 20 oC
0.173 g
1730 g

 1730 ppm
100 g 1000000 g
Concentration and contents
5
Concentration and contents
Stella Artois 11.2 fl. oz, 5.2 % alcohol
Four Loko 23.5 fl. oz, 12 % alcohol
Four Loko is larger 2.0982 times by
volume and has 2.3077 times as much
alcohol as a bottle of Stella Artois.
Thus, the can has 2.3077x2.0982=4.84
as much alcohol
Concentration of Solutions:
Moles and Equivalents
• Chemical equations represent the relative
number of moles of reactants producing
products
• Many chemical reactions occur in solution
where it is most useful to represent
concentrations on a molar basis
6
Molarity
• The most common mole-based
concentration unit is molarity
• Molarity
– Unit usually “M” but properly written mol/l
– Defined as the number of moles of solute per
liter of solution
cM 
moles solute
 solution
Calculating Molarity from
Moles and Volume
• Calculate the molarity of 2.0 l of solution
containing 5.0 mol NaOH
• Use the equation
moles solute
cNaOH 
 solution
• Substitute into the equation:
c NaOH 
5.0 mol solute
2.0 l solution
= 2.5 M or 2.5 mol/l
7
Calculating Molarity from Moles
… x-molar solution
moles of solute
liters of solution
Note in the denominator it is
c
liters of solution, not liters of solvent
3-molar solution
Calculating Molarity From Mass
• If 5.00 g glucose are dissolved in 1.00 x 102 ml of
solution, calculate molarity, cM, of the glucose
solution
• Convert from g glucose to moles glucose
– Molar mass of glucose = 1.80 x 102 g/mol
5.00 g x 1 mol / 1.80 x 102 g
= 2.78 x 10-2 mol glucose
– Convert volume from ml to l
1.00 x 102 ml x 1 l / 103 ml = 1.00 x 10-1 l
• Substitute into the equation:
cglu cos e 
c
moles solute
liters solution
2.78  102 mol glucose
 2.78  101 mol / l
1
1.00  10 l solution
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Dilution
Dilution is required to prepare a less
concentrated solution from a more
concentrated one
– c1 = molarity of solution before dilution
– c2 = molarity of solution after dilution
– V1 = volume of solution before dilution
– V2 = volume of solution after dilution
c
moles solute
 solution
moles solute = (mol/l)(l solution)
Dilution
• In a dilution will the
number of moles of solute
change?
– No, only fewer per unit
volume
• So, c1V1 = c2V2
• Knowing any three terms
permits calculation of the
fourth
9
Calculating Molarity
After Dilution
• Calculate the molarity of a solution made by
diluting 0.050 l of 0.10 mol/l HCl solution to a
volume of 1.0 l
–
–
–
–
c1 = 0.10 mol/l molarity of solution before dilution
c2 = x mol/l molarity of solution after dilution
V1 = 0.050 l volume of solution before dilution
V2 = 1.0 l volume of solution after dilution
• Use dilution expression
c1V1 = c2V2
• X mol = (0.10 mol/l) (0.050 l) / (1.0 l)
0.0050 mol/l HCl
OR
5.0 x 10-3 mol/l HCl
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