Solutions
Dr. Ron Rusay
Spring 2003
Limestone Caves: Solubility of CaCO3
Solutions
1
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Substances can mix together to form
homogeneous mixtures (solutions). The material
present in the larger amount is the solvent and
the other(s) is (are) the solute(s). Together they
form a solution.
The most common solutions are liquids. The
solute can be a solid, liquid or gas which is
dissolved in a liquid solvent.
The most common solvent is water.
© Copyright 1995-2001 R.J. Rusay
DHMO, dihydromonoxide :
“The Universal” Solvent
http://www.dhmo.org
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04_40
H
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2
105
O
H
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Water as a Solvent
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The oil (nonpolar)
and water (polar) mixture don’t mix and
H
are immiscible. If liquids form a homogeneous mixture,
they are miscible.
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04_40

2
105
O
H
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Generally, likes dissolve likes, i.e. polar-polar and
nonpolar-nonpolar. If polar and nonpolar mix , eg.
oil and water:
Salt dissolving in a glass of water
Water dissolving an ionic solid
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Concentration and Temperature
Relative Solution
Concentrations:
Saturated
Unsaturated
Supersaturated
Besides amount, the rate also increases.
What are two other ways of increasing the
solubility of a solid, eg. sugar in coffee?
Solution Types
1
 Solutions
with less solute dissolved than is
physically possible are referred to as
“unsaturated”. Those with a maximum
amount of solute are “saturated”.
 Occasionally there are extraordinary
solutions that are “supersaturated” with
more solute than normal.
A Giant, Single Crystal
and Nuclear Energy
Tooth
Enamel
(Dentyne) &
Fluoride
Treatment
Gas Solubility
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Solubility @ P2 = Solubility @ P1 x [P2 / P1 ]
P is the partial pressure of the gas vapor.
Solubility units (Concentration) are usually: g / 100 ml
Preparation of Solutions
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Solution Concentration
1
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A solution’s concentration is the measure of the
amount of solute dissolved.
Concentration is expressed in several ways. One way
is mass percent.
Mass % = Mass solute / [Mass solute + Mass solvent ]
x100
What is the mass % of 65.0 g of glucose dissolved in
135 g of water?
Mass % = 65.0 g / [65.0 + 135]g x100
= 32.5 %
© Copyright 1995-2001 R.J. Rusay
Solution Concentration
1Concentration
is expressed more importantly as
molarity (M).
Molarity (M) = Moles solute / Liter
Solution
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An important relationship is M x Vsolution= mol
This relationship can be used directly in mass
calculations of chemical reactions.
What is the molarity of a solution of 1.00 g KCl in
75.0 mL of solution?
M = 1.00g KCl x 1mol KCl / 74.55 g KCl x 1/ 75mL x 1000mL / L
= 0.18 mol / L
© Copyright 1995-2001 R.J. Rusay
Acid-Base Titration
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Solution Applications
Neutralization-Titration
1
15.50
mL of vinegar, a solution of acetic acid (aa),
required 20.00 mL of a 0.3000 M (mol/L)
solution of a sodium hydroxide solution to react
completely. Maa = ?
HC2H3O2 (aq) + NaOH(aq)  ? + ?
HC2H3O2 (aq) + NaOH(aq)  1 NaC2H3O2 (aq) + 1H2O (l)
?Maa
= [MNaOHx VNaOH / VHC2H3O2 ] [? molHC2H3O2 / ? molNaOH]
?Maa =
?Maa
0.3000 molNaOH x 0.02000 LNaOH x 1 molHC2H3O2
LNaOH x 0.01550 LHC2H3O2 x 1 molNaOH
= 0.3870 molHC2H3O2 / LHC2H3O2 = 0.3870 MHC2H3O2
Solution Applications
http://ep.llnl.gov/msds/Chem106/vinegar.html
What is the weight percent of the acetic acid
(aa)?
(Density of vinegar = 1.006 g/mL; Molar Mass (aa) = MW
1
�
(aa) in grams/ mole aa = 60.056g aa/mole aa)
(Molarity aa x Volume aa) x (60.056g aa /mole aa)
�
= mass aa = 0.360 gaa
[(mass of Acetic Acid) / (mass of vinegar) ] * 100% = %
Acetic Acid
� mass of vinegar = density x sample volume = 15.59 g
� % = 0.360 gaa / 15.59 g * 100 = 2.3%
Solution Dilution
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