(5) UNIT SOLUTIONS
A solution is a homogeneous mixture of two or more substances.
5.1 Solution Terminology
Solute and Solvent A simple solution has two components, a solute, and a solvent.
The substance in smaller amount is called the solute, the substance in larger amount is
called the solvent. When sugar is dissolved in water to form a solution, sugar is a solute
and water is a solvent.
Aqueous Solutions The most common solutions are those containing water as the
solvent. These solutions are called aqueous solutions.
Concentrated and Dilute A concentrated solution has a relatively large amount of a
solute per unit of solution, and a dilute solution has a relatively small amount of the same
solute per unit of solution.
Miscible and Immiscible are terms limited to solutions of liquids in liquids.
If two liquids dissolve in each other they form a single continuous layer (like water and
alcohol) and are miscible. Liquids that are not soluble in each other form two distinct
layers (like water and oil) and are immiscible.
Saturated and Unsaturated A solution whose concentration is at the solubility limit at
a given temperature is a saturated solution. If the concentration of a solute is less than
the solubility limit the solution is unsaturated.
Common Aqueous Solutions
Solute
Solvent
Example
gas
in
water
carbonated water (CO2 in water)
liquid
in
water
beer (ethanol in water)
ionic solid
in
water
saline solution (NaCl in water)
molecular solid
in
water
sugar solution (C12H22O11 in water)
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5.2 Water as a Solvent
Water is called the universal solvent because of its ability to dissolve or dissociate many
compounds. The properties that make water such a good solvent are its polarity (see unit
3) and its hydrogen bonding capability.
A) Hydrogen Bonding
Recall that the O-H bonds in water are highly polar. This is due to the large
electronegativity difference between oxygen and hydrogen. The oxygen atom has a
partial negative charge (-) and each hydrogen atom has a partial positive charge (+).
As a result, a partially positive H atom in one molecule can be attracted to a partially
negative oxygen atom in another molecule of water (opposites attract). This interaction
between H and O in different molecules (intermolecular) is called a hydrogen bond due
to the involvement of hydrogen.
Hydrogen bond formation is possible whenever two molecules are the same (waterwater) or different (water-alcohol).
Hydrogen bonding is possible between two molecules if one of them contains O, F, or N
atom and the other contains O-H, O-F, or N-H bond.
XH
XH
X = O, F, or N
..
N
HHH
+
+
..
O
H
+
+
F
H
+
H
+
The hydrogen bond between molecules appears as dashed lines to differentiate it from the
covalent bonds within the water molecule.
O
H
H +
O
H
H
B) How Water Dissolves Ionic Compounds
We learned in Unit (3) that ionic compounds are composed of ions and are held together
by ionic bonds. For NaCl, attractions occur between the cations (Na+) and the anions
(Cl-). Water, a polar molecule, is a good solvent for many ionic compounds. When NaCl
crystals are added to water, the ions at the surface of the crystal become surrounded by
water molecules. The cations (Na+) attract the negative ends of water molecules, and the
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anions (Cl-) attract the positive ends of water molecules. These attractive forces between
the adjacent water molecules and the ions in NaCl are sufficient to overcome the
attractions that exist between ions in the solid sodium chloride. The ions then leave the
sodium chloride crystals and enter the solution. Once in solution, Na+ and Cl- ions are
completely surrounded by water molecules, a phenomenon called hydration.
The process of dissolving NaCl in water can be shown as:
H2O
NaCl(s)
+ Cl (aq)
Na+(aq)
The H2O above the arrow represents a large number of water molecules required to
break up the crystal and hydrate the ions but it is not a reactant in the usual sense.
The (aq) indicates that the ions are surrounded by water molecules. Recall (aq) is the
symbol for in water.
C) How Water Dissolves Molecular Compounds
Sugar, C12H22O11, is one example of molecular compound that is very soluble in water.
The reason that sugar is so soluble in water lies in the structure of sugar. A sugar
molecule contains a large number of polar O-H bonds, each of which is capable of
forming a hydrogen bond to any adjacent water molecule. Water molecules are attracted
to the polar sugar molecules and pull the molecules of sugar into the solution. Several
water molecules surround each sugar molecule in the solution.
The process of dissolving sugar in water can be shown as:
C12H22O11(s)
H2O
C12H22O11(aq)
Notice that in aqueous solutions of molecular compounds no ions are present.
5.3 Electrolytes and Nonelectrolytes
Substances whose aqueous solutions are conductors of electricity are called electrolytes.
Substances whose aqueous solutions are not conductors of electricity are called
nonelectrolytes.
In order for a solution to conduct an electric current, ions must be present.
Electrolytes produce ions in aqueous solutions. Nonelectrolytes do not produce ions.
H2O
Electrolyte
NaCl(s)
Na+(aq)
+ Cl (aq)
H2O
Nonelectrolyte
C12H22O11(s)
C12H22O11(aq)
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5.4 Solubility
A good rule of thumb for determining the solubility is like dissolves like.
Polar solvents dissolve polar substances, nonpolar solvents dissolve nonpolar substances.
Solubility is the maximum amount of solute that can dissolve in a given amount of
solvent.
For most solids the solubility increases with increasing temperature. In contrast, gases are
less soluble in water at higher temperature.
5.5 Solution Concentration
When dealing with solutions, it is important to know how much solute is present per
given amount of solution. As a per expression, the concentration has the form of a
fraction.
Concentration =
amount of solute
amount of solution
(Amount of solution = amount of solute + amount of solvent)
There are several ways to express concentration; we will consider mass percent (%m/v),
parts per thousand (ppt), parts per million (ppm), parts per billion (ppb), and molarity.
A) Concentration Based on Mass parts per
g of solute
x 100
Parts per hundred (% m/v) =
mL of solution
g of solute
Parts per thousand (ppt) =
x 1000
mL of solution
g of solute
Parts per million (ppm) =
x 1,000,000
mL of solution
g of solute
Parts per billion (ppb) =
x ( 1 x 109 )
mL of solution
5-4
Worked Example 5-1
Calculate the % (m/v) of a solution containing 8.92 g of KCl in 455 mL of
solution.
Solution
Parts per hundred (% m/v) =
8.92 g
455 mL
x 100 = 1.96% (m/v)
Practice 5-1
A 5.00 mL sample of solution has 3.8 x 10-5 g of calcium ion. Calculate the
concentration in each of the following units: a) %(m/v) b) ppt c) ppm.
Answer
g of solute
a) %(m/v) = ---------------------- x 100
L of solution
3.8 x 10-5 g
%(m/v) = ------------------- x 100 = 7.6 x 10-4 %
5.00 mL
g of solute
b) Parts Per Thousand (ppt) = --------------------- x 1000
mL of solution
3.8 x 10-5 g
ppt = ---------------------- x 1000 = 7.6 x 10-3 ppt
5.00 mL
g of solute
c) Parts Per Million (ppm) =- --------------------- x 106ppm
mL of solution
3.8 x 10-5 g
ppm = ---------------------- x 1,000,000 = 7.6 ppm
5.00 mL
5-5
Percent Concentrations as Conversion Factors
Worked Example 5-2
Calculate the grams of NaCl in 85 mL of a 4.0% (m/v) NaCl solution.
Solution
The percent (m/v) indicates the mass of solute in 100 mL of a solution. Since the
solution concentration is 4.0%, we can write two conversion factors:
4.0 g NaCl
100 mL solution
and
100 mL solution
4.0 g NaCl
The volume of the solution is converted to mass of solute by selecting the first
conversion factor.
85 mL solution x
4.0 g NaC l
100 mL solution
= 3.4 g NaCl
Practice 5-22
Calculate the number of grams of KCl in150.0 mL of a 6.80% (m/v).
Answer
6.80 g KCl
150.0 mL solution x = ---------------------- 10.2 g of KCl
100 mL solution
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Worked Example 5-3
How many mL of 12.0% (m/v) sucrose solution would contain 3.47 g of sucrose?
Solution
Conversion Factors:
12.0 g sucrose
100 mL solution
and
100 mL solution
12.0 g sucrose
100 mL solution
= 28.9 mL solution
3.47 g sucrose x
12.0 g sucrose
Practice 5-32
How many mL of a 3.0% (m/v) H2O2 solution would contain 7.8 g of H2O2?
Answer
100 mL solution
7.8 g H2O2 x
= 260 mL solution
3.0 g H2O2
B) Molarity (Concentration Based on Moles)
The most common way of expressing concentration in chemistry is called molarity.
Molarity (M) is the moles of solute per liter of solution.
Molarity (M) =
moles of solute
liters of solution
mol
=
L
For example if a solution contains 1.20 mol of sodium hydroxide per liter, its molar
concentration is 1.20 M NaOH.
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Worked Example 5-4
What is the molarity of 5.88 g of CaCl2 in 0.265 L of solution?
Solution
5.88 g CaCl2 x
1 mol CaCl2
110.98 g CaCl2
M=
0.0530 mol CaCl2
= 0.0530 mol CaCl2
= 0.200 M
0. 265 L
Practice 5-4
What is the molarity of a solution made by dissolving 22.50 g of Mg(NO3)2 in
enough water to make 450.0 mL of solution?
Answer
1 mol Mg(NO3)2
22.50 g Mg(NO3)2 x
M=
148.33 g Mg(NO3)2
0.1517 mol Mg(NO3)2
= 0.1517 mol Mg(NO3)2
= 0.3371 M
0. 4500 L
5-8
Molarity as a Conversion Factor
Worked Example 5-5
Calculate the number of grams of NaCl required to prepare1.50 L of 0.250 M
NaCl.
Solution
Like percent concentrations, we can write two conversion factors for molarity:
0.250 mol NaCl
0.250 mol NaCl
1 L solution
1.50 L solution x
1 L solution
and
0.250 mol NaC l
1 L solution
= 0.375 mol NaCl
We use the molar mass of NaCl from the periodic table to calculate the mass of
NaCl (1 mol NaCl = 58.44 g NaCl).
58.44 g NaCl
= 21.9 g NaCl
0.375 mol NaCl x
1 mol NaC l
Practice 5-52
Calculate the volume, in liters, of 3.00 M NaOH containing 0.456 mol of NaOH.
Answer
1 L solution
0.456 mol NaOH x ------------------------ = 0.152 L of solution
3.00 mol NaOH
5-9
5.6 Dilution Calculations
Dilution is the process of adding more solvent to a solution in order to lower its
concentration.
A simple relationship exists between the concentration and volume of the concentrated
and diluted solutions.
C1 x V1 = C2 x V2
C1 = concentration before dilution
V1 = volume before dilution
C2 = concentration after dilution
V2 = volume after dilution
Worked Example 5-6
You begin with 45.0 mL of 16% (m/v) of sugar solution and add enough water to
give a final volume of 55.0 mL. What is the new concentration?
Solution
C1 x V1 = C2 x V2
C1 x V1
C2 =
V2
16% x 45.0 mL
C2 =
= 13%
55.0 mL
Practice 5-6
An aqueous solution contains 75 ppm Cu2+. What volume of this solution is
needed to prepare 350.0 mL of a solution that is 18 ppm Cu2+?
Answer
C1 x V1 = C2 x V2
C2 x V2
V1 = ------------C1
18 ppm x 350.0 mL
V1 = ---------------------------- = 84 mL
75 ppm
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Worked Example 5-7
A 25.0-mL sample of KNO3 solution is diluted to a volume of 75.0 mL. If the
concentration of the diluted solution is found to be 0.840 M, calculate the
concentration of KNO3 in the original solution?
Solution
C1 x V1 = C2 x V2
C2 x V2
C1 =
V1
0.840 M x 75.0 mL
C1 =
25.0 mL
= 2.52 M
Practice 5-7
To what volume should you dilute 75.0 mL of a 10.0 M HCl solution to obtain a
1.20 M HCl solution?
Answer
C1 x V1 = C2 x V2
C1 x V1
V2 = ------------C2
10.0 M x 75.0 mL
V2 = ------------------------- = 625 mL
1.20 M
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5.7 Osmosis and Osmotic Pressure
Osmosis
Water movement across cell membranes is an important biological process.
Osmosis is the passage of water through a semi-permeable membrane from a region of
high water concentration to a region of low water concentration.
A Semipermeable membrane is a thin layer of material that allows small molecules such
as water to pass freely through the membrane but prevents larger molecules or ions from
passing through.
Osmotic Pressure
The amount of pressure that is required to prevent the net flow of solvent (water) across
the membrane and into the more concentrated solution is called osmotic pressure.
The greater the concentration differences between the separated solutions, the greater the
magnitude of the osmotic pressure.
Osmolarity
Osmotic pressure of a solution depends on the number of solute particles present.
(Recall: if the solute is ionic, it splits into ions).
Osmolarity is used to compare the osmotic pressures of different solutions.
Osmolarity = molarity x i
i = number of particles
Worked Example 5-8
What is the osmolarity of each of the following solutions?
a) 3 M NaCl
b) 3 M sucrose (table sugar) c) 3 M in both NaCl and glucose
Solution
a) Osmolarity = molarity x i
NaCl is an ionic solute, and it dissociates in solution to give two particles
(Na+ and Cl-). The value of i is 2. (NaCl Na+ + Cl- )
Osmolarity = 3 M x 2 = 6 osmol
b) Sucrose is a molecular solute and therefore nondissociating. The value of i is 1.
Osmolarity = 3 M x 1 = 3 osmol
c) This is a two-solute system. The value of i is 2 for NaCl and 1 for glucose.
Osmolarity = (3 M x 2) + (3M x 1) = 9 osmol
NaCl
sucrose
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Practice 5-8
What is the osmolarity of each of the following solutions?
a) 0.25 M NaBr
b) 0.15 M Na2SO4
c) 0.25 M NaBr + 0.15 M Na2SO4
Answer
Osmolarity = molarity x i
a) 0.25 M x 2 = 0.50 osmol
(NaBr Na+ + Br-)
b) 0.15 M x 3 = 0.45 osmol
(Na2SO4 2Na+ + SO42- )
c) (0.25 M x 2) + (0.15 M x 3) = 0.95 osmol
Isotonic, Hypertonic, and Hypotonic Solutions
Isotonic Solution:
A solution with an osmotic pressure that is equal to that within cells. Isotonic solutions do
not change cell volume because water is entering and leaving the cell at the same rate.
Normal saline (0.9% NaCl) is isotonic to red-blood cells.
Hypertonic Solution:
A solution with an osmotic pressure that is higher than that within cells. Hypertonic
solutions cause cells to shrink (crenation) because there is a net movement of water out of
the cell.
Hypotonic Solution:
A solution with an osmotic pressure that is lower than that within cells. Hypotonic
solutions cause cells to burst (lysis) because there is a net movement of water into of the
cell.
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Homework Problems
5.1 Diagram a hydrogen bond between:
a. two molecules of NH3.
b. two molecules of HF.
c. water molecule and CH3-OH molecule.
5.2 Write an equation for the dissociation of each of the following ionic compounds
in water.
a. LiOH
b. K2CrO4
c. CaCl2
d. (NH4)3PO4
e. Mg(C2H3O2)2
f. AlBr3
g. Na2HPO4
h. Fe(NO3)2
5.3 Calculate the following concentrations in a given units:
a. A teardrop with a volume of 0.5 mL contains 50 mg NaCl.
What is the %(m/v) concentration?
b. A 250 mL sample of orange juice contains 75 mg vitamin C.
What is the concentration in ppt?
c. An 8-Ounce (237 mL) sample of green tea contains 15 mg of caffeine.
What is the concentration in ppm?
d. The concentration of arsenic in water was found to be 0.02 mg per 100 mL.
Express the concentration in ppb.
5.4 Calculate the mass of solute in each of the following:
a. 85.0 mL of 4.5% (m/v) sucrose solution
b. 275 mL of 10.5% (m/v) NaOH solution
c. 1.80 L of 22.0% (m/v) LiNO3 solution.
5.5 Calculate the volume, in liters, of solution required to supply each of the
following:
a. 0.85 mol of Na2CO3 from a 0.74 M Na2CO3
b. 1.35 mol of MgCl2 from a 0.62 M MgCl2
c. 150.0 g of glucose (C6H12O6) from a 2.75 M glucose solution.
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5.6 Dilution Problems
a. What volume of a 3.6% (m/v) NaOH solution can be prepared by diluting 15.0
mL of 8.5% (m/v) NaOH solution?
b. What is the molarity of a solution prepared when 25.0 mL of 6.0 M HCl is
diluted to a volume of 85.0 mL?
c. How many milliliters of water should be added to 75.0 mL of 2.60 M AgNO3
to produce a 2.15 M solution?
5.7 A solution was prepared by dissolving 4.65 g of AlCl3 in enough water to give
250.0 mL of AlCl3 solution:
a. calculate the molarity of the solution.
b. calculate the osmolarity of the solution.
5-15