Transcript Solutions II: Concentrations and Colligative Properties

SOLUTIONS II
Concentrations and Colligative
Properties
Solution Concentrations
• Always:
concentration =
amount of solute_____
amount of solvent or solution
• Expressed as relative amounts:
dilute
or concentrated
unsaturated or saturated
• Most often expressed with specific
amounts of each component
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Percent Solution
volume
of
solute
• By volume: % v/v =
x 100
volume of solution
(volume must have the
same units)
• By mass:
mass of solute
% m/m =
x 100
mass of solution
(mass must have the
same units)
• By mass/volume:
mass of solute
(generally grams
% m/v =
x 100
and mL)
volume of solution
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Percent Solution
1. What is the % 1. % m/v = mass of solute x 100
m/v of a solution
volume of solution
containing 35 g
35 g
of sodium sulfate % m/v = 500 mL = 7.0 % Na2SO4
in 500 mL of
water solution?
2. Since the nature of
2. What would the
the substance does not
% be if the
solute was
play a part in the
potassium
calculation the % value
acetate?
does not change.
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Percent Solution
• How many grams
mass of solute
% m/v =
x 100
of copper (II)
volume of solution
chloride are
required to make
345 mL of a 25% mass = % m/v x volume of solution
100
solution (m/v)?
25 % x 345 mL
mass =
= 86 g
100
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Molarity (again)
• Because many reactions happen between
substances in solution, it is important to
have a concentration based on moles.
moles of solute
Molarity (M) =
1 liter of solution
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Molarity
• How many grams of sodium phosphate
are required to make 250 mL of a 0.140
M solution?
mol
M =
L
to find grams we need moles
mol = M x L
mol = 0.140 M x 0.250 L = 0.0350 mol Na3PO4
164.0 g Na3PO4
0.0350 mol Na3PO4 x
 5.74 g Na3PO4
1 mol mol Na3PO4
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Reactions with Solutions
How many mL of 0.30 M NaCl are needed to
react with 20.0 mL of 0.12 M Pb(NO3)2?
2 NaCl (aq) + Pb(NO3)2 (aq)  PbCl2 (s) + 2 NaNO3 (aq)
mol = M x V = 0.12 M x 0.0200 L = 0.0024 mol Pb(NO3 )2
2 mol NaCl
0.0024 mol Pb(NO3 )2 x
= 0.0048 mol NaCl
1 mol Pb(NO3 )2
1 L NaCl
0.0048 mol NaCl x
= 0.016 L NaCl = 16 mL NaCl
0.30 mol NaCl
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Figure 3.11 Converting a concentrated solution to a dilute solution.
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Dilution Calculations
• For dilutions use the formula:
Ci x V i = Cf x V f
Where Ci = initial concentration
Vi = initial volume
Cf = final concentration
Vf = final volume
You can use any type of concentration or
volume unit as long as they are the same.
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Dilution Calculations
• How many mL of 0.50 M KCl are
required to make 1.5 L of 0.20 M KCl?
Ci x Vi = Cf x Vf
Cf x Vf
0.20 M x 1.5 L
Vi =
=
Ci
0.50 M
Vi = 0.60 L = 6.0 x 10 mL
2
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Colligative Properties
• Properties that change depending on the
number of particles dispersed in a
solvent
• Vapor pressure lowering
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Figure 13.11
The effect of a solute on the vapor pressure of a solution.
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Colligative Properties
• Properties that change depending on the
number of particles dispersed in a
solvent
• Vapor pressure lowering
• Boiling point elevation
• Freezing point depression
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Figure 13.12
Phase diagrams of solvent and solution.
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Boiling and Freezing Point
Changes
• ΔTb = Kb x m where
m= moles solute
kg solvent
Kb = boiling pt constant
• ΔTf = Kf x m
Kf = freezing point constant
• Applications…
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Table 13.5
Solvent
Molal Boiling Point Elevation and Freezing Point
Depression Constants of Several Solvents
Boiling
Point (0C)*
Acetic acid
Kb (0C/m)
Melting
Point (0C)
Kf (0C/m)
117.9
3.07
16.6
3.90
Benzene
80.1
2.53
5.5
4.90
Carbon disulfide
46.2
2.34
-111.5
3.83
Carbon tetrachloride
76.5
5.03
-23
Chloroform
61.7
3.63
-63.5
4.70
Diethyl ether
34.5
2.02
-116.2
1.79
Ethanol
78.5
1.22
-117.3
1.99
Water
100.0
0.512
0.0
1.86
*at 1 atm.
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30.
Colligative Properties
• Properties that change depending on the
number of particles dispersed in a
solvent
• Vapor pressure lowering
• Boiling point elevation
• Freezing point depression
• Osmotic pressure – movement of solvent
particles across a semipermeable
membrane
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Figure 13.13
The development of osmotic pressure.
osmotic pressure
pure
solvent
semipermeable
membrane
solution
net movement of solvent
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Applied pressure
needed to prevent
volume increase
Osmotic Pressure
• Remember PV = nRT?
PV=nRT
P =nRT
V
Π = M RT
Where Π = osmotic pressure
M = molarity
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Osmosis and cells
a. When a cell is placed in a solution with a hypertonic (higher solute
concentration but less water) water flows out of the cell and the cell
shrinks (crenation).
b. When a cell is placed in a hypotonic solution (less solute but more water),
water flows into the cell and the cell swells and may burst (hemolysis).
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Colligative Properties of Electrolyte Solutions
For electrolyte solutions, the compound formula
tells us how many particles are in the solution.
The van’t Hoft factor, i, tells us what the
“effective” number of ions are in the solution.
van’t Hoff factor
(i)
i
=
measured value for electrolyte solution
expected value for nonelectrolyte solution
For vapor pressure lowering:
P = i(solutex P0solvent)
For boiling point elevation:
Tb = i(bm)
For freezing point depression:
Tf = i(fm)
For osmotic pressure :
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 = i(MRT)
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