Chapter 17. Properties of Solutions. Chapter 17: Properties of Solutions. 17.1 Solution Composition 17.2 The Thermodynamics of Solution Formation 17.3 Factors Affecting Solubility 17.4 The Vapor Pressure of Solutions 17.5 Boiling-Point Elevation and Freezing-Point Depression
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Properties of Solutions
17.1 Solution Composition
17.2 The Thermodynamics of Solution Formation
17.3 Factors Affecting Solubility
17.4 The Vapor Pressure of Solutions
17.5 Boiling-Point Elevation and Freezing-Point Depression
17.6 Osmotic Pressure
17.7 Colligative Properties of Electrolytic Solutions
The left beaker contains copper sulfate solution. In the right beaker, ammonia is being added to create a precipitate of copper(II) hydroxide.
Solute- The smaller (in mass) of the components in a solution, the
material dispersed into the solvent.
Solvent - The major component of the solution, the material that the
solute is dissolved into.
Solubility - The maximum amount that can be dissolved into a
particular solvent to form a stable solution at a specified
Miscible - Substances that can dissolve in any proportion, so that it is
difficult to tell which is the solvent or solute!
weight percent Mass percent = x 100%
Mole fraction: symbolized by the Greek letter, chi = X
mole fraction of component A = XA =
Molarity: M ; (chapter 4)
grams of solute
grams of solution
nA + nB
moles of solute
liter of solution
moles of Solute
Kg of solvent
Problem: Predict which solvent will dissolve more of the given solute.
(a) Sodium Chloride in methanol (CH3OH) or in propanol
(b) Ethylene glycol (HOCH2CH2OH) in water or in hexane
(c) Diethyl ether (CH3CH2OCH2CH3) in ethanol (CH3CH2OH) or in
Plan: Examine the formulas of each solute and solvent to determine
which forces will occur. A solute tends to be more soluble in a solvent
which has the same type of forces binding its molecules.
(a)Methanol - NaCl is an ionic compound that dissolves through ion-
dipole forces. Both methanol and propanol contain a polar hydroxyl
group, and propanol’s longer hydrocarbon chain would form only weak
forces with the ions, so it would be less effective at replacing the ionic
attractions of the solvent.
(b) Water Ethylene glycol molecules have two -OH groups, and the
molecules interact with each other through H bonding. They would be
more soluble in water, whose H bonds can replace solute H bonds better
than can the dispersion forces in hexane.
(c) Ethanol Diethyl ether molecules interact with each other through
dipole and dispersion forces and could form H bonds to both water and
ethanol. the ether would be more soluble in ethanol because the solvent
can form H bonds and replace the dispersion forces in the solute,
whereas the H bonds in water must be partly replaced with much weaker
Solar ponds are shallow bodies of salt water ( a very high salt content)
designed to collect solar energy as it water the water in the ponds, and
then it can be used for heating, or converted into other forms of energy.
Sufficient salt must be added to establish a salt gradient in a pool
2-3 meters deep, with a dark bottom. A salt gradient will be established
in which the upper layer, called the conductive layer, has a salt content
of about 2% by mass. The bottom layer, called the heat storage layer has
a salt content of about 27%. The middle layer called the nonconvective
layer has an intermediate salt content, and acts as an insulator between
the two layers. The water in the deepest layer can reach temperatures of
between 90 and 100oC, temperatures as high as 107oC have been
reported. A 52 acre pond near the Dead sea in Israel can produce up to
5 mega watts of power.
Step #1 :
Breaking up the solute into individual components:
(expanding the Solute)
Step #2 :
Overcoming intermolecular forces in the solvent to make room
for the solute: (expanding the solvent)
Step #3 :
Allowing the solvent and solute to interact and form the solution.
Figure 17.2: (a) Enthalpy of solution Hsoln has a negative sign (the process is exothermic) if Step 3 releases more energy than is required by Steps 1 and 2. (b) Hsoln has a positive sign (the process is endothermic) if Steps 1 and 2 require more energy than is released in Step 3.
NaCl(s) Na+(g) + Cl –(g) Ho1 = 786 kJ/mol
H2O(l) + Na+(g) + Cl –(g) Na+(aq) + Cl –(aq)
Hohyd = Ho2 + Ho3 = _____________ kJ/mol
Hhyd = enthalpy (heat) of hydration
Hosoln = 786 kJ/mol – _______ kJ/mol = _______ kJ/mol
The dissolving process is positive, requiring energy. Then why is
NaCl so soluble? The answer is in the Gibbs free energy equation
from chapter 10, G = H – T S The entropy term – T S is
Negative, and the result is that G becomes negative, and as a
Result, NaCl dissolves very well in the polar solvent water.
Source: AP/Wide World Photos
Step 1: Solute separates into Particles - overcoming attractions
Step 2: Solvent separates into Particles - overcoming intermolecular
attractions Therefore -- Endothermic
solvent (aggregated) + heat solvent (separated)
Step 3: Solute and Solvent Particles mix - Particles attract each other
Therefore -- Exothermic
solute (separated) + solvent (separated) solution + heat
The Thermochemical Cycle
Hsolution = Hsolute + Hsolvent + Hmix
If HEndothermic Rxn< HExothermic Rxn solution becomes warmer
If HEndothermic Rxn> HExothermic Rxn solution becomes colder
A Fat – Soluble Vitamin A Water – Soluble Vitamin
A Hydrophobic Vitamin A Hydrophilic Vitamin
Figure 17.4: (a) a gaseous solute in equilibrium with a solution. (b) the piston is pushed in, which increases the pressure of the gas and the number of gas molecules per unit volume. (c) greater gas
P = kHX
P = Partial pressure
of dissolved gas
X = mole fraction
of dissolved gas
kH = Henry’s Law
Problem: The lowest level of oxygen gas dissolved in water that will
support life is ~ 1.3 x 10 - 4 mol/L. At the normal atmospheric pressure of
oxygen is there adaquate oxygen to support life?
Plan: We will use Henry’s law and the Henry’s law constant for oxygen
in water with the partial pressure of O2 in the air to calculate the amount.
The Henry’s law constant for oxygen in water is 1.3 x 10 -3 mol
and the partial pressure of oxygen gas in the atmosphere is 21%,
or 0.21 atm.
Soxygen = kH x PO2 = 1.3 x 10 -3 mol x ( 0.21 atm)
SOxygen = mol O2 / liter
This is adaquate to sustain life in water!
on Solubility - I
Problem: From the following information, predict whether the
solubility of each compound increases or decreases with an increase in
(a) CsOH Hsoln = -72 kJ/mol
(b) When CsI dissolves in water the water becomes cold
(c) KF(s) K+(aq) + F -(aq) + 17.7 kJ
Plan: We use the information to write a chemical reaction that includes
heat being absorbed (left) or released (right). If heat is on the left, a
temperature shifts to the right, so more solute dissolves. If heat is on
the right, a temperature increase shifts the system to the left, so less
(a) The negative H indicates that the reaction is exothermic, so when
one mole of Cesium Hydroxide dissolves 72 kJ of heat is released.
on Solubility - II
CsOH(s) Cs+(aq) + OH -(aq) + Heat
A higher temperature (more heat) decreases the solubility of CsOH.
(b) When CsI dissolves, the solution becomes cold, so heat is absorbed.
CsI(s) + Heat Cs+(aq) + I -(aq)
A higher temperature increases the solubility of CsI.
(c) When KF dissolves, heat is on the product side, and is given off
so the reaction is exothermic.
KF(s) K+(aq) + F -(aq) + 17.7 kJ
A higher temperature decreases the solubility of KF
Figure 17.6: The solubilities of several gases in water as a function of temperature at a constant pressure of 1 atm of gas above the solution.
Source: Visuals Unlimited
Problem: Calculate the vapor pressure lowering when 175g of sucrose
is dissolved into 350.00 ml of water at 750C. The vapor
pressure of pure water at 750C is 289.1 mm Hg, and it’s
density is 0.97489 g/ml.
Plan: Calculate the change in pressure from Raoult’s law using the
vapor pressure of pure water at 750C. We calculate the mole
fraction of sugar in solution using the molecular formula of
sucrose and density of water at 750C.
molar mass of sucrose ( C12H22O11) = 342.30 g/mol
= 0.51125 mol sucrose
350.00 ml H2O x 0.97489g H2O = 341.21g H2O
341.21 g H2O
= ______ molH2O
moles of water + moles of sucrose
0.51125 mole sucrose
18.935 mol H2O + 0.51125 mol sucrose
Xsurose = = 0.2629
P = Xsucrose x P 0H2O = 0.2629 x 289.1 mm Hg = ________ mm Hg
A solution was prepared by adding 40.0g of glycerol to 125.0g of water
at 25.0oC, a temperature at which pure water has a vapor pressure of
23.76 torr. The observed vapor pressure of the solution was found to be
22.36 torr. Calculate the molar mass of glycerol!
Roults Law can be rearranged to give:
XH2O = = = 0.9411 =
mol H2O = = 6.94 mol H2O
mol gly = = 0.4357 mol
mol gly + mol H2O
mol gly + 6.96 mol
6.94 mol – (6.94 mol)(0.9411)
(MMglycerol = 92.09 g/mol)
H O H
H H H H H H
H H H H H H Hexane
H H +
Acetone + Water
A solution is prepared by dissolving 62g of sucrose in 150.0g of water
the resulting solution was found to have a boiling point of 100.61oC.
Calculate the molecular mass of sucrose.
T = kbmsolute kb = 0.51
T = 100.61oC – 100.00 oC = 0.61oC
msolute = = = 1.20 mol/Kg
Msolute = mol solute = (0.150 kg)(1.2 mol/kg)
mol solute = 0.18 mol MM = = _________g/mol
(MMsucrose = 342.18 g/mol)
What mass of ethanol (C2H6O) must be added to 20.0 liters of water to
keep it from freezing at a temperature of -15.0oF?
oC = (oF – 32)5/9 = (-15 – 32)5/9 = -26.1oC
T = kf msolute msolute = = = 14.0 mol/kg
14.0 mol/kg(20 kg H2O) = 280 mol ethanol
Ethanol = 2x12.01 + 6x 1.008 + 1x16.0 = 46.07
280 mol ethanol (46.07 g ethanol/mol) = ___________ kg ethanol
Point Depression of an Aqueous Solution
Problem: We add 475g of sucrose (sugar) to 600g of water. What will
be the Freezing Point and Boiling Points of the resultant solution?
Plan: We find the molality of the sucrose solution by calculating the
moles of sucrose and dividing by the mass of water in kg. We then apply
the equations for FP depression and BP elevation using the constants
from table 12.4.
Solution: Sucrose C12H22O11 has a molar mass = 342.30 g/mol
= 1.388 mole sucrose
1.388 mole sucrose
0.600 kg H2O
molality = = 2.313 m
Tb = Kb x m = (2.313 m)= 1.180C BP = 100.000C + 1.180C
BP = 101.180C
Tf = Kf x m = (2.313 m) = 4.300C FP = 0.000C - 4.300C = -4.300C
Point Depression of a Non-Aqueous Solution
Problem: Calculate the effect on the Boiling Point and Freezing Point of
a chloroform solution if to 500.00g of chloroform (CHCl3) 257g of
napthalene (C10H8, mothballs) is dissolved.
Plan: We must first calculate the molality of the cholorform solution by
calculating moles of each material, then we can apply the FP and BP
change equations and the contants for chloroform.
Solution: napthalene = 128.16g/mol chloroform = 119.37g/mol
molesnap = =2.0053 mol nap
molarity = = = 4.01 m
Tb = Kb m = (4.01m) = 14.560C normal BP = 61.70C
new BP = ______0C
Tf = Kfm = (4.01m) =18.850C normal FP = - 63.50C
new FP = _______0C
Figure 17.16: A pure solvent and its solution (containing a nonvolatile solute) are separated by a semipermeable membrane through which solvent molecules (blue) can pass but solute molecules (green) cannot.
Calculate the osmotic pressure generated by a sugar solution made up of
5.00 lbs of sucrose per 5.00 pints of water.
5.00 lbs ( ) = 2.27 kg
Molar mass of sucrose = 342.3 g/mol 2,270g
5.00 pints H2O ( )( ) = 2.36 liters
P = MRT = ( )(0.08206 )(298 K) = __________ atm
= 6.63 mol sucrose
To determine the molar mass of a certain protein, 1.7 x 10-3g of the
protein was dissolved in enough water to make 1.00 ml of solution. The
osmotic pressure of this solution was determined to be 1.28 torr at 25oC.
Calculate the molar mass of the protein.
P = 1.28 torr( ) = 1.68 x 10-3 atm
T = 25 + 273 = 298 K
M = = 6.87 x 10-5 mol/L
1.70 g xg
6.87 x 10-5mol mol x = _______________g/mol
1.68 x 10-3 atm
0.08206 L atm(298 K)
Osmotic Pressure - I
Problem: A physician studying a type of hemoglobin formed during a
fatal disease dissolves 21.5 mg of the protein in water at 5.00C to make
1.5 ml of solution in order to measure its osmotic pressure. At
equilibrium, the solution has an osmotic pressure of 3.61 torr. What is
the molar mass(M) of the hemoglobin?
Plan: We know the osmotic pressure (),R, and T. We convert from
torr to atm and T from 0C to K and use the osmotic pressure equation to
solve for molarity (M). Then we calculate the moles of hemoglobin from
the known volume and use the known mass to find M.
P = 3.61 torr x = 0.00475 atm
Temp = 5.00C + 273.15 = 278.15 K
0.082 L atm (278.2 K)
M = = = 2.08 x 10 - 4 M
Finding moles of solute:
2.08 x 10 - 4 mol
n = M x V = x 0.00150 L soln = 3.12 x 10 - 7 mol
Calculating molar mass of Hemoglobin (after changing mg to g):
3.12 x 10-7 mol
M = = ________________ g/mol
What concentration of sodium chloride in water is needed to produce
an aqueous solution isotonic with blood (p = 7.70 atm at 25oC).
P = MRT M =
M = = 0.315 mol/L
Since sodium chloride gives two ions per molecule, the concentration
would be ½ that value, or 0.158 M
NaCl Na+ + Cl-
(0.08206 L atm) (298 K)
Source: Recovery Engineering, Inc.
Figure 17.20: Residents of Catalina Island off the coast of southern California are benefiting from a desalination plant that can supply 132,000 gallons of drinkable water per day, one-third of the island's daily needs.
From Raoult’s law, we know that:
Psolvent = Xsolvent x P0solvent and Psolute = Xsolute x P0solute
Let us look at a solution made up of equal molar quantities of acetone
and chloroform. Xacetone = XCHCl3 = 0.500, at 350C the vapor pressure of
pure acetone = 345 torr, and pure chloroform = 293 torr. What is vapor
pressure of the solution, and the vapor pressure of each component. What
are the mole fractions of each component?
Pacetone = Xacetone x P0acetone = 0.500 x 345 torr = 172.5 torr
PCHCl3 = XCHCl3 x P0CHCl3 = 0.500 x 293 torr = 146.5 torr
From Dalton’s law of partial pressures we know that XA =
172.5 + 146.5 torr
Xacetone = = = 0.541
Total Pressure = 319.0 torr
172.5 + 146.5 torr
XCHCl3 = = = 0.459
I ) Vapor Pressure Lowering - Raoult’s Law
II ) Boiling Point Elvation
III ) Freezing Point Depression
IV ) Osmotic Pressure
For ionic solutions we must take into account the number of ions present!
i = van’t Hoff factor = “ionic strength”, or the number of ions present
For vapor pressure lowering: P = i XsoluteP 0solvent
For boiling point elevation: Tb = i Kb m
For freezing point depression: Tf = i Kf m
For osmotic pressure: = i MRT
i = ______ This is less than the value expected of 4 so there must be
some ion paring occurring in the solution.
(0.10 )(0.08206 )(298 K)
Source: Stock Boston
A Cottrell Precipitator for Removing Particulates from Industrial Flue Gases