1 Chapter 14 Physical Properties of Solutions alloy Concentration Units Concentration Units Concentration Units Concentration Units Molarity (M) Molarity (M) Molarity (M) Molarity (M) = moles solute / Liters of solution Percent by Mass (weight) Percent by Mass (weight) Percent by Mass (weight) Percent by Mass (weight) Percent by Volume Percent by Volume Percent by Volume Percent by Volume Parts per Parts per Parts per Parts per million - (for dilute aqueous solutions 1ppm ≅ 1mg/L, since the density is close to that of water at 1.00g/mL) Parts per billion - (1ppb ≅ 1μg/L) Parts per trillion - (1ppt ≅ 1ng/L) These units can be in terms of numbers of particles or mass (usually mass for liquids, particles for gases) Example: 1. Show that for dilute solutions, 1ppm is approximately equal to 1mg/L Example: 1. Show that for dilute solutions, 1ppm is approximately equal to 1mg/L Result: At 1.00g/mL for water there are 1,000g/L In one milligram there is .001g .001g / 1,000g = 1x10 -6 = 1 / 1,000,000 = 1ppm Example: 2. What is the mass percent sucrose (table sugar) in the solution obtained by mixing 225g of an aqueous solution that is 6.25% sucrose by mass with 135g of an aqueous solution that is 8.20% sucrose by mass?
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
1
Chapter 14
Physical Properties of
Solutions alloy
Concentration UnitsConcentration UnitsConcentration UnitsConcentration Units
b. Molality = .49376mol CH3OH / .0786kg CH3CN = 6.28m
c. Molarity = .49376mol CH3OH / .120L solution = 4.11M
4
Physical Properties of
Solutions
Solution Energy
Energy of solution is
higher than before
Energy of solution is
equal to before
Energy of solution is
less than before
∆Hsoln = ∆ H1 + ∆ H2 + ∆ H3
Force of attraction
between solute
molecules (∆H2)
Force of attraction
between solute and
solvent molecules
(∆H3)
Force of attraction
between solvent
molecules (∆H1)
There are four possibilities in solution formation:
1. All intermolecular forces are of comparable strength in
which case ∆Hsoln = 0. (We also expect that ∆Vsoln = 0)
This type of solution is known as an ideal solutionideal solutionideal solutionideal solution.
2. Intermolecular forces between solute and solvent
molecules are stronger than the separate intermolecular
forces.
∆Hsoln < 0 exothermic
and ∆Vsoln < 0
This is a This is a This is a This is a nonidealnonidealnonidealnonideal solutionsolutionsolutionsolution
Image: The sum of two 50.0mL
volumes of water and ethyl
alcohol is less than 100.0mL
3. Intermolecular forces between solute and solvent
molecules are somewhat weaker than the separate
intermolecular forces.
∆Hsoln > 0 endothermic
A solution may or may not form depending on other
conditions (entropy) and is considered a nonideal solutionnonideal solutionnonideal solutionnonideal solution
4. Intermolecular forces between solute and solvent
molecules are much weaker than the separate
intermolecular forces.
∆Hsoln >> 0 endothermic
The enthalpy of the solution is so positive that a solution
will not form and instead creates a heterogeneous mixture.
Special NoteSpecial NoteSpecial NoteSpecial Note: As a general rule, remember like dissolves like, meaning that polar substances tend to dissolve in polar substances and nonpolar substances dissolve in
nonpolar substances.
Question: Identify the substances and arrange them in the expected
order of increasing solubility in water.
a b
cd
5
c (hexane) < b (hexanol) < d (butanoic acid) < a (ethanoic (acetic) acid)
Based on molecular length and polarity
a b
cd
Aqueous Solutions of Ionic Compounds
Ions dissolved in aqueous solution have a certain number of water molecules
surrounding them. They are said to be hydrated.
The solubility of an ionic compound is determined largely by the competition between interionic attractions that hold ions in a crystal and the ion-dipole
attractions that pull them into solution.
It is difficult to predict the solubility of some ionic compounds. Hydrated ion
Ion – Dipole
attraction
Heat of SolutionHeat of SolutionHeat of SolutionHeat of Solution
Heat related to the solvation process with respect to ionic
compounds in water
Energy must be supplied to separate the ions in the lattice
against their attractive forces (- ∆Hlattice)
Energy is evolved when the individual ions are transferred
into water (∆Hhydration)
Overall: ∆Hsoln = - ∆Hlattice + ∆Hhydration
General Rule: To be soluble, salts must have ∆Hsoln that is
exothermic or only slightly endothermic
(Enthalpy and entropy (chp 19) must be taken into account)
Note that the energy quantities are affected by ion sizes
and charges.
Question:
Given ∆H0f(s) = -425.9kJ/mol and ∆H
of(aq, 1m)
= -469.2kJ/mol for sodium hydroxide, calculate the enthalpy
Equilibrium in Solution FormationEquilibrium in Solution FormationEquilibrium in Solution FormationEquilibrium in Solution Formation
Liquids that mix in all proportions are said to be miscible.
Dynamic equilibriumDynamic equilibriumDynamic equilibriumDynamic equilibrium of solution concentration occurs when the
rate of crystallization rate of crystallization rate of crystallization rate of crystallization = rate of ions = rate of ions = rate of ions = rate of ions leaving the crystalleaving the crystalleaving the crystalleaving the crystal. The
solution is said to be saturatedsaturatedsaturatedsaturated.
95% of ionic compounds have aqueous solubilities that increase
with temperature. Some change very little and an even smaller
portion becomes less soluble with increased temperature.
Solubility is expressed on a solubility curve which plots solubility
in g solute per 100 g water, as a function of temperature. (p672)
6
Solubilities of some ionic substances (per 100g of water) at different temperatures
Carefully cooling a saturated solution can lead to
supersaturation, a condition in which more solute is dissolved in solution than is normally possible. The addition of a seed crystal
causes the oversaturated solute to quickly crystallize.
Sodium Acetate
Superstaurated solution with a “seed
crystal” dropped in causes immediate
crystallization
Fractional Crystallization
of aqueous potassium
nitrate solution containing
impure cupric sulfate
This method can be used
to purify the potassium
nitrate in greater quantity
The solubility of KNO3 and
CuSO4 are different.
The presence of these
ions will also have an
effect on the freezing
point of the solution.
Gas Solubility
Most gases become less soluble with temperature.
External pressure has very little effect on the solubility of solids and
liquids but increases the solubility of gases with increased pressure.
At constant temperature, the solubility (Sg) of a gas is directly
proportional to the partial pressure of the gas (Pg) in equilibrium
with the solution.
Henry's LawSg = kHPg
kH = constant (at a given temperature) that is dependent upon the gas and solvent
(in mg/g atm, M/mmHg, etc. (i.e. some concentration unit over pressure))
Henry’s Law Constants (25oC)
Gas kH (M/mmHg)
N2 8.42x10-7
O2 1.66x10-6
CO2 4.48x10-5
Effect of gas pressure on aqueous
solubilities of gases (at 20oC)
Henry's Law Sg = kHPg
The concentration of the
gas molecules both
above and in the
solution increases
The slope
of the line
is k
7
Effect of temperature on solubility of gasesEffect of temperature on solubility of gases
Oxygen only makes up about 23% of
air by mass but constitutes about 35%
of the air dissolved in water, due to
greater solubility
Question:
At 25oC and 1atm gas pressure, the solubility of CO2(g) is
149mg/100g water. When air at 25oC and 1 atm is in
equilibrium with water, what is the concentration of
dissolved CO2, in mg/100g water? Air contains 0.037
mole% CO2(g)
At 25oC and 1atm gas pressure, the solubility of CO2(g) is 149mg/100g water. When
air at 25oC and 1 atm is in equilibrium with water, what is the concentration of
dissolved CO2, in mg/100g water? Air contains 0.037 mole% CO2(g)?