Embed Size (px)
Citation preview
Properties of Solutions
Properties of Solutions
AP ChemistryChapter 13
JMS
AP ChemistryChapter 13
JMS
What is a solutionWhat is a solution
A solution is a homogeneous mixture of one or more pure substances (solutes) which disperse uniformly throughout another pure substance (solvent).
IMF’s determine the interactions between solute and solvent known as solvation.
When water is the solvent, solvation is usually referred to as hydration.
A solution is a homogeneous mixture of one or more pure substances (solutes) which disperse uniformly throughout another pure substance (solvent).
IMF’s determine the interactions between solute and solvent known as solvation.
When water is the solvent, solvation is usually referred to as hydration.
Hydration of NaClHydration of NaCl
In the Ion-dipole attractions, we see the positive hydrogen ends of the water attracting the Cl- ions, while the negative Oxygen ends of water attract the Na+ ion.
In the Ion-dipole attractions, we see the positive hydrogen ends of the water attracting the Cl- ions, while the negative Oxygen ends of water attract the Na+ ion.
CrystallizationCrystallization
The reverse process of solvation is crystallization, where solids come out of solution.
The reverse process of solvation is crystallization, where solids come out of solution.
Enthalpy and SolvationEnthalpy and Solvation
Sometimes addition of a solute releases heat, and sometimes it absorbs heat.
Sometimes addition of a solute releases heat, and sometimes it absorbs heat.
Enthalpy and NatureEnthalpy and Nature
Why does oxygen form diatomic molecules?
Nature favors more stable compounds; this means that lower enthalpy states are more likely to occur.
So nature prefers exothermic reactions.
Why does oxygen form diatomic molecules?
Nature favors more stable compounds; this means that lower enthalpy states are more likely to occur.
So nature prefers exothermic reactions.
Entropy and NatureEntropy and Nature
So why would water evaporate? (evaporation is endothermic).
Nature also favors higher entropy.Entropy is a measure of disorder or randomness.
By evaporating, the molecules of water (structured with hydrogen bonding) can become much more random as a gas.
So why would water evaporate? (evaporation is endothermic).
Nature also favors higher entropy.Entropy is a measure of disorder or randomness.
By evaporating, the molecules of water (structured with hydrogen bonding) can become much more random as a gas.
Entropy or ChaosEntropy or Chaos
In the solution process, solids become solute particles.
Is this an increase or decrease in entropy?
In the solution process, solids become solute particles.
Is this an increase or decrease in entropy?
SpontaneitySpontaneity
In chemistry, spontaneity means a process can occur without added energy.
It does not mean “instantaneous”.If I put a glass of water out at room temperature, will it evaporate on its own?
Yes eventually. This is a spontaneous process at room temperature.
If I put a lump of salt into water, will it dissolve spontaneously?
In chemistry, spontaneity means a process can occur without added energy.
It does not mean “instantaneous”.If I put a glass of water out at room temperature, will it evaporate on its own?
Yes eventually. This is a spontaneous process at room temperature.
If I put a lump of salt into water, will it dissolve spontaneously?
SpontaneitySpontaneity
Since nature prefers exothermic reactions and more random states:A reaction that decreases in enthalpy and increases in entropy is always spontaneous.
A reaction that increases in enthalpy and decreases in entropy is never spontaneous.
Any other reaction may be spontaneous, depending on the room temperature.
Since nature prefers exothermic reactions and more random states:A reaction that decreases in enthalpy and increases in entropy is always spontaneous.
A reaction that increases in enthalpy and decreases in entropy is never spontaneous.
Any other reaction may be spontaneous, depending on the room temperature.
Key TermsKey Terms
Concentration - the amount of solute in a solvent.
Solubility - the maximum amount of solute that can be dissolved in a given amount of solvent.
Concentration - the amount of solute in a solvent.
Solubility - the maximum amount of solute that can be dissolved in a given amount of solvent.
Unsaturated - a solution under its solubility.
Saturated - a solution at its solubility.
Supersaturated - a solution above its solubility.
Unsaturated - a solution under its solubility.
Saturated - a solution at its solubility.
Supersaturated - a solution above its solubility.
How can a solution be above its maximum?
How can a solution be above its maximum?
Solubility varies with temperature.
At 298 K, 91 g of salt dissolved.
At 323 K, over 100 g can dissolve.
If we dissolve 100 g at 298 K, then lower the temperature, the solution is now supersturated.
Solubility varies with temperature.
At 298 K, 91 g of salt dissolved.
At 323 K, over 100 g can dissolve.
If we dissolve 100 g at 298 K, then lower the temperature, the solution is now supersturated.
MiscibleMiscible
Liquids that mix in all proportions
Dry Gas is added to gasoline to eliminate any moisture in the line. The alcohol based substance is miscible in water and gasoline.
Liquids that mix in all proportions
Dry Gas is added to gasoline to eliminate any moisture in the line. The alcohol based substance is miscible in water and gasoline.
ImmiscibleImmiscible
Liquids that do not mix.
Oil and water are the common example of immiscible liquids.
Liquids that do not mix.
Oil and water are the common example of immiscible liquids.
Solubility vs Temperature
Solubility vs Temperature
At different temperatures, the solubility of a given solute/solvent varies.
For example, most solids are more soluble in water at higher temperatures.
At different temperatures, the solubility of a given solute/solvent varies.
For example, most solids are more soluble in water at higher temperatures.
Solubility of SolidsSolubility of Solids
In this case, the solubility is on the amount of solute in a saturated solution based on 100 g (or 100 ml) of water.
In this case, the solubility is on the amount of solute in a saturated solution based on 100 g (or 100 ml) of water.
Solubility of GasesSolubility of Gases
Generally, the solubility of gases decreases as the temperature increases.
Why?
Generally, the solubility of gases decreases as the temperature increases.
Why?
Calculating ConcentrationCalculating
ConcentrationMolarity(M)=moles of solute/liters of solution. The unit is Molars.
molality(m)=moles of solute/kilograms of solvent. The unit is molals.
Mole Fraction()=moles of solute/total moles of solution. There is no unit.
Mass percent (%)=mass of solute/total mass of solution. This is expressed as a percent.
Parts per Million (ppm) = (mass of solute/total mass of solution)*1,000,000. The units are ppm.
Molarity(M)=moles of solute/liters of solution. The unit is Molars.
molality(m)=moles of solute/kilograms of solvent. The unit is molals.
Mole Fraction()=moles of solute/total moles of solution. There is no unit.
Mass percent (%)=mass of solute/total mass of solution. This is expressed as a percent.
Parts per Million (ppm) = (mass of solute/total mass of solution)*1,000,000. The units are ppm.
Sample ProblemSample Problem
A solution is prepared from 117 g of NaCl in 900 g of water. The volume of the solution was 1.00 liter. Calculate the:
MolaritymolalityMole fractionMass percent
A solution is prepared from 117 g of NaCl in 900 g of water. The volume of the solution was 1.00 liter. Calculate the:
MolaritymolalityMole fractionMass percent
MolarityMolarity
Molarity = moles of solute/liter of solution.
117 g of NaCl/(58.5 g/mol) = 2.00 mol NaCl.
Molarity = 2.00 mol NaCl/ 1.00 liter = 2.00 M
Molarity = moles of solute/liter of solution.
117 g of NaCl/(58.5 g/mol) = 2.00 mol NaCl.
Molarity = 2.00 mol NaCl/ 1.00 liter = 2.00 M
molalitymolality
molality = moles of solute/kilograms of solvent
We already established that there are 2.00 moles NaCl.
900 g H2O = .900 kg H2Omolality = 2.00 moles NaCl / .900 kg H2O = 2.22 m
molality = moles of solute/kilograms of solvent
We already established that there are 2.00 moles NaCl.
900 g H2O = .900 kg H2Omolality = 2.00 moles NaCl / .900 kg H2O = 2.22 m
Mole FractionMole Fraction
= moles of solute/total number of moles
We already established that there are 2.00 moles NaCl.
900 g H2O / (18 g/mol) = 50 mol H2O
X = (2 moles NaCl)/(2 moles NaCl + 50 moles H2O) = .0385
= moles of solute/total number of moles
We already established that there are 2.00 moles NaCl.
900 g H2O / (18 g/mol) = 50 mol H2O
X = (2 moles NaCl)/(2 moles NaCl + 50 moles H2O) = .0385
Mass percentMass percent
Mass percent = mass of solute/total moles of solution
Mass percent = (117 g NaCl)/(117 g NaCl + 900 g H2O) = .115=11.5%
Mass percent = mass of solute/total moles of solution
Mass percent = (117 g NaCl)/(117 g NaCl + 900 g H2O) = .115=11.5%
Colligative PropertiesColligative Properties
Certain properties of solutions are independent on the type of solute molecules, but are instead dependent on the amount of solute particles. These are called colligative properties.
Certain properties of solutions are independent on the type of solute molecules, but are instead dependent on the amount of solute particles. These are called colligative properties.
Analyzing Colligative Properties
Analyzing Colligative Properties
Let’s say we put some sugar in water. Now we want to boil the water. For the water to boil, it has to leave the liquid phase and become random gaseous particles.
However, the sugar particles are attracted to multiple water molecules, holding the water molecules together with additional force.
The more sugar particles present, the more attractions to water, and the harder it will be to boil the water.
Let’s say we put some sugar in water. Now we want to boil the water. For the water to boil, it has to leave the liquid phase and become random gaseous particles.
However, the sugar particles are attracted to multiple water molecules, holding the water molecules together with additional force.
The more sugar particles present, the more attractions to water, and the harder it will be to boil the water.
So Why Doesn’t the Type of Solute Matter?
So Why Doesn’t the Type of Solute Matter?First of all we have to understand “like dissolves like”. This means that a polar solvent will best dissolve a polar solute and a nonpolar solvent will best dissolve a nonpolar solute.
Water is polar as is sugar (sucrose).
First of all we have to understand “like dissolves like”. This means that a polar solvent will best dissolve a polar solute and a nonpolar solvent will best dissolve a nonpolar solute.
Water is polar as is sugar (sucrose).
So?So?
Since both water and sugar are polar, they form a solution. This means there are already strong attractions between solute and solvent.
What is more important now is the number of sugar particles bonded to water molecules. The more sugar particles present, the harder it will be to boil.
Since both water and sugar are polar, they form a solution. This means there are already strong attractions between solute and solvent.
What is more important now is the number of sugar particles bonded to water molecules. The more sugar particles present, the harder it will be to boil.
Boiling Point Elevation
Boiling Point Elevation
Tb = kbim Tb is the change in boiling point.
kb is the boiling point constant for the solvent. Each solvent has its own constant. For water, kb = .51 Co/m.
i is the vant hoff factor.m is the molality.
Tb = kbim Tb is the change in boiling point.
kb is the boiling point constant for the solvent. Each solvent has its own constant. For water, kb = .51 Co/m.
i is the vant hoff factor.m is the molality.
Vant Hoff FactorVant Hoff Factor
We said that colligative properties depend on the number of solute particles.
Well, strong electrolytes tend to dissociate in water, thus giving more particles.
The vant hoff factor is a relative number indicating how many particles are present per molecule of solute.
We said that colligative properties depend on the number of solute particles.
Well, strong electrolytes tend to dissociate in water, thus giving more particles.
The vant hoff factor is a relative number indicating how many particles are present per molecule of solute.
For exampleFor example
For sucrose (a nonelectrolyte) there is no dissociation so there is 1 particle in every molecule. i =1.
For NaCl (a strong electrolyte), there is dissociation, so there are 2 particles in every molecule. i = 2.
For PbCl2 (a poor electrolyte), there is only slight dissociation, so the number of particles is harder to determine. 1<i<3
For sucrose (a nonelectrolyte) there is no dissociation so there is 1 particle in every molecule. i =1.
For NaCl (a strong electrolyte), there is dissociation, so there are 2 particles in every molecule. i = 2.
For PbCl2 (a poor electrolyte), there is only slight dissociation, so the number of particles is harder to determine. 1<i<3
Try ThisTry This
424 g of K3PO4 are dissolved in 1.00 liter of water. At what temperature will the water boil?
424 g of K3PO4 are dissolved in 1.00 liter of water. At what temperature will the water boil?
The processThe process
First find the molality.424 g K3PO4 / (212 g/mol) = 2.00 mol K3PO4.
1.00 liter H2O = 1.00 kg H2Om = 2.00 mol/1.00 kg = 2.00 m
Now solve: Tb = kbimTb = (.51 Co/m)(4)(2.00 m) = 4.08 Co
Since water normally boils at 100 oC, it would now boil at 104.08 oC.
First find the molality.424 g K3PO4 / (212 g/mol) = 2.00 mol K3PO4.
1.00 liter H2O = 1.00 kg H2Om = 2.00 mol/1.00 kg = 2.00 m
Now solve: Tb = kbimTb = (.51 Co/m)(4)(2.00 m) = 4.08 Co
Since water normally boils at 100 oC, it would now boil at 104.08 oC.
Freezing Point Depression
Freezing Point Depression
To freeze water, we have to force the liquid molecules to line up in orderly bonds. This is harder to do when solute particles hold the water in irregular patters.
Tf = kfimkf is the freezing point constant for the solvent. Each solvent has its own constant. For water, kf = 1.86 Co/m.
To freeze water, we have to force the liquid molecules to line up in orderly bonds. This is harder to do when solute particles hold the water in irregular patters.
Tf = kfimkf is the freezing point constant for the solvent. Each solvent has its own constant. For water, kf = 1.86 Co/m.
Now Try ThisNow Try This
424 g of K3PO4 are dissolved in 1.00 liter of water. At what temperature will the water freeze?
424 g of K3PO4 are dissolved in 1.00 liter of water. At what temperature will the water freeze?
The Same ProcessThe Same Process
First find the molality.424 g K3PO4 / (212 g/mol) = 2.00 mol K3PO4.1.00 liter H2O = 1.00 kg H2Om = 2.00 mol/1.00 kg = 2.00 m
Now solve: Tf = kfimTf = (1.86 Co/m)(4)(2.00 m) = 14.88 Co
Since water normally freezes at 0 oC, it would now freeze at -14.88 oC.
First find the molality.424 g K3PO4 / (212 g/mol) = 2.00 mol K3PO4.1.00 liter H2O = 1.00 kg H2Om = 2.00 mol/1.00 kg = 2.00 m
Now solve: Tf = kfimTf = (1.86 Co/m)(4)(2.00 m) = 14.88 Co
Since water normally freezes at 0 oC, it would now freeze at -14.88 oC.
Vapor Pressure Depression
Vapor Pressure Depression
Like we discussed with boiling, molecules have a harder time leaving the liquid phase when a solute is present.
The Boiling Point increased because more heat is needed to vaporize the liquid.
At any given temperature, there will be less gas molecules above the liquid, so the vapor pressure will have decreased.
Like we discussed with boiling, molecules have a harder time leaving the liquid phase when a solute is present.
The Boiling Point increased because more heat is needed to vaporize the liquid.
At any given temperature, there will be less gas molecules above the liquid, so the vapor pressure will have decreased.
Calculating Vapor Pressure DepressionCalculating Vapor
Pressure DepressionPA = XA*PA
o PA is the vapor pressure above the solution.
XA is the mole fraction of the solvent.
PAo is the vapor pressure of the
pure solvent at the same temperature.
PA = XA*PAo
PA is the vapor pressure above the solution.
XA is the mole fraction of the solvent.
PAo is the vapor pressure of the
pure solvent at the same temperature.
Osmotic Pressure Elevation
Osmotic Pressure Elevation
Osmosis is the net movement of solvent through a semipermeable membrane toward the solution with greater solute concentration.
= T*R*i*M is the osmotic pressure.M is the molarity of the solution.R is the ideal gas law costant (.0821 latm/molK)
T is the absolute temperature.
Osmosis is the net movement of solvent through a semipermeable membrane toward the solution with greater solute concentration.
= T*R*i*M is the osmotic pressure.M is the molarity of the solution.R is the ideal gas law costant (.0821 latm/molK)
T is the absolute temperature.
Last ProblemLast Problem
What is the osmotic pressure of a saline solution made from 117 g of NaCl in 1.00 liter of solution at 298 K?
What is the osmotic pressure of a saline solution made from 117 g of NaCl in 1.00 liter of solution at 298 K?
The answer is:The answer is:
= i*M*R*T117 g NaCl /(58.5 g/mol) = 2.00 mol
M = 2.00 mol/ 1.00 liter = 2.00 M
= (2)*(2.00 M)*(.0821 latm/molK)*(298 K)
= 97.9 atm
= i*M*R*T117 g NaCl /(58.5 g/mol) = 2.00 mol
M = 2.00 mol/ 1.00 liter = 2.00 M
= (2)*(2.00 M)*(.0821 latm/molK)*(298 K)
= 97.9 atm
Here’s Another ProblemHere’s Another Problem
What is the vapor pressure above a solution made from dissolving 360 g of glucose (C6H12O6) in 144 g of water (H2O) at 20 oC?
The vapor pressure of pure water at 20 oC is 17.5 Torr.
What is the vapor pressure above a solution made from dissolving 360 g of glucose (C6H12O6) in 144 g of water (H2O) at 20 oC?
The vapor pressure of pure water at 20 oC is 17.5 Torr.
This Process is Different
This Process is Different
First find the mole fraction of water360 g of glucose/(180 g/mol) = 2 mol144 g of water/(18 g/mol) = 8 molXA = 8 mol H2O/(8 mol H2O + 2 mol C6H12O6) = .800
PA = XA*PAo = (.800)(17.5 Torr)
PA = 14.0 Torr
First find the mole fraction of water360 g of glucose/(180 g/mol) = 2 mol144 g of water/(18 g/mol) = 8 molXA = 8 mol H2O/(8 mol H2O + 2 mol C6H12O6) = .800
PA = XA*PAo = (.800)(17.5 Torr)
PA = 14.0 Torr
ColloidsColloids
Although they often appear to be solutions, there are heterogeneous mixtures called colloids in which larger particles appear to dissolve.
Although they often appear to be solutions, there are heterogeneous mixtures called colloids in which larger particles appear to dissolve.
Colloidal SuspensionColloidal Suspension
Yeah I know it’s a lava lamp, but its also a colloidal suspension.
A colloidal suspension is a colloid in which the particles are visible to the unaided eye.
Yeah I know it’s a lava lamp, but its also a colloidal suspension.
A colloidal suspension is a colloid in which the particles are visible to the unaided eye.
The Tyndall EffectThe Tyndall Effect
Colloids can appear similar to a solution until we shine a beam of light in them.
In the solution, the beam of light is coherent.
In the colloid, the beam of light bounces off the particles, scattering the light.
Colloids can appear similar to a solution until we shine a beam of light in them.
In the solution, the beam of light is coherent.
In the colloid, the beam of light bounces off the particles, scattering the light.
There are many types of Colloids
There are many types of Colloids
Hydrophobic ColloidsHydrophobic ColloidsHydrophobic colloids do not mix well with water; they form clusters within the water and hang there (thus the term suspension).
They are usually particles that are charged or become charged in water and use their own localized attractions to avoid the water.
Hydrophobic colloids do not mix well with water; they form clusters within the water and hang there (thus the term suspension).
They are usually particles that are charged or become charged in water and use their own localized attractions to avoid the water.
Hydrophilic ColloidsHydrophilic Colloids
These “water loving” colloids disperse well with water, generally due to the similar IMF attractions.
These “water loving” colloids disperse well with water, generally due to the similar IMF attractions.
Associative (Amphipathic) Colloids
Associative (Amphipathic) Colloids
These chemicals are usually “designed” to have a hydrophobic end and a hydrophillic end. One end allows it to attract to nonpolar substance, the other end attracts polar substances, like water.
Sometimes these colloids are called surfactants.
These chemicals are usually “designed” to have a hydrophobic end and a hydrophillic end. One end allows it to attract to nonpolar substance, the other end attracts polar substances, like water.
Sometimes these colloids are called surfactants.
