Download Unit 6: Solution Chemistry Content Outline: Basic Solution Chemistry

Unit 6: Solution Chemistry
Content Outline: Basic Solution Chemistry (6.1)
I.
Solution
A. Term referring to a homogenous (same) mixture of two or more substances in a single phase of
matter. ( one substances dissolves in the other)
1. Each sample of a homogenous solution would possess the exact same chemical composition and
same chemical properties…hence homogenous (same).
B. Most solutions tend to be of the liquid state, but they can also be mixtures of gases (such as O2 and
N2 in the atmosphere) or solids such as 14 K gold (gold, silver and Copper blended together while in
the molten state…remember the previous unit.)
II. Solvent
A. This term refers to the substance or medium (liquid base) that will “perform the dissolving”
(separating of atoms/molecules) of the other substance.
For example: water – the universal solvent as it can dissolve most substances.
1. The solvent for a solution is usually water, but it can be other substances such as ethanol (an
alcohol) or a gas such as Nitrogen in the atmosphere.
III. Soluble
A. Refers to a substance that is capable of being dissolved (separated apart at the atomic/molecular
level) into another substance.
For example: salt or sugar in water OR water in ethanol.
IV. Solute
A. This is the term referring to the substance that is “being dissolved” (separated).
B. Terms referring to the amount of solute in a solution.
1. Unsaturated
a. The solution contains less solute than the maximum amount at that temperature.
b. Basically, you can add more to the solution.
2. Saturated
a. The solution contains the maximum amount of solute at that temperature.
b. Basically, you cannot add more, unless you change the temperature.
For example: you add sugar to cold unsweet tea, but you can see the sugar sitting in the
bottom of the glass.
c. Solution equilibriumi.
This is the state of a solution when dissolving rate = re-crystallization rate.
3. Supersaturated
a. A solution that contains more than the maximum amount of solute at that temperature.
b. This is achieved by heating a solution, so that more can be dissolved into it. Then it is
allowed to cool.
i.
The heating breaks the intermolecular bonds (in water…think hydrogen bonds
breaking) so that there are more free solvent molecules to grab and hold separated
solute molecules.
For example: To make sweet tea, you first brew (hot) tea. Then add in the sugar. Stir. And
then cool down and add ice to enjoy on a hot afternoon.
ii.
The solute tends to stay suspended in the solvent as long as the substance is not
disturbed (shaken).
iii.
Once shaken the solute recrystallizes and “falls out” of the solution.
This is how rock candy pops (sugar on a stick) are made. You heat up water. Add sugar until
you can not dissolve anymore. Let it cool in an undisturbed area. Then come back in about
10 minutes and stick a clean wooden stick in the container and wait. Also, honey is a super
saturated solution. If you put honey in the refrigerator it will crystallize.
V. Solubility
A. The amount of a substance required to form a saturated solution with a specific amount of solvent
at a specific temperature.
B. These values are usually represented in a Solubility Table showing various temperatures.
1. Warmer temperatures usually hold more solute.
a. More “free solvent molecules” to grab solute molecules because of broken intermolecular
bonds.
2. Colder temperatures usually hold less solute.
a. Less “free solvent molecules” to grab solute because they are bound inter-molecularly.
For example, cold tea is much harder to sweeten than hot tea.
3. Gases also require a specified pressure…remember that temperature directly affects pressure.
For example, cold water holds more dissolved Oxygen gas than warm water. The amount of
dissolved Oxygen gas determines the types of life forms that can live in that environment. Also,
a cold Coke has less fizz when opened that a hot Coke.
VI. Factors that affect the rate of dissolving:
A. Increasing the Surface Area of a solid by “crushing it”.
Take two ice cubes and two glasses of warm water (same temperature remember). Put one in a
paper towel. Wrap the cube so it doesn’t go all over your kitchen. Crush it with a hammer or mallet.
Now put the crushed in one cup and the intact cube in the other. Which one melted first?
1. “Frees up” solute molecules to have greater interaction with the solvent molecules more
collisions.
B. Agitation (stirring the solution)
Repeat the above, but do not crush one cube. Put one cube in each glass of warm water. Stir the
water of one glass with a spoon. Which melted first?
1. Stirring the solvent increases the Kinetic Energy of the solvent molecules…more collisions
quicker.
C. Heating the solvent
1. Heating causes the intermolecular bonds of the solvent to break and “free up” the molecules to
grab solute.
2. In other words, there are more collisions between solvent molecules and solute molecules.
I.
The 6 Solutions compositions by “states” of matter:
A. Gas into another Gas (such as air… Oxygen gas (O2) dissolved in Nitrogen gas (N2)
B. Gas into a Liquid (such Oxygen or CO2 into blood or CO2 into a soda)
1. Henry’s Law
a. The British Chemist, William Henry in 1803, proposed this law.
b. The law states, “The solubility of an unreactive gas dissolved in a liquid solvent is directly
proportional to the partial pressure of that gas on the surface of that liquid at a constant
temperature.”
Basically it states: If pressure goes up so will the dissolved amount of gas at a constant
temperature.
c. Equilibrium (equal rates)
i.
Amount/rate of a gas entering the solution will equal the amount/rate of gas leaving
the solution. Both processes are occurring at equal rates… hence the term
equilibrium.
ii.
If pressure on the gas increases, then the amount/rate of dissolving into the solution
will increase.
iii.
If the amount/rate of dissolving into the solution increases, the amount/rate of gas
leaving the solution increases.
iv.
Until a new equilibrium is established between the gas and liquid at the surface.
d. Effervescence (“fizzing”)
i.
This is the rapid release of a dissolved gas from a solvent liquid.
For example: the fizzing sound and small gas bubbles leaving your soft drink when
you open it. Keeping the lid on helps your soda not go flat as the gas cannot escape
into the air. The gas is CO2.
ii.
Adding Kinetic energy in the form of heat or shaking will increase the effervescence
(by making the gas less soluble in the liquid).
Gas molecules moving faster (greater Kinetic energy), makes it easier for them to
escape the liquid’s attractive forces.
C. Liquid into another Liquid (Such as an alcohol into water)
1. Miscible
a. This term refers to a liquid being freely dissolvable, in any proportion, to another liquid.
2. Immiscible
a. This term refers to a liquid that will not dissolve into another liquid.
Like dissolves like… not different.
Polar liquids dissolve other polar liquids.
Non-polar liquids dissolve other non-polar liquids.
Polar liquids do not dissolve non-polar…for example water and oil.
How do you clean greasy dishes and pan with water? You use colored highly concentrated salt
water (dish washing soap). The salt ions act as a “connection” between the two substances.
D. Solid into a Liquid (Such sugar or salt into water)
1. Homogenous (Pure)
a. Equal distribution of the solid throughout giving it the same chemical properties and
composition.
b. To separate the two would require heating (evaporating) one out, like the water.
2. Suspension
a. This term refers to a heterogeneous solution with particles that are too big so that the liquid
cannot suspend them. Gravity “drags” them to the bottom of the container.
b. Agitating (“stirring”) can resuspend them for a short period of time.
Why you want to shake or stir Italian dressing before you put it on your salad? So you get
the good spices and not just oil.
c. Can be separated by filtering.
3. Colloid
a. A heterogeneous solution with particle of intermediate size in comparison to Pure solutions
and Suspensions.
b. They will appear to be homogenous (dispersed phase), but over time gravity will separate
the particles into differing layers based upon molecular density.
c. Milk is a good example. If left undisturbed, the proteins and sugars will separate into
different layers. Why you should always shake your milk before pouring or drinking in the
carton?
d. Colloids can display what is called the Tyndall Effect
 John Tyndall, an Irish Physicist, proposed in 1850.
i.
The scattering of light passing through because of the moving suspended particles.
It appears “hazy” instead of clear like water.
ii.
This effect is used to tell the difference between a true solution and a colloid.
iii.
The motion of the rapidly moving particles is called Brownian Motion.
 Robert Brown, British Botanist proposed in 1827.
e. Can be separated by complex filtering or by putting it in a centrifuge (“spinner”).
4. Electrolytes & Non-Electrolytes
a. Electrolytes – Solid dissolves into ions. The ions allow electricity (flowing electrons) to be
conducted through the solution. For example: salt in water.
b. Non-electrolytes – Solid does not create ions. So there is no ability to conduct electricity.
For example: sugar in water.
c. Hydration Shell – the surrounding of ions by water molecules using Hydrogen Bonds in
between them.
d. Hydrates - dried Ionic crystals that contain water.
α. These are written chemically like the following: CuSO4* 5 H2O – called Copper (II) Sulfate
Pentahydrate.
E. Liquid into a solid (not common but an example is the material used to fill teeth cavities, called
Dental Amalgam)
F. Solid into a solid
1. Must be in the molten (super-cooled) phase so as to mix uniformly. These “blended” metals are
called alloys.
a. They have a mixture of chemical properties of the composing metals.
b. Can be separated by heating to the different melting points of each metal.
Content Outline: Calculating Enthalpy of Solutions (6.3)
I.
Enthalpy (H)
A. This is the total amount of Energy within a system.
1. This energy is found in the intermolecular bonds between atoms/molecules.
2. Also found within the molecular motion (Kinetic Energy) …also known as heat (Temperature).
3. The change in Enthalpy (due to a reaction or phase change) is represented as: ΔH.
II. Enthalpy of Solution (Energy change of the “mixing”)
A. This term refers to the net outcome of Energy absorbed or released, as heat, by the solution when a
specific amount of solute is dissolved in a solvent, at constant Pressure.
B. Exothermic or Spontaneous
1. Energy is released from the system to the surroundings.
2. These are expressed as – ΔHSOL.
C. Endothermic or Non-spontaneous
1. Energy is absorbed by the system from the surroundings.
2. These are expressed as + ΔHSOL.
III. Solvated
A. This term refers to the separating and surrounding of a solute atom/molecule by solvent molecules
making a solvent shell.
B. If water is the solvent, you called it a Hydration Shell.
C. This separating and surrounding keeps the solute “dissolved”.
IV. Calculating ΔH for solutions:
A. These problems must be thought in terms of 3 different processes:
1. Breaking Solute –Solute intermolecular attractions within the solute. (E required…Endo)
2. Breaking Solvent-Solvent intermolecular attractions within the solvent. (E required…Endo)
3. Mixing Solvent –Solute intermolecular attractions take place between the two components.
(E released…Exo)
4. Each of these will each have its own change in enthalpy (ΔH).
ΔHSOL = ΔHSOLUTE + ΔHSOLVENT + ΔHMIX
V. Enthalpy and Dissolving
A. Exothermic reactions dissolve readily.
B. Endothermic reactions may dissolve. If too much energy is needed, no dissolving will occur.
C. You must remember, that “like dissolves like” (Outline 6.2)
For example, the strong Ionic attractive forces, of a solute, cannot be over come by weak attractive
forces of a non-polar solvent, so it will not dissolve…without added energy.
Basically, the greater the release; (Exothermic) the more readily the dissolving is accomplished.
VI. Solution versus a chemical Reaction
A. In a solution, the solute can be recovered in a filtering/dehydration process.
B. In a chemical reaction, a new product is formed and the solute cannot be recovered.
C. This is an extremely important difference you must remember.
Content Outline: Making Solutions of Varying Concentrations (6.4)
I.
Concentration
A. A measurement of the amount of solute in a specified amount of solvent or solution.
B. This can be expressed in two different ways:
1. Molarity (M) (Think volume)
a. The number of moles in 1 L of solvent/solution.
b. Molarity (moles/L) = Amount of solute (mol)
Volume of Solvent (L)
2. Molality (m) (Think mass)
a. The number of moles per kg of solvent.
b. Molality (moles/kg) = Amount of solute (mol)
Mass of solvent (kg)
c. Molality is mainly used when studying properties of solutions related to vapor pressure
(gas) and temperature changes.
i. Remember, that temperature changes can affect volume (Charles’ Law) but not mass.
II. Dilution/Dilute
A. This is the process of adding water to a stock solution to lower the concentration of a solution.
B. Stock Solution
1. Solutions of high concentration, used often in labs, that are prepared or purchased from
Industrial Chemical Supply companies to save money. These solutions are then diluted to
desired amount.
C. Moles before Dilution = Moles after Dilution
1. Since matter is neither created nor destroyed… just transformed and transferred, the number of
mole did not change, BUT the volume of solvent changed and thereby gave a new concentration.
This is a very important concept to remember.
D. Calculating Dilutions:
M1V1 = M2V2
(M1 and V1 are the values before dilution.)
(M2 and V2 are the desired values after dilution.)
Step 1: Enter the given information into the formula above…making sure your units match!
 If they do not match, you will need to convert one so that they do match.
Step 2: Isolate the missing Unit, by dividing on both sides with the given for that side.
Step 3: Calculate by multiplying the top and then dividing using the bottom value.
 If needed, convert the answer to the desire unit.
III. Making Molar solutions
A. Step 1: Calculate the Molar Mass (MM) of the solute.
Step 2: Convert the Moles/Molar Mass desired to grams by multiplying.
Remember, Unit Given (mol) X Unit Wanted (grams) = Unit wanted (grams)
Unit given (mol)
Step 3: Weigh out the desired number of grams on a scale.
Step 4: Dump the grams of solute into a volumetric flask, using a funnel.
Step 5; Add some of the solvent to initially dissolve the solute. Be sure to rinse the funnel as you
pour the solvent in to prevent loss of solute.
Step 6: Stopper the flask and swirl.
Step 7: Fill to the desired volume.
Step 8: Re-stopper…swirl (holding the stopper on the flask) and invert 10 times (still holding the
stopper).
Step 9: Ready to use.
IV. Making Molality solutions
A. Step 1: Calculate the Molar Mass (MM) of the solute.
Step 2: Convert the Moles/Molar Mass desired to grams by multiplying.
Remember, Unit Given (mol) X Unit Wanted (grams) = Unit wanted (grams)
Unit given (mol)
Step 3: Weigh out the desired number of grams on a scale.
Step 4: Weigh out the desired mass of solvent in kg.
 If water, 1 mL = 1mg… so 1kg of water = 1Liter.
Step 5: Combine in a flask or beaker and stir.
Step 6: Ready to use.
Content Outline: Ionic Solutions (6.5)
I.
Ion
A. An Ion is a charged particle that is the end result of an ionic or molecular (covalent bonds)
compound being broken apart and an electron transfer taking place.
B. The electron transfer creates the charges on the particles.
C. Two types of ions exist:
1. Anion
a. These particles carry a negative charge. (Think angry…negative thoughts.)
b. The negative charge is from the extra electron(s), which outnumber the protons.
c. This particle was reduced by reduction. (“gaining” an electron… Think “RIG”)
2. Cations
a. These particles carry a positive charge. (Think cats… cats make some people
happy/positive.)
b. The positive charge is from having more protons than electrons.
c. This particle was oxidized by oxidation. (“loss” of electrons… Think “OIL”)
3. Law of Conservation of Matter
a. Matter is neither created nor destroyed; just transferred and transformed.
In this case it is the electron that was transferred between atoms and that resulted in a
charge transformation of those particles.
b. They always go together, so think “OIL RIG” for a Redox reaction.
II. Dissociation (Sounds like “dis- associate.)
A. The dissolving of an Ionic compound, such as NaCl, into water causing the ionic bonds to break and
release “free” ions.
For example salt in water (the water is understood by the term aqueous (aq).
NaCl  Na+(aq) + Cl-(aq)
1. Aqueous (aq)
a. Substance dissolve in water.
B. Hydration Shell
1. These form around the “free ions” as they are separated
2. They form because the water is polar and attracted to the charged ions.
a. Hydrogen Bonds form as a result of the attraction.
3. Keeps the ionic particles separated and dissolved.
C. Not all ionic compounds dissociate into ions.
1. Solubility Rules for Ionic Compounds: You must memorize these, as they are important!
a. Sodium (Na), Potassium (K), and Ammonium (NH4) compounds are soluble in water.
Think “Yes on SPA in H2O”.
b. Most Carbonates, Phosphates, and Silicates are insoluble, EXCEPT those containing Sodium
(Na), Potassium (K), and Ammonium (NH4).
Think “No on CPSates, except with SPA.”
c. Most Chlorides are soluble in hot water, EXCEPT those of Silver (Ag), Mercury I (Hg), and
Lead II (Pb).
Think “Yes on Hot Water Cides, except SMiLii (sounds like …“smily”)”
d. Most Sulfides are insoluble, EXCEPT those with Calcium (Ca), Strontium (Sr), Sodium (Na),
Potassium (K), and Ammonium (NH4).
Think “No on Sides, except CSSPA.”
e. Most Sulfates are soluble, EXCEPT those with Barium (Ba), Strontium (Sr), Lead (Pb),
Calcium (Ca), and Mercury (Hg).
Think “Yes on Sates, except BLSCaHG (Sounds like..“Blues Catalog”)
f. Nitrates (NO3), Acetates (CH3COOH), and Chlorates (CO3) are soluble.
Think “Yes on NACates.
2. These rules can be useful in predicting the outcome of mixing two different solutions containing
soluble compounds.
a. If the “mixing” results in a combinations of ions that forms an insoluble compound, then a
precipitate will form as a result of the Double Replacement Reaction.
i.
Precipitates are insoluble solids present in a solution.
ii.
The precipitate occurs when the inter-molecular attractions between the ions is
greater than the inter-molecular attractions between the ions and surrounding
water molecules.
III. Ionization (The “making” of ions.)
A. Ions are “formed” from a molecular compound (covalent bonds) solute by the action of the polar
solvent separating the component atoms/molecules.
Basically; “Creating ions where there were none originally.”
B. They are Hydrated Ions, as are all ions in aqueous solutions.
C. The energy released, as heat (exothermic), during the hydration (bonds forming between ions and
water) supplies the energy to break the covalent bonds of the molecular compound (solute).
1. The covalent bonding strength is inversely proportional to the amount of ionization that will
occur.
2. Stronger Covalent bonds = less ionization will occur.
3. Weaker Covalent bonds = more ionization will occur.
For example, (weak covalent bonds) HCl  H+(aq) + Cl-(aq)… when in water.
(strong covalent bonds) C6H12O6 stays intact in water.
IV. Hydronium Ion (H3O+)
A. This ion is “created” when a “free” H+ (proton) attaches to a water molecule (H2O) by a covalent
bond.
1. The H+ comes from a molecular compound containing a Hydrogen atom(s), such as an Acid like
HCl.
2. These H+ have very strong attractive forces, and so are rarely found alone in solution.
B. The energy released (Exothermic) by “creating” the Hydronium Ion powers the breaking of the
covalent bond associated with the Hydrogen atom from the molecule.
Content Outline: Net Ionic Equations & Electrolytes (6.6)
I.
Electrolytes
A. A substance that yields (produces) ions when it dissolves and dissociates in a solvent and thus
produces a solution that conducts electrical current (transferring a charge) through the solution.
For example: Hydrogen Halides (Group 17 on Periodic Table) HF, HCl, HBr, HI. You should
recognize these as “acids” by your naming rules.
II. Non-electrolytes
A. A substance that does not yield (produce) ions when dissolved (does not dissociate) in a solvent
and therefore cannot conduct electrical current.
For example: sugars, such as Sucrose (Table Sugar) in water.
III. Strengths of electrolytes
A. This concept of “strength” is directly proportional to the ability of a solute to form ions (percent
dissociation) when dissolved.
1. Strong Electrolytes - 100% dissociation –all of the substance is dissociated in to ions 
Strong Conductors
2. Weak electrolytes  <100% dissociation – not all of substance is dissociated in to ions 
Weak conductors
B. Strong Electrolytes (Good Conductors of electricity…electron flow)
1. Any compound that has close to 100% ion formation in the aqueous state. Strong Acids, Strong
Bases and Soluble Salts.
For example: HCl, HBr, HI (acids), NaOH, KOH (bases) or NaCl, KCl, AgNO3 (salts)
C. Weak Electrolytes (Poor Conductors of electricity…electron flow)
1. Any compound that has little ion formation in the aqueous state. Weak Acids, Weak Bases, and
Insoluble Salts
For example: HF (this acid has a very strong bond) or Acetic Acid. Also sugars are weak
electrolytes.
D. Non-Electrolytes
1. Any compound that has no dissociation and does not conduct electric current.
E. Please do not confuse the two above terms with the terms dilute and concentrated.
1. Strong and Weak – refer to the degree of dissociation OR ionization.
2. Dilute and Concentrated – refer to the amount of solute dissolved in the solvent (Molarity or
%Concentration)
IV. Writing Net Ionic EquationsA. These equations are different than chemical reaction equations.
1. These are just focused on those Ions or compounds in the reaction that undergo a chemical
change.
B. All soluble compounds (Strong Electrolytes) are shown as being dissociated or in ion form and in the
aqueous state (aq).
Hope you remember your solubility rules.
1. If a precipitate is formed, it is shown as a solid(s).
2. Spectator Ions
a. These are ions found on both sides of the equation arrow and are not involved in the actual
chemical reaction that is occurring.
It is like you are watching a Pro Football Game at the stadium. You are watching (spectating)
the game. You are not actually involved in playing the game (reaction).
C. Writing
Step 1: Make sure your original chemical reaction is balanced.
Step 2: Separate all soluble compounds into the component ions/molecules for reactants & products.
Step 3: Label any precipitates as solid (s). Know those solubility rules!
Step 4: Identify and remove any spectator ions from the equation. They must be the same on both
sides of the equation.
Step 5: Rewrite all that remains in a new equation that only shows those items that underwent
chemical change.
These steps will be very important later in electrochemistry.
For example: CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + 2NaCl(aq)
Step 1: Balanced reaction…yes.
Step 2: Ca2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + CO32-(aq)  CaCO3(s) + 2 Na+(aq) + 2 Cl-(aq)
Step 3: CaCO3(s) is the precipitate.
Step 4: Ca2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + CO32-(aq)  CaCO3(s) + 2 Na+(aq) + 2 Cl-(aq) Spectator ions
Step 5: Ca2+(aq) + CO32-(aq)  CaCO3(s) . This is the net ionic equation.
Content Outline: Colligative Properties: Vapor Pressure Lowering (6.7)
I.
Colligative Properties
A. These are the properties of solutions that are dependent upon the concentration of solute, but not on
the solution identity (name).
1. Concentration for colligative properties is molality (m). Think mass.
a. Molality is the number of moles of solute/kg (mass) of solvent.
II. Volatile Substance
A. This refers to a substance that easily evaporates from a liquid to a gas.
This is a term from Unit 5. The example we used was liquid Bromine (Br2). (A member of “Hydrogen
and the Diatomic 7”)
III. Non-volatile Substance
A. A substance that has little to no tendency to evaporate to a gas under existing conditions.
B. A Phase diagram is helpful in understanding the pressure and temperature relationship for “states”
of matter and the conditions for equilibrium between the gas and liquid “states”.
IV. Vapor Pressure Lowering
A. This is the outcome when a non-volatile, non-electrolyte solute is dissolved in a solvent. The solute
will cause the vapor pressure (solvent as a gas above the liquid solution) to be lower than it would
be for the pure solvent.
In other words, the solvent is “taking” the needed energy trying to dissolve the solute to become a
gas. Less are “free” to evaporate away as a gas at that temperature.
1. Vapor pressure is the pressure created on top of a liquid’s surface as molecules are in
equilibrium (same rate) going between liquid  gas and also gas  liquid.
2. The solute being added to the pure solvent lowers the vapor pressure at that temperature
because the “free” solvent molecules are now “busy” trying to dissolve, making solvent shells,
the solute and therefore cannot evaporate away.
a. Equal molality values lower vapor pressures equally.
Basically, if any solute added has the same molality, it will cause the same amount of
lowering because the value for the pure solvent never changes.
3. A change in the vapor pressure will change the boiling points and freezing points of the
solution.
1. This change always affects the liquid phase range. It almost always increases the range.
2. Boiling Point and Freezing Point are also Colligative Properties.
V. Calculating the Vapor Pressure Lowering
A. Raoult’s Law
1. The French Physicist, Francois Marie Raoult, proposes this Law in 1882.
2. The Law states “The partial pressure exerted by solvent vapor above the solution, equals the
product (multiplication) of the mole fraction of the solvent times the vapor pressure of the pure
solvent.”
Basically, (moles of pure solvent/moles of solution) X (Pressure constant of pure solvent).
3. Mathematically expressed as:
ΔPSOLUTE = (XSOLUTION) X (POSOLVENT)
P = pressure
X = mole fraction (see blue above)
For example: Glycerin (C3H8O3), a non-volatile substance, has a density of 1.26 g/mol at 25.0OC.
Calculate the vapor pressure lowering at 25.0OC of a solution made from adding 50 mL
of Glycerin to a 500.0 mL sample of pure water (H2O). Pure water has a vapor
pressure constant of 23.8 torr at 25.0OC and a density of 1.00 g/mol.
Remember: Unit Given X Unit wanted = Unit wanted
Unit Given
Step 1: We need moles so calculate the Molar mass of solute and solvent.
Molar Mass of Glycerin: C = 12.0 AMUs X 3 = 36.0 AMUs
H = 1.0 AMUs X 8 = 8.0 AMUs
O = 16.0 AMUs X 3 = 48.0 AMUs
Total = 92.0 AMUs
Molar Mass of Water: H = 1.0 AMUs X 2 = 2.0 AMUs
O = 16.0 AMUs X 1 = 16.0 AMUs
Total = 18.0 AMUs
Step 2: Using the Molar Mass, find the number of moles present for each…solvent and solute.
Assume at this point, 1 mL = 1g
Moles of Glycerin:
50.0 (g) C3H8O3 X 1 mol C3H8O3 = 0.543 mol of C3H8O3
92.0 g C3H8O3
Moles of Water:
500.0 (g) H2O X 1 mol H2O = 27.8 mol of H2O
18.0 g H2O
Step 3: As we need molality (mol/kg), we need to multiply each mol value by the density for each
substance. This will correct our assumption in step 2 as 1 ml of each substance has
different mass in reality.
Molality of Glycerin: (0.543 mol of C3H8O3) X (1.26 g/mol) = 0.684 mol
Molality of Water: (27.8 mol of H2) X (1.00 g/mol) = 27.8 mol
Step 4: Determine your mole fraction for the Solution.
Mole Fraction = Mol of Pure Solvent =
Mol of Solution
27.8 mol
(27.8 mol of H2O + 0.684 mol of C3H8O3)
= 27.8 = 0.975
28.5
Step 5: Multiply the Mole Fraction by the Vapor pressure constant for water (solvent) at 25.0OC.
0.975 X 23.8 torr = 23.2 torr
Step 6: To find the amount of vapor pressure decrease, subtract the final from the original.
23.8 torr - 23.2 torr = 0.6 torr decrease due to the adding of Glycerin to pure water.
Content Outline: Colligative Properties: Freezing Point Depression (6.8)
VI. Colligative Properties
B. These are the properties of solutions that are dependent upon the concentration of solute, but not on
the solution identity (name).
2. Concentration for colligative properties is molality (m). Think mass.
b. Molality is the number of moles of solute/kg (mass) of solvent.
VII. Freezing Point Depression
A. The difference between the Freezing points of the pure solvent and a solution containing a nonvolatile, non-electrolyte solute in the same solvent.
B. Freezing Point Depression is directly proportional to the molality concentration of the solution.
1. If Molality increases the amount of Depression increases (the temperature drops).
C. Mathematically expressed as:
Δtf = Kfm
Kf = Freezing Depression Constant for a pure solvent (OC/molality)
M = Molality of the solution (moles of solute/ kg solvent)
tf = Numeric Amount of Temperature change (OC)
VIII.
Molal Freezing Point Constant (Kf)
A. The Freezing Point Depression of the solvent when a 1-molal solution is created by dissolving 1 mole
of non-volatile, non-electrolyte solute in 1 kg of a solvent at 1 atm.
For example: The normal Freezing Point of Water is 0OC. The Freezing Point Constant for water is
-1.86 OC. So if a 1 molal solution is dissolved in water, the new Freezing Point is -1.86O C. If a 2-molal
solution is dissolved in water, the new Freezing Point becomes – 3.72OC (2 x -1.86) and so forth. The
Freezing Point is being depressed (dropping).
1. The depression is the result of more solvent molecules being “actively” involved in making
solvent bubbles. They cannot interact with each other to make the liquid “fix” into a solid.
B. Each individual solvent has its own unique molal freezing point constant (Kf).
IX. Calculating Freezing Point Depression of non-electrolytes
A. The equation is: Δtf = Kfm
B. If looking for Kf = Δtf/m
If looking for m = Δtf/Kf
For example, Antifreeze (Ethylene Glycol) is C2H6O2. It is a non-volatile, non-electrolyte solute that is added to pure
water to lower (depress) the freezing point of water so that it does not freeze in your car/truck engine and crack
the engine block (metal) or split the rubber hosing. If a 250.0 gram amount of antifreeze is added to 750.0 mL of
H2O. What is the new Freezing Point?
Remember, for water 1 mL = 1 gram
Step 1: Determine the molar Mass for each substance.
For Antifreeze: C = 12.0 AMUs x 2 = 24.0 AMUs
H = 1.0 AMU x 6 = 6.0 AMUs
O = 16.0 AMUs x 2 = 32.0 AMUs
Total = 62.0 AMUs
For water: H = 1.0 AMU x 2 = 2.0 AMUs
O = 16.0 AMUs x 1 = 16.0 AMUs
Total = 18.0 AMUs
Remember: Unit Given X Unit wanted = Unit wanted
Unit Given
Step 2: Calculate the molarity for Antifreeze.
250.0 (g) C3H8O3 X 1 mol C2H6O2 = 4.03 mol of C2H6O2
62.0 g C2H6O2
Step 3: Calculate the molality for the amount of solvent provided.
a.
moles/.750 kg = 5.37 molal
Step 4: Take the molality and plug it into the equation. As the solvent is water, use the Kf of water:
(-1.86).
Δtf = Kfm = (-1.86) x (5.37) = 9.99OC
Step 5: Subtract the depression value from the pure freezing point value for the solvent.
0.0OC (water) – 9.99OC = -9.99OC (New Freezing Point of water)
X. Calculating Freezing Point Depression for Electrolytes(Ionic solutions)
A. Follow Steps 1 – 5 from above but in step 4 use the molal solution of the ions in place of the molal
concentration for the non-electrolyte.
For example, a 0.100 molal solution of K2SO4  0.2 for K+ (as the are double) and 0.1 for SO4So… (-1.86) (0.3) = - 0.558OC (New Freezing Point of ionized water solution)
Content Outline: Colligative Properties: Boiling Point Elevation (6.9)
XI. Colligative Properties
C. These are the properties of solutions that are dependent upon the concentration of solute, but not on
the solution identity (name).
3. Concentration for colligative properties is molality (m). Think mass.
c. Molality is the number of moles of solute/kg (mass) of solvent.
XII. Boiling Point Elevation
D. The difference between the Boiling points of the pure solvent and a solution containing a nonvolatile, non-electrolyte solute in the same solvent.
E. Boiling Point is where the vapor pressure of the liquid is equal to the atmospheric pressure.
1. If you change the atmospheric pressure, you change the boiling point of a liquid.
For example, water boils at a lower temperature in Denver, Colorado (the “Mile High City”… 1
mile above sea level) than at Mobile, Alabama. Remember, as the altitude increases, the
pressure decreases. If the pressure decreases, the boiling point drops.
F. Boiling Point Elevation is directly proportional to the molality concentration of the solution.
2. If Molality increases the amount of Elevation increases (the temperature rises).
This elevation is referring to temperature; not altitude.
G. Mathematically expressed as:
Δtb = Kbm
Kb = Boiling Point Constant for a pure solvent (OC/molality)
M = Molality of the solution (moles of solute/ kg solvent)
Tb = Numeric Amount of Temperature change (OC)
XIII.
Molal Boiling Point Constant (Kb)
C. The Boiling Point Elevation of the solvent when a 1-molal solution is created by dissolving 1 mole of
non-volatile, non-electrolyte solute in 1 kg of a solvent at 1 atm.
For example: The normal Boiling Point of Water is 100.0OC. The Boiling Point Constant for water is
0.51OC. So if a 1 molal solution is dissolved in water, the new Boiling Point is 0.51OC higher. If a 2molal solution is dissolved in water, the new boiling Point becomes 1.02OC (2 x 0.51) and so forth.
The Boiling Point is being elevated (rising).
1. The elevation is the result of more solvent molecules being “actively” involved in making
solvent bubbles. They cannot break the inter-molecular attractions with the solute to “free
from” the liquid and “escape” as a gas.
3. The solute being dissolved in the solvent also causes the vapor pressure to drop. If the pressure
drops, it will require more energy (heat) to get the solution to boil. You should be able to see the
boiling point line on a phase diagram drops, and thus increases the boiling point on the graph.
a. Solute in solvent lowers vapor pressure  requires more energy input  raises Boiling Pt.
D. Each individual solvent has its own unique molal boiling point constant (Kb).
XIV.
Calculating Freezing Point Depression
C. The equation is: Δtb = Kbm
D. If looking for Kb = Δtb/m
If looking for m = Δtb/Kb
Let’s use the same example as for freezing point as it is the same thought process in calculating.
For example, Antifreeze (Ethylene Glycol) is C2H6O2. It is a non-volatile, non-electrolyte solute that is added to pure
water to elevate (increase) the boiling point of water so that it does not overheat in your car/truck engine and
crack the engine block (metal) or spit the rubber hosing. If a 250.0 gram amount of antifreeze is added to 750.0 mL
of H2O. What is the new Boiling Point?
Remember, for water 1 mL = 1 gram
Step 1: Determine the molar Mass for each substance.
For Antifreeze: C = 12.0 AMUs x 2 = 24.0 AMUs
H = 1.0 AMU x 6 = 6.0 AMUs
O = 16.0 AMUs x 2 = 32.0 AMUs
Total = 62.0 AMUs
For water: H = 1.0 AMU x 2 = 2.0 AMUs
O = 16.0 AMUs x 1 = 16.0 AMUs
Total = 18.0 AMUs
Remember: Unit Given X Unit wanted = Unit wanted
Unit Given
Step 2: Calculate the molarity for Antifreeze.
250.0 (g) C3H8O3 X 1 mol C2H6O2 = 4.03 mol of C2H6O2
62.0 g C2H6O2
Step 3: Calculate the molality for the amount of solvent provided.
a.
moles/.750 kg = 5.37 molal
Step 4: Take the molality and plug it into the equation. As the solvent is water, use the Kf of water:
(-1.86).
Δtf = Kbm = (0.51) x (5.37) = 2.74OC
Step 5: Add the elevation value to the pure boiling point value for the solvent.
100.0OC (water) + 2.74OC = 102.74OC (New Boiling Point of water)
XV. Calculating Boiling Point Elevation for Electrolytes (Ionic solutions)
B. Follow Steps 1 – 5 from above but in step 4 use the molal solution of the ions in place of the molal
concentration for the non-electrolyte.
For example, a 0.100 molal solution of K2SO4  0.2 for K+ (as the are double) and 0.1 for SO4So… (0.51) (0.3) = 0.15OC
100.0OC + 0.15OC = 100.15 (New Boiling Point of ionized water solution)