4.10.12 •  Welcome back! •  What is the mole fraction of 37.3 g of KCl in 3000

g of aqueous solution?

•  Objectives: •  Odds and ends from ch 12: calculate mole fraction

and qualitatively describe the enthalpy of solution •  Movin’ on: write equations to describe the

dissolution of ionic compounds as well as precipitation reactions

But first -

•  Homework •  Questions on molarity/molality? •  No test on ch 12, we will include ch 12

material on the unit test for ch 13

Mole fraction •  Describes a solution in terms of moles of

either a solute or solvent in terms of the total number of moles of solute(s) + solvent

•  What is the mole fraction of NaCl in a 1.0 M solution?

1. what is the mass of this solution? 2. What is the mass of the solvent? 3. How many moles of solvent do you have? 4. How many total moles are there? 5. Compute!

Enthalpy of solution

•  Enthalpy is a measure of the energy of a system

•  When a solution is formed, solute particles interact with solvent particles

•  If ions are produced in aqueous solution, water molecules coordinate on the ion and give up some of their kinetic energy

•  This can lead to a change in energy, which means change in what?

Thinking about enthalpy •  The energy of a system •  How is kinetic energy measured? •  Recall: all bond forming reactions release

energy, bond breaking reactions require input of energy

•  In pure water, what bonds exist? •  In an ionic compound, what bonds exist? •  What then is the nature of dissolution in

terms of bond breaking/forming?

So in terms of enthalpy •  The energy required to break bonds is

always positive (energy must be added to break bonds)

•  If new bonds are formed during the reaction, energy is released (so if a substance loses energy, this is subtraction)

•  The net enthalpy change is the energy required to break bonds MINUS the energy released from the formation of new bonds

•  The net enthalpy change may be positive or negative

“Ideal” enthalpy of solution

An endothermic enthalpy of solution

An exothermic enthalpy of solution

Some values of enthalpy of solution (kj/mol solute at 298 K)

Substance Enthalpy of solution

Substance Enthalpy of solution

AgNO3(s) +22.59 KOH(s) -57.61 CH3COOH(l) -1.51 MgSO4(s) +15.9 HCl(g) -74.84 NaCl(s) +3.88 KCl(s) +17.22 NaOH(s) -44.51 KNO3(s) +34.89 HI(g) -81.67

Some assumptions here: the solvent is water, the energy is from 1.0 mole of solute; what would happen is the temperature was something other than 298 K?

Which brings us to

•  Properties of ions in aqueous solution! (a.k.a. chapter 13)

What happens when a compound dissolves

•  A change in enthalpy •  Production of ions •  When we say production, think “product”,

which means reactions, which means equations!!

Dissociation

•  This is the separation of an ionic compound via dissolution by a solvent into its constituent ions

•  Example: NaCl(s) ---H2O--> Na+(aq) + Cl-(aq)

In words: one mole of NaCl dissolved in water would yield one mole each of Sodium and Chloride ions

4.11.12 •  In a beaker of water, you add some ammonium

phosphate and some potassium nitrate. What do you end up with inside the beaker?

•  HW – page 443 #1-5

•  TYGAGT – write out net ionic equations, balance double replacement reactions using solubility rules

Precipitation reactions

•  In general, all ions are (technically) soluble in water

•  However, some are more soluble than others; others are so insoluble that they may be considered totally insoluble for practical purposes

Solubility rules

The double replacement reaction

•  This is two ionic compounds in aqueous solution – does an actual reaction happen?

•  Example 1: ammonium nitrate + potassium chlorate

•  Reaction? Yes or no? •  All combinations are totally soluble, so no

distinct product would form

How do you know if you have a reaction?

•  One of the products must be: insoluble (a precipitate), water, or a gas

•  Example 2: •  Ammonium sulfide + cadmium nitrate

•  CdS would be insoluble, so it would separate itself from the other components of the solution, so we WOULD call this a double replacement reaction

More examples (assume all are aq)

•  Lead nitrate and potassium acetate •  Silver nitrate and ammonium chlorate •  Ammonium acetate and sodium carbonate •  Sodium carbonate and calcium nitrate •  Calcium nitrate and ammonium carbonate

4.12.12

•  Write out a complete and net ionic equations for aqueous calcium nitrate and aqueous ammonium carbonate.

•  Objectives today: •  Write out net ionic equations •  Explain how polar solutes may be ionized

in water and produce electrolyte solutions

Net ionic equations

•  This is a way that reactions of ions in aqueous solution are written out

•  It includes only those ions that undergo a chemical change

•  Ions that do not undergo chemical change are called spectator ions

steps Example: (NH4)2S(aq) + Cd(NO3)2(aq)

2NH4NO3(aq) + CdS(s) 1.  Convert chemical equation into an overall ionic

equation 2NH4

+(aq) + Cd2+(aq) + S2-(aq) + 2NO3-(aq)

CdS(s) + 2NO3-(aq) + 2NH4

+(aq) 2. Cancel out the spectator ions 3. Rewrite showing just the participant ions

Cd2+(aq) + S2-(aq) CdS(s) 4. This reaction will be true for ANY and all reactions

involving the actual ions

example

•  In the reaction of zinc nitrate and ammonium sulfide, write the equation for the double replacement reaction, the overall ionic equation, and then the net ionic equation.

ionization

•  Some molecular (non-ionic) compounds can form ions in solution (ALL compounds examined thus-far have either been ionic or composed of polyatomic ions)

•  Highly polar bonds (large differences in electronegativity) can result in ionization

•  Examples: •  HCl(aq) H+(aq) + Cl-(aq)

A new ion to know •  Hydronium, H3O+

•  In the ionization of an acid, such as HCl(aq), H+ ions are produced

•  These ions are attracted to the lone pair electrons on a water molecule

•  This forms a new compound, Hydronium •  Because the compound is between a water

molecule and a Hydrogen ion, charge is conserved, hence the (+) charge on Hydronium

More on Hydronium

•  Water molecules do ionize in aqueous solution (think about this for a minute) to a small extent

•  So does a quantity of water become positively charged?

Review: electrolytes

•  Strong versus weak electrolytes •  Ionization of molecular compounds is

reversible •  The factors that lead to formation of strong

versus weak electrolytes amount to how easily it is to remove an ionizable Hydrogen from a compound

•  Examples: HF, HCl, HBr, HI, acetic acid, sugar

4.13.12 •  Which pair contains the larger total concentration

of ions? a. 0.10 M HCl or 0.05 M HCl b. 0.10 M HCl or 0.10 M HF c. 0.10 M HCl or 0.10 M CaCl2

•  HW – review questions on page 451

•  Today – calculate changes to boiling point and freezing point for non-electrolyte solutes

•  Finish labs on Monday (too many field trips today)

Properties of matter

•  Intensive – density, concentration (M or m) •  Extensive – mass and volume (properties

that depend on size) •  Colligative – properties that depend on the

concentration of solute particles but not their identity

Review: boiling point

•  The temperature at which the vapor pressure exceeds the air pressure above the liquid

•  If vapor pressure exceeds air pressure, phase change occurs

•  Freezing point is the point at which a liquid turns into a solid (assumes a regular crystalline formation)

Pure water

Colligative properties

•  Lowering vapor pressure •  non-volatile substance – substance that

has little chance to become a gas under existing conditions

•  Addition of a non-volatile solute to water will change the freezing point and melting points

•  Why?

Boiling point elevation •  The attraction between ions and polar

molecules is stronger than the Hydrogen bonds that form between water molecules

•  Also, if you add a solute, what happens to the mole fraction of water?

•  Adding a solute diminishes the amount of water that can escape the solution

•  So adding a solute makes it more difficult for water to leave the system, so more energy is needed to induce a phase change

Effect of non-electrolytes on BP

•  Changes to BP and FP are due to molar quantities of solutes, not the identity of the solute itself (definition of colligative property!)

•  Example: – 1 m glucose lowers vapor pressure by

5.5 X 10-4 atm – 1 m sucrose lowers vapor pressure by

5.5 X 10-4 atm

Boiling point elevation

•  Molal boiling point constant: Kb •  The boiling-point elevation of the solvent in

a 1-m solution of a non-volatile, non-electrolyte solute.

•  The boiling point elevation of a 1-m solution of any non-electrolyte solute in water is 0.51°C/m

•  This means that Kb for water is also 0.51°C/m!

Determining Δtb

•  Δtb is change in boiling point •  Δtb = Kbm

•  Kb is the boiling point constant for the solvent, and m is the molal concentration of the solute

For solvents other than water •  Other solvents, there

are going to be different values for Kb

•  All of the substances listed are mostly non-polar

•  Why would their Kb be higher for a non-electrolyte solute than water’s?

Solvent Kb (°C/m)

Acetic acid CH3COOH

2.93

Benzene C6H6 2.67

Carbon tetrachloride CCl4

5.02

Chloroform CHCl3 3.85

Ethanol CH3CH2OH

1.20

Water H2O 0.51

example

•  What is Δtb for a solution made from 20.1 g of a non-electrolyte solute and 400.0 g of water? The molar mass of the solute is 62.0 g.

Freezing point depression

•  We use Δtf to represent this •  In order for water to freeze, it must be able

to assume a regular crystalline structure (ice). Solute particles get in the way of this

•  Their getting in the way keeps the water molecules moving around, which means you must remove MORE energy from the system in order to make water freeze

Calculating Δtf •  A 1.0-m concentration of any non-volatile non-

electrolyte solute changes the freezing point by -1.86°C/m

•  So the molal freezing point constant (Kf) for water is therefore _______?

•  The freezing point depression Δtf is the difference between the freezing point of the pure solvent and a solution of a non-electrolyte in that solvent

•  It is directly proportional to the molal concentration of the solution Δtf = Kfm

Other solvents

•  Just like water, other solvents will have different values for Kf

•  Again, why is the value for water so much different than the others?

Solvent Kf (°C/m)

Acetic acid 3.90 Benzene 5.12 Naphthalene (C10H8)

7.00

Chloroform CHCl3 4.68 Camphor (C10H16O)

40.0

Water 1.86

example

•  What is the freezing point depression of water in a solution of 17.1 g of sucrose (C12H22O11), in 200. g of water? What is the actual freezing point of the solution?

•  Remember: Δtf = Kfm

4.17.12 •  A 1.0-m NaCl solution changes the freezing

point twice as much as a 1.0-m sucrose solution. Why?

•  HW – p 456 1-4

•  Calculate ΔTf and ΔTb for electrolytes •  Calculate osmotic pressure given •  Finish labs from other day

The Van’t Hoff Factor

VH factor

•  The van’t Hoff factor (i) is an integer that describes the number of ions a compound will dissociate into

•  Examples: •  Sucrose i = 1 •  NaCl i =2 •  H2SO4 i = 3

application

•  We plug i into the equations we are already using to calculate changes to boiling point/freezing point

•  ΔTb = iKbm

•  ΔTf = ikfm

example

•  What is the change in boiling point/freezing point from the addition of 1.0 mol of sodium hydroxide to water? What are the new freezing and boiling points?

•  Recall: •  Kb for water is 0.51°C/m •  Kf for water is -1.86°C/m

Osmotic pressure

•  Some fancy Mr. K artwork… •  Consider a U-shaped tube •  Remember, a solute gets surrounded by

water molecules dramatically increasing its effective size

•  Used in dialysis machines, for example

Hyper/hypo/iso