(Chapter 13 Andrew Meade Mitchell Korotkin Kayla Devin)

(Chapter 13Andrew Meade

Mitchell KorotkinKayla Devin)

-The presence of certain ions may cause a compound to be soluble or insoluble.

- The solubility rules are a list of ions and whether they make a compound soluble or insoluble.

-If there are no solids (all products are soluble), then there is no reaction

Solubility Rules:

ION Soluble or insoluble?

Exceptions?

NO3- (Nitrate) Soluble None

C2H3O2- (Acetate) Soluble AgCH3COO

ClO3- (Chlorate) Soluble None

Cl- (Chloride) Soluble AgCl, Hg2Cl2, PbCl2

Br- (Bromide) Soluble AgBr, PbBr2, Hg2Br2, HgBr2

I- (Iodide) Soluble AgI, Hg2I2, HgI2, PbI2

SO4 2- (Sulfate) Soluble BaSO4, PbSO4, Hg2SO4, CaSO4, Ag2SO4, SrSO4

Alkali metal ions, NH4+ Soluble None

H+ Soluble None

CO32- (Carbonate) Insoluble IA elements, NH4 +

CrO42- (Chromate) Insoluble IA elements, NH4 +, CaCrO4, SrCrO4

OH- (Hydroxide) Insoluble IA elements, NH4 +, Ba(OH)2, Sr(OH)2, Ca(OH)2

PO4 3- (Phosphate) Insoluble IA elements, NH4 +

SO3 2- (Sulfite) Insoluble IA elements, NH4 +

S 2- (Sulfide) Insoluble IA elements, IIA elements, NH4 +

Potassium chromate- K2CrO4

Example:

Normally, chromates tend to be insoluble. However, one of the exceptions is the IA elements, and because potassium(K) is an IA element, Potassium chromate is soluble.

CrO42- (Chromate) Insoluble IA elements, NH4 +, CaCrO4, SrCrO4

ION Soluble or insoluble?

Exceptions?

Answer: SOLUBLE

1. Look at solubility rules for chromate.

Use the solubility rules to predict the reaction products:

AuCl (aq) + NaNO3 (aq)

Example:

AuNO3(aq) + NaCl(aq)

**There is no reaction because both of the products are aqueous

Complete Ionic Equations

** Notice the Spectator Ions circled- these are ions that do not take part in a chemical reaction and are found on both sides of the reaction **

-Include only the compounds that undergo a chemical change in a reaction in an aqueous solution

-To do this simply omit the spectator ions from the complete ionic equation

Net Ionic Equations

Scaffolded ExampleWrite the complete and net ionic equation for the molecular equation:

HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l)

H+ + Cl- + Na+ + OH- Na+ +Cl- + H2O

H+ + Cl- + Na+ + OH- Na+ +Cl- + H2O

H+ + OH- H2ONet ionic equation:

2. Cancel out spectator ions

1. Split up aqueous reactants and products into ions.(complete ionic equation)

PROBLEM 1Write the complete and net ionic equation for the molecular equation:

NaCl(aq) + AgNO3 (aq) AgCl(s) + NaNO3(aq)

Na+ + Cl- + Ag+ + NO3- AgCl(s) + Na+ + NO3

-

Problem 1: ANSWERWrite the complete and net ionic equation for the molecular equation:

NaCl(aq) + AgNO3 (aq) AgCl(s) + NaNO3(aq)

Na+ + Cl- + Ag+ + NO3- AgCl + Na+ + NO3

-

Ag+ + Cl- AgCl(s)Net ionic equation:

1. Split up aqueous reactants and products into ions.(complete ionic equation)

2. Cancel out spectator ions

Problem 2Write the complete and net ionic equation for the molecular equation:

KI(aq) + AgClO3 (aq) AgI(s) + KClO3(aq)

K+ + I- + Ag+ + ClO3- AgI(s) + K+ + ClO3

-

Problem 2: ANSWERWrite the complete and net ionic equation for the molecular equation:

KI(aq) + AgClO3 (aq) AgI(s) + KClO3(aq)

K+ + I- + Ag+ + ClO3- AgI + K+ + ClO3

-

Ag+ + I- AgI(s)Net ionic equation:

1. Split up aqueous reactants and products into ions.(complete ionic equation)

2. Cancel out spectator ions

Colligative Properties- The physical properties of a solution are different

from those of the pure solvent.- Some of these differences are due to the mere

presence of solute particles in the solution.- Colligative properties depend on the number

of particles dissolved in a given mass of solvent. - They do not depend on the chemical nature of the solute or solvent

The 4 Colligative Properties Are: 1. Vapor Pressure Lowering2. Freezing Point Depression3. Boiling Point Elevation4. Osmotic Pressure

Vapor Pressure Lowering-Vapor pressure occurs because some molecules of

a pure liquid leave the liquid state and enter the gaseous state (vaporization)

-At the same time, molecules from the gaseous state return to the liquid state (condensation)

-Equilibrium is established when the rate of vaporization and condensation becomes equal

-The gas pressure resulting from the vapor molecules over the liquid is the vapor pressure

-Vapor Pressure of a solvent containing a nonvolatile solute is lower than the vapor pressure of the pure solvent

Osmotic Pressure- Solvents are able to move through a

semipermeable membrane, where they move from a high to a low concentration.

-Through the movement of the solute, the levels of the solution becomes uneven.

-The osmosis of the solvent will stop when the pressure difference becomes large and the levels are very uneven.

-The pressure difference is called osmotic pressure.

Boiling Point Elevation

-The molal boiling point constant (Kb) is the boiling point elevation of the solvent in a 1-molal solution of a nonvolatile, nonelectrolyte solute

-The boiling point elevation is the difference between the boiling points of the pure solvent and a non electrolyte solution of that solvent and is directly proportional to the molal concentration of the solution -Can be calculated using the formula:

Δ T b = K b (i) (m)Boiling Point

Elevation

Boiling Constant

Molality

Van't Hoff Constant

Boiling Point ElevationExplanation

Boiling point is essentially the amount of energy it takes to make the vapor pressure of a solution equal to the external pressure.

With normal water, the amount of energy it takes for the vapor pressure to equal the external pressure is reached at 100 ºC. 100 ºC

However, when a solute like NaCl is added, the vapor pressure of the solution is lowered, and it takes more energy to equalize the vapor and external pressure, causing a higher boiling point.

Vapor pressure

Boiling Point

1. What is the boiling point elevation when 35.0 g of NaCl is

dissolved in 750 g of water?

(Kb for water is .52 ºC/m)

Scaffolded ExampleΔTb = Kb (m)(i)

Kb = .52 ºC/mm =i =

Step 1: Find molality

Step 3: Plug in values into the formula

Step 1: Find molality

Step 3: Plug in values into the formula

Step 2: Determine Van't Hoff FactorStep 2: Determine Van't Hoff Factor

1 mole NaCl 58.43 g NaCl

35.0g NaCl X = .599 moles NaCl

STEP 1: FIND MOLALITY

750g H2O X 1 kg H2O 1000 g H2O = .750 kg H2O

molality = moles solute kg solvent

m= .599 moles NaCl.750 kg H2O

molality = .799 m

.799 m

2.00STEP 2: Van't Hoff Factor

NaCl

Na+ Cl-

2 ions = Van't Hoff Factor of 2

STEP 3: Plug in values

ΔTb = Kb (m)(i)

ΔTb = .52ºC/m(.799m)(2.00)

ΔTb = .831 ºC

Practice 11. What is the boiling point elevation when 15 g of ammonia (NH3) is dissolved in 250 g of H2O?

(Kb for water is .52 ºC/m)

Δ Tb = Kb(i)(m)

Practice 1: AnswerΔ Tb = Kb (i)(m)

1. 15 g NH3 x

2. m=

3. Δ Tb = .52ºC/m(1)(3.52m)

4. Δ Tb = 1.8 ºC

1 mole NH3 17.024 g NH3

= .88 mole NH3

.88 moles NH3

.25 kg H20 3.52 m

Practice 21. What is the new boiling point when 23 g of ammonia

(NH3) is dissolved in 180 g of H2O?(Kb for water is .52 ºC/m)

Δ Tb = Kb(i)(m)

Practice 2: AnswerΔ Tb = Kb (i)(m)

1. 23.0 g NH3 x

2. m=

3. Δ Tb = .52ºC/m(1)(7.5m)

4. Δ Tb = 3.5ºC

5. 100ºC+3.5ºC = 103.5ºC

1 mole NH3 17.024 g NH3

= 1.35 mole NH3

1.35 moles NH3

.18 kg H20 7.5 m

-Freezing point depression is the ability of a dissolved solute to lower the freezing point of its solution.

-When a solute is added to a solvent, more kinetic energy must be withdrawn from the solution for it to solidify.

The equation for freezing point depression is:

ΔTf = Kf (m)(i)

Freezing Point Depression

Freezing Point Depression

FreezingConstant Molalit

y

Van't Hoff Constant

Freezing Point Depression Explanation

With pure water, the water molecules move freely around, and it is easy for the water to freeze into a lattice structure as solid ice.

Free moving molecules in liquid water.

Water freezing

However, when a solute like NaCl is added, the additional movement from the solute particles impairs the freezing process and makes the freezing process require more energy, lowering the freezing point.

Water and NaCl molecules

Water partially freezing

1. What is the freezing point depression when 65.0 g of NaCl is dissolved in 1250 g of water?(Kf for water is -1.86 ºC/m)

Scaffolded ExampleΔTf = Kf (m)(i)

Kf = -1.86 ºC/mm =i =

Step 1: Find molality

Step 3: Plug in values into the formula

Step 1: Find molality

Step 3: Plug in values into the formula

Step 2: Determine Van't Hoff FactorStep 2: Determine Van't Hoff Factor

1 mole NaCl 58.43 g NaCl

65.0g NaCl X = 1.11 moles NaCl

STEP 1: FIND MOLALITY

1250g H2O X 1 kg H2O 1000 g H2O = 1.250 kg H2O

molality = moles solute kg solvent

m= 1.11 moles NaCl1.250 kg H2O

molality = .888 m

.888 m

2.00STEP 2: Van't Hoff Factor

NaCl

Na+ Cl-

2 ions = Van't Hoff Factor of 2

STEP 3: Plug in values

ΔTf = Kf (m)(i)

ΔTf = -1.86ºC/m(.888m)(2.00)

ΔTf = -3.31 ºC

PRACTICE 11. What is the new freezing point when

90.0 g of oxygen gas is dissolved in 1750 g of water?

(Kf for water is -1.86 ºC/m)

ΔTf = Kf (m)(i)

PRACTICE 1: ANSWERSΔTf = Kf (m)(i)

1. 90.0 g O2 x

2. m=

3. Δ Tf = -1.86ºC/m(1.60m)(1)

4. Δ Tf = -2.98 ºC5. 0 - 2.98 = -2.98 ºC

1 mole O2 32.00 g O2

= 2.80 moles O2

2.80 moles O2

1.75 kg H201.60 m

PRACTICE 21. What is the freezing point depression

when 48.0 g of oxygen gas is dissolved in 2300 g of water?

(Kf for water is -1.86 ºC/m)

ΔTf = Kf (m)(i)

PRACTICE 2: ANSWERSΔTf = Kf (m)(i)

1. 48.0 g O2 x

2. m=

3. Δ Tf = -1.86ºC/m(.652m)(1)

4. Δ Tf = -1.21ºC

1 mole O2 32.00 g O2

= 1.50 moles O2

1.50 moles O2

2.30 kg H20.652 m