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Chapter 12 - Solutions PowerPoint PPT Presentation


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Chapter 12 - Solutions. A) Sapling HW 12 Due by 11:50 pm Monday, 11/4/2013 B) End-of-Chapter Problems. C) Final exam on Wednesday, December 11 from 8-10 am. Final exam includes all the chapters & lectures – including my lab reviews. D) Quiz #2: Chapters 11&12 on Monday, November 4, 2013 - PowerPoint PPT Presentation

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Chapter 12 - Solutions

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Chapter 12 solutions l.jpg

Chapter 12 - Solutions

A) Sapling HW 12 Due by 11:50 pm Monday, 11/4/2013

B) End-of-Chapter Problems.

C) Final exam on Wednesday, December 11 from 8-10 am. Final exam includes all the chapters & lectures – including my lab reviews.

D) Quiz #2: Chapters 11&12 on Monday, November 4, 2013

Exam #2: Chapters 11&12 on Wednesday, November 6, 2013


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End-of-Chapter ProblemsPages 526 - 532

3 - 9 11 15 16 17 19 20 21 22

23 24 28 32 35 40 - 43 47 49

51 56 59 69 73 75 77 79 81 99


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I. Solutions A. Types

Definitions:

Solution – Homogeneous mixture of two or more substances.

Solvent – Single solution component present in greatest amount.

Solute(s) – Solution component(s) present in smaller amounts.

Solubility – Maximum amount of solute that can dissolve at a given temperature (Note: Is a dynamic equilibrium)

Solid Solute Dissolved Solute

Saturated Solution – Solution with maximum amount of dissolved solute.

Hydration - Attraction of solute ions for water molecules.

Solvation -Attraction of solute ions for any solvent molecules.

Miscible - Fluids that completely dissolve in each other.


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I. Solutions A. Types

Types:SoluteSolventExamples

1GasGasAir (O2 Ar CO2 in N2 )

2GasLiquidSoda Water (CO2 in H2O)

3LiquidLiquidEthyl Alcohol in H2O

4SolidLiquidNaCl in H2O

5LiquidSolidHg in Ag – old dental filling

6SolidSolidBrass – Zn, Sn, Pb, Fe in Cu


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I. Solutions B. Rules

  • General rule for solutions is “like dissolves like” in terms of polarity. Polar solvent can dissolve polar solutes and non-polar solvent can dissolve non-polar solutes.

  • Examples:

    • Polar CH3OH will dissolve as a molecule in polar H2O.

    • Ionic KI will dissolve as ions in polar H2O.

    • Non-polar C6H14 will not dissolve in polar H2O.

    • Non-polar C6H14 will dissolve in non-polar CCl4.


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I. Solutions B. Some Solubility Rules for IONIC compounds in water (Know These Chem 121 Rules)

  • All group IA (1), ammonium, acetate & nitrates are soluble. Examples:

    NaI NH4Cl Ca(C2H3O2)2 Pb(NO3)2 K3PO4 = soluble

  • All Chlorides - Iodides are soluble except silver, mercury & lead. Examples:

    CrCl3 = soluble PbBr2 = insoluble AgCl = insoluble

  • All sulfides are insoluble except for first rule above. Examples:

    (NH4)2S = soluble CdS= insoluble Ag2S = insoluble

  • All hydroxidesinsoluble except for first rule above & Ca, Sr, Ba. Examples:

    Ni(OH)2 = insoluble Al(OH)3 = insoluble Ba(OH)2 = soluble


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I. Solutions B. Solubility Rules for ionic compounds in H2O cont .

- Why are some ionic compounds insoluble; why are some dissolutions in water endothermic & others exothermic?

- Dissolving involves two factors:

A - Break lattice structure = Energy required/added.

B- Solvation of released parts = Energy released.

  • Dissolving most compounds is exothermic:B > A

  • Many ionic compounds are insoluble : A > B

    Examples: 1)CuSO4 orH2SO4(caution)in water = Exothermic (B >> A)

    2) NH4NO3in water = Endothermic (A > B)

    3) AgCl, BaSO4, CuS are insoluble ionic compounds (A >> B)

    Note: Heating the solvent for an insoluble compound usually increases its solubility; can be used for purification – recrystallization.


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I. Solutions B. General Rules

  • Polar CH3OH dissolves in H2O; energy released from H bonds formed.

  • AgCl is insoluble in water; hydration energy not enough to overcome high lattice energy.

  • Ionic LiF dissolves in H2O since polar H2O solvates ions (Hydration) which releases enough energy to pull ions out of LiF lattice.

F-


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I. Solutions B. General Rules

- Ivory Soap in water: CH3 - (CH2)16 - C – O- Na+ =

O

- 1) Water dissolves ionic/polar compounds & 2) Non-polar part of soap dissolves non-polar substances like grease. Soap forms micelles.


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I. Solutions B. General Rules – Le Chatelier’s Principle

Le Chatelier’s Principle: A system at equilibrium will rearrange to relieve any stress. (Know this)

A saturated gaseous solution is in equilibrium as illustrated:

CO2 (aq) CO2 (gas)

Apply pressure & above equilibrium will rearrange to remove extra pressure (dissolve gas; go to left).


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I. Solutions B. Rules - Henry’s Law

  • In 1803 William Henry studied the effect of pressure on gas solubility.

  • Henry’s Law: Solubility of a gas (S) α to the partial pressure of the gas above the solution: S1 = k x P1 or S2 = P2(Know these)

    S1 P1

  • Important to a scuba diver; Henry's Law tells us how much N2(g) will be absorbed in our blood at various pressures.

  • Example: A diver is at 3.0 atm (P1) where solubility of N2 in blood is 53 ug/L (S1). What is the solubility (S2) of N2 in blood when the diver surfaces at 1.0 atm (P2)?

    S2=P2S2 = P2x S1 = 1.0 atm x 53 ug/L = 18 ug/L (18 ppb)

    S1 P1 P1 3.0 atm

    Note: 53 – 18 = 35 ug N2(g) released per liter of blood; get small N2 bubbles in blood.


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II. Solution Concentration Terms A. Review - (know all of these)

1. W / W % = g solute x 100 % (also have W / V & V / V %)

g solution

- 18.0 g NaCl is added to 200. g of water. What is the mass % NaCl?

% = 18 g NaCl x 100 % = 8.26 %

218 g soln

2. ppm, ppb, ppt ppm = ug/mLppb = ng/mLppt = pg/mL

(Note: g = mL for dilute, aqueous solutions; so, ppm also = ug/g, etc)

- 4.2 x 10-8 g of Pb+2 is dissolved in 1000 mL H2O. What is ppt Pb+2 ?

4.2x10-8 g 1012 pg = 42 pg = 42 ppt

1000 mL g mL


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II. Solution Concentration Terms A. Review - know

3. Mole Fraction, X = Xa / XtotalNote: X is always ≤ 1.00

- 0.12 mol N2 is mixed with 0.33 mol O2 and 0.01 mol Ar. What is X of N2

XN2 = 0.12 mole (N2) = 0.26 (units cancel)

0.12 + 0.33 + 0.01 mole Total

Note: Sum of all Xa’s in a solution = 1.0000

4.Molarity, M = moles solute / L solution M = m / L

- 0.80 g of NaOH is dissolved in 200. mL. Calculate the M

M = 0.80 g NaOH x 1 mol / 40. g NaOH = 0.10 mol = 0.10 M

0.200 L L

Note: Use M1 x V1 = M2 x V2 only for dilution problems (moles initially = moles after)


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II. Solution Concentration Terms B. Molality, m

5. molality, m = moles Solute

kg Solvent

- This concentration term is strange, but useful for colligative property calculations; it’s the only one dealing with amount / “SOLVENT.”

Example: 5.7 g of glucose (MW=180) is in 30.9 g of solution. Calculate m.

- 5.7 g glucose x (1 mol / 180 g) = 0.032 mol glucose (solute)

- 30.9 g soln - 5.7 g solute = 25.2 g solvent = 0.0252 kg solvent

m = moles glucose= 0.032 mol glucose = 1.3 m

kg solvent 0.0252 kg solvent


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III. Colligative Properties A. Vapor Pressure Lowering (VPL)

- Colligative Property = Property that depends only on # of solute particles (not on the type of the solute particles).

- Examples: VPL, BP elevation, FP depression, Osmosis

- VPL = a nonvolatile solute lowers the vapor pressure of a solvent.

- VP relationship = Raoult’s Law: Pa = Psa x Xa(Know this)

Pa is VP of solvent(in the solution)

Psa is VP of pure solvent

Xa is the mole fraction of the solvent.

- Note: are other forms of Raoult’s Law.


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III. Colligative Properties A. Vapor Pressure Lowering (VPL)

Example: What is VP & change in VP of 1.00 mol H2O that contains 0.100 mole NaI at 20. oC? Given: VP of H2O (Psa) at 20. oC = 17.5 Torr

Pa= Psa x Xa = 17.5 torr xXaNeed mole fraction of water:

**0.100 mol NaI yields 0.200 mols of particles (Na+ & I-) when it dissolves.

**Total mols of particles = 1.00 mol H2O + 0.200 mol ions = 1.20 mols

Xa = 1.00 mole H2O / 1.20 mole = 0.833[mole fraction of water]

Pa = PsaxXa

Pa = 17.5 Torr x 0.833 = 14.5 Torr

ΔVP = 17.5 - 14.5 = 3.0 Torr


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III. Colligative Properties B. BP Elevation & FP Depression

  • an added nonvolatile solute lowers MP & raises BP of a solvent; the change in BP or MP depends on the type of solvent and the number of particles of solute dissolved (colligative property).

    For MP depression:ΔTf = Kfxm

    (know these)

    For BP elevation:ΔTb = Kbxm

    - ΔTf & ΔTb are changes in BP or MP & m = molality of SOLUTE particles.

    - Kf & Kb are constants in units of oC/m; are unique for each solvent:

    Kf = 1.86 oC/m for water Kf = 40.0 oC/m for camphor

    Kb = 0.512 oC/m for water Kb = 3.12 oC/m for acetic acid


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III. Colligative Properties B. BP Elevation & FP Depression

Example: Calculate the freezing point of a 50.0 % (W/W) ethylene glycol (C2H6O2 or EG) solution in your car radiator? 50.0% - assume 100. g of ethylene glycol is added to 100. g (0.100 kg) of H2O. Kf = 1.86 oC/m for H2O

ΔTf = Kfx m = (1.86 oC/m) x ?m

1) Need m of ethylene glycol (MW = 62.0 g/mol) in 50.0 % aqueous soln:

- 100. g C2H6O2 x 1 mol EG / 62.0 g EG = 1.61 moles C2H6O2

- m = 1.61 moles C2H6O2 / 0.100 kg H2O = 16.1 mols EG/kg water = 16.1 m

2) ΔTf = Kfxm = (1.86 oC/m) x16.1 m= 30.0 oC .

MP = (0.0-30.0)oC = -30.0oC. Car engine protected to -30 oC (-22 oF)


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III. Colligative Properties B. BP Elevation & FP Depression

  • FP or BP changes can be used to determine MW of low MW molecules.

  • Example: 0.99 g sorbitolin 0.010 kgwater has a MP of -1.0 oC. Calculate the MW of sorbitol.

    - Problem solving method: MW = g / mol. You are given g; get mol.

    ΔTf = KfxmGiven: Kf = 1.86 oC/m for water

    m = ΔTf / Kf = 1.0 oC / 1.86 oC/m = 0.54 m(mol sorbitol per kg water)

    - (0.54 mol sorbitol / kg water) x 0.010 kgwater = 0.0054 mol sorbitol

    - MW = g/mol = 0.99 g sorbitol / 0.0054 mol sorbitol = 180 g / mol (2 SF)


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III. Colligative Properties B. BP Elevation & FP Depression

  • Question: An aqueous 1.0 molal solution of ethanol (CH3-CH2-OH) lowers the mp of water by 1.86 oC, but an aqueous 1.0 molal solution of NaCl lowers the mp of water by 3.72 oC (twice that of ethanol). Why?

  • Answer: In water: 1.0 mol of ethanol yields 1.0 mols of particles; 1.0 mol of NaCl (ionic) yields 2.0 mols of particles (1.0 mol of Na+ & 1.0 mol of Cl-).

  • Better equation: ΔTf = ixKfxm(know)

  • i = van’t Hoff factor (pg 517): i= # moles of ions per mole of compound i = 1.0 for nonelectrolyte (CH3OH)

    i > 1.0 for electrolyte (2.0 for NaCl)

    Question:Why is i = 1.02 for 1.00 M acetic acid?


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III. Colligative Properties C. Osmosis

- Jacobus Van’t Hoff received 1st nobel prize

in chem. in 1901 for work on solns. Definition of

Van’t Hoff Factor: i = # moles of ions per mole

of compound. (Know this)

- Summary of Colligative Property Equations using the Van’t Hoff factor:

- Note: assume i = 1.00 unless problem dictates another value.

ΔTf = i x Kfxm

ΔTb = i x Kbxm

π = i xMx R x T (for osmotic pressure) (Know These)

Note: i is automatically included in Raoult’s Law.


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III. Colligative Properties C. Osmosis

- Osmosis is the net flow of solvent molecules from a pure solvent through a semipermeable membrane into a solution.

- Osmotic Pressure (π) is a colligative property & equals the pressure that, when applied to the solution, just stops osmosis.


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III. Colligative Properties Examples & Notes

Note: During osmosis, water molecules go from the purest water to the lesspure water – from less concentrated to more concentrated.

  • Skin gets “pruney” when one swims in ocean or takes a long bath.

  • Red blood cells puff up & burst when placed in deionized water.

  • Osmotic pressure can be very large

  • Can use π to obtain MW of large molecules.

  • We can use reverse osmosis to purify sea water.


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III. Colligative Properties Osmosis & Colloids

- π = i x M x R x T (Know this)

M = Molarity (mols / Lsoln) R = Gas Constant = 0.0821 La/Kmol T in K

- Example: Whatis the π of a 0.2 M (i=1.0) soln. at 300 K?

  • π = 1.0 x 0.2 mol/L x [0.0821 Latm/Kmol] x 300 K = 5 atm

    (5 atm can support a 1.0 in2 column of water over 150 feet high)

    - π can be used to determine M of dilute solutions due to large values of π. Useful for determining MW of dilute solutions of high MW molecules - proteins, carbohydrates, nucleic acids


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III. Colligative Properties Osmosis – MW example

  • Problem:Hemocyanin (Hy) is a protein found in crabs. If 1.5 g of Hy dissolved in 0.250 L of water has a π of 0.00370 atm at 277 K, then what is the MW of Hy? Given i = 1.1 for Hy.

  • π = i x M x R x T MW Hy = g/mol = 1.5g/molM = π / ( i x R x T)

    1)M of Hy = (0.00370 a) = 1.48 x 10-4 mol/L

    1.1 x (0.0821 La/Km) x 277 K

    2)1.48 x 10-4 mol/L x 0.250 L= 3.70 x 10-5 mol Hy

    3)MW = g / mole = 1.5 g / 3.70 x 10-5 mol Hy = 41000 g/mol

    Note: If used mp or bp, the ΔTof this same solution would have been ~10-4oC; much too small to measure accurately.


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IV - Colloids

- Colloid = dispersion (not solution) of particles of one substance throughout another substance. (Know this)

- Colloid differs from solutions in that dispersion particles are usually between 1 and 200 nm in diameter – they are not in solution, but they are small enough to stay in suspension.

- Colloids will give Tyndall effect = the scattering of a light beam by the colloidal particles.(Know this)

-Three names for colloids:

1) aerosols = liquid/solid dispersion in a gas;

2) emulsion = liquid dispersed in a liquid;

3) sol = solid dispersed in a liquid


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IV – Colloids; common names

1) Aerosols= liquids or solids dispersed in a gas

Examples: Fog (liquid water in air)

Dust in air (solid dirt in air)

2) Emulsions= liquids dispersed in a liquid

Examples: Mayonnaise (oil in water)

Milk (fat in water)

3) Sols= solids dispersed in a liquid

Examples:Jell-O (collagens in water)

Milk (casein in water)


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IV . Colloids - Tyndall Effect