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
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
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
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.
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
NaI NH4Cl Ca(C2H3O2)2 Pb(NO3)2 K3PO4 = soluble
CrCl3 = soluble PbBr2 = insoluble AgCl = insoluble
(NH4)2S = soluble CdS= insoluble Ag2S = insoluble
Ni(OH)2 = insoluble Al(OH)3 = insoluble Ba(OH)2 = soluble
- 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.
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.
- Ivory Soap in water: CH3 - (CH2)16 - C – O- Na+ =
- 1) Water dissolves ionic/polar compounds & 2) Non-polar part of soap dissolves non-polar substances like grease. Soap forms micelles.
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).
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.
1. W / W % = g solute x 100 % (also have W / V & V / V %)
- 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
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)
5. molality, m = moles Solute
- 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
- 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.
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
For MP depression:ΔTf = Kfxm
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
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)
- 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)
i > 1.0 for electrolyte (2.0 for NaCl)
Question:Why is i = 1.02 for 1.00 M acetic acid?
- 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.
- 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.
Note: During osmosis, water molecules go from the purest water to the lesspure water – from less concentrated to more concentrated.
- 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?
(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
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.
- 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
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)