II III I I. The Nature of Solutions (p. 401 - 410, 425 - 433) Ch. 13 & 14 - Solutions.
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Transcript of II III I I. The Nature of Solutions (p. 401 - 410, 425 - 433) Ch. 13 & 14 - Solutions.
II
III
I I. The Nature of Solutions
(p. 401 - 410, 425 - 433)
Ch. 13 & 14 - Solutions
A. Definitions
Solution - Solution - homogeneous mixture
Solvent Solvent - present in greater amount
Solute Solute - substance being dissolved
A. Definitions
Solute Solute - KMnO4 Solvent Solvent - H2O
B. Solvation
Solvation – Solvation – the process of dissolving
solute particles are separated and pulled into solution
solute particles are surrounded by solvent particles
B. Solvation
StrongElectrolyte
Non-Electrolyte
solute exists asions only
- +
salt
- +
sugar
solute exists asmolecules
only
- +
acetic acid
WeakElectrolyte
solute exists asions and
molecules DISSOCIATION IONIZATION
View animation online.
B. Solvation
DissociationDissociation• separation of an
ionic solid into aqueous ions
NaCl(s) Na+(aq) + Cl–(aq)
B. Solvation
IonizationIonization• breaking apart of
some polar molecules into aqueous ions
HNO3(aq) + H2O(l) H3O+(aq) + NO3–(aq)
B. Solvation
Molecular Molecular SolvationSolvation• molecules
stay intact
C6H12O6(s) C6H12O6(aq)
B. Solvation
NONPOLAR
NONPOLAR
POLAR
POLAR
““Like Dissolves Like”Like Dissolves Like”
B. Solvation
Soap/DetergentSoap/Detergent• polar “head” with long nonpolar “tail”• dissolves nonpolar grease in polar water
C. Solubility
SATURATED SOLUTION
no more solute dissolves
UNSATURATED SOLUTIONmore solute dissolves
SUPERSATURATED SOLUTION
becomes unstable, crystals form
concentration
C. Solubility
SolubilitySolubility• maximum grams of solute that will
dissolve in 100 g of solvent at a given temperature
• varies with temp• based on a saturated soln
C. Solubility
Solubility CurveSolubility Curve• shows the
dependence of solubility on temperature
C. Solubility
Solids are more soluble at...Solids are more soluble at...• high temperatures.
Gases are more soluble at...Gases are more soluble at...• low temperatures &• high pressures
(Henry’s Law).• EX: nitrogen narcosis,
the “bends,” soda
II
III
I II. Concentration
(p. 412 - 418)
Ch. 13 & 14 - Solutions
A. Concentration
The amount of solute in a solution.
Describing Concentration• % by mass - medicated creams• % by volume - rubbing alcohol• ppm, ppb - water contaminants• molarity - used by chemists• molality - used by chemists
A. Concentration
SAWS Water Quality Report - June 2000
B. Molality
solvent ofkg solute of moles(m)molality
mass of solvent only
1 kg water = 1 L waterkg 1mol0.25 0.25m
B. Molality
Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.
75 g MgCl2 1 mol MgCl2
95.21 g MgCl2
= 3.2m MgCl2
0.25 kg water
kgmolm
B. Molality
How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water?
0.500 kg water 1.54 mol NaCl
1 kg water
= 45.0 g NaCl
58.44 g NaCl
1 mol NaCl
kg 1mol1.5 1.5m
2211 VMVM
C. Dilution
Preparation of a desired solution by adding water to a concentrate.
Moles of solute remain the same.
C. Dilution
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
GIVEN:M1 = 15.8MV1 = ?
M2 = 6.0MV2 = 250 mL
WORK:M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3
D. Preparing Solutions 500 mL of 1.54M NaCl
500 mLwater
45.0 gNaCl
• mass 45.0 g of NaCl• add water until total
volume is 500 mL• mass 45.0 g of NaCl• add 0.500 kg of water
500 mLmark
500 mLvolumetric
flask
1.54m NaCl in 0.500 kg of water
D. Preparing Solutions
Copyright © 1995-1996 NT Curriculum Project, UW-Madison(above: “Filling the volumetric flask”)
D. Preparing Solutions
Copyright © 1995-1996 NT Curriculum Project, UW-Madison(above: “Using your hand as a stopper”)
D. Preparing Solutions
250 mL of 6.0M HNO3 by dilution• measure 95 mL
of 15.8M HNO3
95 mL of15.8M HNO3
water for
safety
250 mL mark
• combine with water until total volume is 250 mL
• Safety: “Do as you oughtta, add the acid to the watta!”
Solution Preparation Lab Turn in one paper per team. Complete the following steps:
A) Show the necessary calculations.B) Write out directions for preparing the solution.C) Prepare the solution.
For each of the following solutions:1) 100.0 mL of 0.50M NaCl2) 0.25m NaCl in 100.0 mL of water3) 100.0 mL of 3.0M HCl from 12.1M concentrate.
II
III
I III. Colligative Properties
(p. 436 - 446)
Ch. 13 & 14 - Solutions
A. Definition
Colligative PropertyColligative Property• property that depends on the
concentration of solute particles, not their identity
B. Types
Freezing Point DepressionFreezing Point Depression (tf)• f.p. of a solution is lower than f.p. of
the pure solvent
Boiling Point ElevationBoiling Point Elevation (tb)• b.p. of a solution is higher than b.p. of
the pure solvent
B. Types
View Flash animation.
Freezing Point Depression
B. Types
Solute particles weaken IMF in the solvent.
Boiling Point Elevation
B. Types
Applications• salting icy roads• making ice cream• antifreeze
• cars (-64°C to 136°C)• fish & insects
C. Calculations
t: change in temperature (°C)k: constant based on the solvent (°C·kg/mol)m: molality (m)n:# of particles
t = k · m · n
C. Calculations # of Particles# of Particles
• Nonelectrolytes (covalent)• remain intact when dissolved • 1 particle
• Electrolytes (ionic)• dissociate into ions when dissolved• 2 or more particles
C. Calculations At what temperature will a solution that is composed
of 0.73 moles of glucose in 225 g of phenol boil?
m = 3.2mn = 1tb = kb · m · n
WORK:m = 0.73mol ÷ 0.225kg
GIVEN:b.p. = ?tb = ?kb = 3.60°C·kg/mol
tb = (3.60°C·kg/mol)(3.2m)(1)
tb = 12°C
b.p. = 181.8°C + 12°C
b.p. = 194°C
C. Calculations Find the freezing point of a saturated solution of
NaCl containing 28 g NaCl in 100. mL water.
m = 4.8mn = 2tf = kf · m · n
WORK:m = 0.48mol ÷ 0.100kg
GIVEN:f.p. = ?tf = ?
kf = 1.86°C·kg/mol
tf = (1.86°C·kg/mol)(4.8m)(2)
tf = 18°C
f.p. = 0.00°C - 18°C
f.p. = -18°C