Unformatted text preview: Chem 210 / WANG / Chapter 13A
Properties of Solution Chapter 13 Properties of Solutions
13.1 The Solution Process
13.2 Saturated Solutions and Solubility
13.3 Factors Affecting Solubility
13.4 Ways of Expressing Concentration
13.5 Colligative Properties solvent – major component
solute – minor component
(may be more than one.) 13.6 Colloids Which
Which of these pictures is:
A “solution”? A “compound”?
Mixture - (diatomic)
element heterogeneous C A compound “Solution” vs. “Compound”
“Compound” Definition of a Solution
A solution A homogeneous mixture of two
or more components. Like a compound, which has at least two
different elements; a solution also has at
least two different components (solvent
plus one or more solutes).
Unlike compounds, in which atoms are
joined through chemical bonds, solutes
and solvent stay together in solutions
through intermolecular forces.
Compound has definite composition and
properties; solution has variable
composition, and its properties change
depending on solute concentration. The major types of intermolecular forces in solutions. Physical states of solutions A solutiton =
mixture Solutions can exist in any of the three states.
dental fillings, 14K gold, sterling silver
saline, vodka, vinegar, sugar water Intermolecular forces keep
solute and solvent together. The
The arrangement of atoms in two types of alloys. brass - a substitution alloy carbon steel -an interstitial alloy Alloys are “solid” solutions! Fall 2009 UM-SJTU JI Solution terms:
dissolving, solvation and hydration
All describe how solute disperses in solvent.
Dissolving - Macroscopic description
• Solute disappearing from sight.
Solvation - Molecular description
• Solvent molecules surrounding solute
particles and stabilizing them in solution
Hydration - Molecular description
• = solvation when water is the solvent.
• H2O surrounding solute particles and
stabilizing them in solution. Gaseous solutions
the atmosphere, anesthesia gases Dissolving, solvation, hydration
Figure 13.2 Figure 13.9 Hydrated ions Page 1 Chem 210 / WANG / Chapter 13A
Properties of Solution Hydrates vs. anhydrous
• Hydration shells
aqueous ion. • A “hydrate” forms when
water of hydration stays
in a crystalline solid
obtained from aqueous
When water is completely
driven from the crystal, it
becomes “anhydrous.” Chromium (III) chloride
• Anhydrous CrCl3 (purple)
• Hydrated CrCl3.6H2O (dark green)
http:\\wwwchem.uwimona.edu.jm:1104/courses/chromium.html Enthalpy of Solution = ∆H3 +∆H2 +∆H1 Instant cold packs looks like a
plastic bag filled with liquid. You
hit it, shake it up and it gets
• The liquid inside the cold pack
is water. In the water is another
plastic bag or tube containing
ammonium nitrate, NH4NO3(s) .
• When you hit the cold pack, it
breaks the tube so that the
water mixes with solid NH4NO3.
• The mixing process is an
endothermic reaction, heat is
absorbed. Figure 13.4 Heat of hydration Heat of hydration Used in
Temperature of the solution falls to about 35 F
for 10 to 15 minutes.
MgSO4 ∆Hsoln = -91.2 kJ/mol http://home.howstuffworks.com/refrigerator7.htm NH4 NO3 ∆Hsoln = +26.4 kJ/mol Dissolving of (a) NaCl in water
(b) CH3OH (methanol) in water
methanol) in Like
Like dissolves like:
solubility of Energy
Energy changes in the
Separation of solute
∆H1 > 0 endothermic
Separation of solvent
∆H2 > 0 endothermic
∆ H3 < 0
Exothermic Like – Dissolves – Like Rule
• When solvent and solute
molecules have similar
intermolecular forces, they are
alike (polar ↔ polar or nonpolar
• Heat required to separate
molecules ~ Heat released when
they come together.
• ∆Hsoln = ∆H1 + ∆H2 + ∆H3~ 0
• Mixing occurs spontaneously,
since natural tendency is toward
increasing disorder (entropy ↑).
disorder O Vit. C C HO C HO H3C
C C O
C HO Vit. B C Fall 2009 UM-SJTU JI C A solution of methanol in
CH2 CH2 N
H C H OH
2 C CH2 CH3
C CH CH O CH3 CH C
CH CH CH3 CH3 C Vit. A C OH
2 Hint: Polar bonds
such as O-H make
polar molecules. H H
CH OH O HO CH3 1) Name the forces that must be overcome to break up
2) Name the forces that attract each solute to dissolve. HC
H CH2 methanol Figure 13.6 Apply the like-dissolves-like rule to
determine if each vitamin is water soluble.
vitamin water water methanol
in water. Figure 13.3 C CH3
CH2 CH2 CH CH3
2 Vit. D CH2 CH CH3
CH2 CH 2 CH2 CH
2 CH3 Ref. page 540 Textbook Page 2 Chem 210 / WANG / Chapter 13A
Properties of Solution LikeLike-dissolves like rule & Compds of double nature
Compds double Figure 13.9 Lecithin – the
found in all cell
has polar and
in tis structure. • One end of the molecule is nonpolar and the
other end is polar.
• It brings polar and nonpolar molecules
solubility For Most SOLIDS… increases
has little effect.
Why? Figure 13.17 -OOC
-OOC - COO
COO -OOC COO -OOC COO - - Temperature and solubility
Why? Figure 13.18 250
0 Pressure compresses
gas molecules into
liquid solutions. Pressure has little effect on
solids / liquids, which are
not compressible. Nonpolar ‘tail’ (long
chain of C and H)
dissolves in grease/dirt.
Polar ‘head’ (-COO-1)
(stays in water.
Micelles cause soapy
water to appear cloudy,
Rinse water carries
micelles away, along
with the grease/dirt
trapped in them. For most GASES… Water solubility Higher T
have more kinetic
Stay in the gas phase. Higher T
Molecules have more
kinetic energy to
•Break away from a solid.
• Mix with water molecules. How soap works Solubility
(g/100ml water) Explain the trend of solubility in Water and (b) Hexane. (Table 13.3) 20 40 60 80 100 Temperature (oC)
Point out two major differences between SO2 and the rest.
Explain how KNO3 and sucrose differ from the others. Henry’s Law
Gas solubility in a liquid is
directly proportional to the
partial pressure of the gas
above the liquid when
temperature is held constant. Sgas = kPgas
S = Solubility of gas in solution
Pgas = partial pressure of gas
k = Henry’s constant at certain T. Fall 2009 UM-SJTU JI Gas
Gas Solubility: Another way to
see Henry’s Law
Solubility (S) of a gas in a liquid is directly
proportional to its partial pressure (P). S = constant
P2 temperature must be held constant! The table below shows the solubility of
several gases at P = 1 atm. Estimate the
concentration of O2 in river water.
Since air is composed of 20% O , if air =
[O2] = 1.2 x 10-3 M 1.0 atm, partial pressure of O = 2P =
@ PO2 = 1 atm
0.20 atm S1
[O2] x 1.0
= 0.20 x 1.2 x10-3
[O2] = 2.4x10-4 M Page 3 Chem 210 / WANG / Chapter 13A
Properties of Solution Solubility, saturated, unsaturated
and supersaturated Solubility depends on a dynamic equilibrium
between dissolving and returning of solutes.
Ref. Fig 13.9 Differ in terms of the amount of solute in a
Unsaturated - Dissolving rate > Returning rate Supersaturated solution is not stable.
Figure 13.10 Saturated – Dissolving rate = Returning rate.
Supersaturated • Temporarily have dissolved too much.
• An unstable state – solute settles out when
disturbed or over time. Solubility = Solute concentration in a
saturated solution, a solution in equilibrium.
(undissolved) Different concentration
• Mass percentage
• Parts per million
• Parts per billion
• Mole Fraction
• Molality solute (dissolved) Solution
Concentration All units specify:
• quantity of solute
• quantity of solvent
or solution. Calculate volume needed in dilution
Prepare 350 mL of 4.5 M Hydrogen peroxide solution
from 8.8 M concentrated solution.
M1V1 = M2V2
Step 1: calculate
M1 = 4.5 M
M2 = 8.8 M
V1 = 350 mL
V2 = ?
4.5 x 350 = 8.8 x ?
? = 178.98 = 180 mL
I need 180 mL of concentrated solution.
Step 2: go to the lab and obtain a 350-mL
Step 3: Transfer the 180 mL concentrated
solution into the flask with a burette
(more precise than graduated cylinder).
Step 4: Add distilled water to the etched mark.
etched Fall 2009 UM-SJTU JI Sodium acetate crystals rapidly form when a
seed crystal is added to a supersaturated
solution of sodium acetate. Molarity
M= moles solute
liters of solution mol
• Recognizes that compounds have different
formula weights. • A 1 M solution of glucose contains the
same number of molecules as 1 M ethanol. • [ ] - special symbol which means molar
( mol/L ) or M Molarity
(a) Calculate the molarity of 1.33 g KOH in 2.5L
= (1.33 gKOH ÷ 56.1 g/mol) / 2.5 L
= 0.0095 M
(b) Calculate the grams of solute in 500.0 mL
of 0.0119 M NaCl
gNaCl = 0.0119 MNaCl x (500.0 mL ÷ 1000)
= 5.95 x10-3 mole x 58.5 g/mol
= 0.348 g NaCl Mass Percentage (%), ppm, ppb
% by mass = Mass solute
Mass of solution x 10 use g for both Parts per million (ppm)= Part (Solute) x 106
Total (solution) Use the same units (either mass or volume) for both Parts per billion (ppb)= Part (Solute) x 109
Total (solution) Use the same units (mass or volume) for both Metric prefixes
Changing the prefix alters the size of a unit.
0.000001 Page 4 Chem 210 / WANG / Chapter 13A
Properties of Solution Express these aqueous solutions in
ppb and ppm
NOTE: For dilute aqueous solutions,
1 g = 1 mL = 1 cm3
1 kg = 1000 g = 1000 mL = 1 L
(a) 5 µg solute per L solution
L = kg
= 5 µg / 1 kg
x 109 = 5 ppb
-6g /103 g) = 5 x10-9
= (5 x 10
x 106 = 0.005 ppm
(b) 3.5 mg per kg
= 3.5 mg / 1 kg
x 109 = 3500 ppb
= (3.5 x 10-3g /103g) = 3.5 x10-6 x 106 = 3.5 ppm
(c) 50 g per m3
= 50 g / (102 cm)3 = 50 g / (106 cm3)
= (50 g / 106 g) = 50 x 10-6 x 109 = 50,000 ppb
x 106 = 50 ppm
PROBLEM 2 Mass % versus Molarity What is the molarity of a 3.0% by mass H2O2
solution, density = 1.05 g/mL. 3% = 3.0 g H2O2 and 100 g solution
? Mole H2O2 /1 L of solution
3.0 g H2O2
100 g soln mole = 3.0 ÷ 34.0 = 0.088 mole H2O2
100g ÷ 1.05g/mL = 95.2 mL = 0.0952 L Molarity = moles ÷ Liters
= 0.088 mole ÷ 0.0952 L
= 0.92 M Molality (m)
Molality, m = mole(Solute)
Kg (solvent) SAMPLE PROBLEM 1(a) 40.5 mg Ca x Fall 2009 UM-SJTU JI g
x 106 3.5 g = 1.16x104 (b) The label on a 0.750-L bottle of
Italian Chianti indicates “11.5%
alcohol by volume”. How many
milliliters of alcohol (C2H5OH)
does the wine contain?
0.750 L wine x ppm Ca 11.5 L alcohol
100 L wine = 0.0862 L alcohol × Chianti wine in a
traditional fiasco. 1000 mL
1L = 86.2 mL alcohol
PROBLEM 3 Mole Fraction ( mf ) Compare M, %, ppm and ppb Which NaCl solution is most concentrated?
1 x10-3 M
0.001 mole NaCl
= 0.0585 g NaCl
= 5.85 x 10-5 g/mL
Since:1 g ~ 1mL
= 5.85 x 10-5 0.01 g NaCl
= 1 x 10-4 1 g NaCl
= 1 x 10-6 1000 g NaCl
= 1 x 10-6 This is the
These are the
Both very low. SUMMARY: Concentration Definitions
Concentration Term Ratio Molarity (M)
Molality (m) moles of solute
mass (kg) of solvent Mole fraction (mf): XA = Parts by mass
Parts by volume Both of the same unit: mole SAMPLE PROBLEM 13.4 What is the molality of a solution prepared by
dissolving 32.0 g of CaCl2 in 271 g of water? Mole fraction
(m.f.) 32.0 g CaCl2 x mass of solute
mass of solution
Volume of solute
volume of solution
mole solute + mole solvent mole(Solute)
mole (solution) What is the mole fraction of each component in a
solution of 36% HCl by mass?
Sum of mf of all components must = 1
36% HCl = 36 g HCl and 64 g H2O
36 g HCl ÷ 36.5 g/mole = 0.986 mole HCl
64 g H2O ÷ 18.0 g/mole = 3.56 mole H2O
XHCl = 0.986 ÷(0.986 + 3.56) = 0.22
XH2O = 1.00-0.22 = 0.78 moles of solute
volume (L) of solution Moles of solute per Kg of solvent The ocean contains 19,000 ppm of chloride ion.
What is the molality of this ion?
19000 ppm Cl-1 = 19000 g Cl-1 / 106 g solution
Solute Cl-1= 19000g ÷ 35.5 g/mole = 535 mole Cl-1
Solvent H2O = 106 g – 19000 g Cl-1
= 9.8 x 105 g H2O
= 9.8 x 102 kg H2O
mCl- = 535 mole Cl-1 / 9.8 x 102 kg H2O = 0.55 m
Cl- SAMPLE PROBLEM 1(b) (a) Find the concentration of calcium (in ppm) in a
3.50-g pill that contains 40.5 mg of Ca. molality = mole CaCl2
110.98 g CaCl2 = 0.288 mole CaCl2
0.288 mole CaCl2
271 g H2O x 103 g
= 1.06 m CaCl2 Page 5 Chem 210 / WANG / Chapter 13A
Properties of Solution SAMPLE PROBLEM 13.5 SAMPLE PROBLEM 13.6 A sample of rubbing alcohol contains 142 g of isopropyl
alcohol (C3H7OH) and 58.0 g of water. What are the
mole fractions of alcohol and water?
moles ethylene glycol = 142 g mole = 2.36 mol
moles water = 58.0g 18.02 g = 3.22 mol H2O Hydrogen peroxide is a powerful oxidizing agent used
in concentrated solution in rocket fuels and in dilute
solution a a hair bleach. An aqueous solution H2O2 is
30.0% by mass and has a density of 1.11 g/mL.
(a) Molality SAMPLE PROBLEM 13.7 Compare M, mf and m Which NaCl solution is most concentrated?
1 x10-2 M
1000 ppm (b) Mole fraction of H2O2 2.36 mol C3H8O
2.36 mol C2H8O + 3.22 mol H2O C3H8O m.f. = 0.423
m.f. 0.423 3.22 mol H2O
2.36 mol C3H8O + 3.22 mol H2O (c) Molarity H2O m.f. = 0.577
m.f. 0.577 Fall 2009 UM-SJTU JI Page 6 ...
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- Spring '09
- Solubility, Chem