UNIT VIII – SOLUTIONS
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I. Mixtures – non-chemical combinations of 2 or more substances; composition can vary
* can be easily separated without changing the components
1) homogeneous mixture – one that appears the same throughout
* solutions
2) heterogeneous mixture – one in which the separate components are visible
* colloids – only one example; many other types of heterogeneous sol’ns
Matter
Mixtures
Homogeneous
Substances
Heterogeneous
Compounds
Elements
II. Types of Colloids
* colloid – a suspension of tiny particles (but bigger than atoms, ions, or molecules) in some
medium – important note: colloids don’t settle like suspensions
Examples
Fog, aerosol sprays
Smoke, airborne bacteria
Whipped cream, soap suds
Milk, mayonnaise
Paint, clays, gelatin
Marshmallow, polystyrene foam
Butter, cheese
Ruby glass
Dispersing medium
(“solvent”)
gas
gas
liquid
liquid
liquid
solid
solid
solid
Dispersed substance
(“solute”)
liquid
solid
gas
liquid
solid
gas
liquid
solid
Colloid type
aerosol
aerosol
foam
emulsion
sol
solid foam
solid emulsion
solid sol
III. Solution – a homogeneous mixture in which the substances are separated and mixed at an atomic level
* 5 properties:
1) Particles are spread evenly.
2) Particles do not settle out of solution.
3) A liquid solution is clear and transparent Tyndall Effect – light is diffracted by a
colloid
4) Particles can’t be filtered out.
5) Solution is in a single phase.
IV. Solution Terminology
A. solute – the substance being dissolved
B. solvent – the medium in which the solute dissolves
C. aqueous solution – a solution in which the solvent is water
D. soluble – a substance that is able to dissolve in another substance (e.g. NaCl is soluble in
water)
- in aqueous solution = 1g solute/100g water
E. insoluble – a substance that is not able to dissolve in another
UNIT VIII – SOLUTIONS
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V. Types of Solutions
Component Component
1
2
State of
Solution
Gas
Gas
Gas
Gas
Liquid
Liquid
Gas
Solid
Solid
Liquid
Liquid
Liquid
Solid
Liquid
Liquid
Solid
Solid
Solid
Examples
air
HCl, soda
H2 gas in Palladium metal
ethanol in water, antifreeze (ethylene glycol in
water)
saltwater, sugar in water
brass (Cu/Zn), bronze (Cu/Ni)
VI. A Molecular View of the Solution Process
* “piranha and cow theory” – think of it as requiring three processes:
1) separation of solute (endothermic)
2) separation of solvent (endothermic)
3) combination of solute and solvent (exothermic)
* ∆H for most solutions: #1 + #2 > #3 result? endothermic (in most cases, only slightly)
* in a few cases (KOH, NaOH): #1 + #2 << #3 result? exothermic (very much in these cases)
* nature favors many solution processes because they create more disorder (increase entropy)
VII. Solutions of Liquids in Liquids
* miscible liquids are able to dissolve into each other
- this is because their bond strengths are similar
- CCl4 and C6H6 – both nonpolar (dispersion forces), so they are miscible
- H2O and C2H5OH (ethanol) – both polar with H-bonding, so they are miscible
- “like dissolves like” – H2O is immiscible with CCl4
- acetic acid (CH3COOH) is miscible in both polar and nonpolar things
VIII. Solutions of Solids in Liquids
A) Ionic Crystals – generally soluble only in polar substances (water) because they have forces
that are comparable to ionic forces (if ionic bond is too strong, substance becomes
less soluble often precipitates if formed quickly in water)
B) Covalent Crystals – bonds are generally too strong don’t dissolve in anything
C) Molecular Crystals – varies based on nature of substance
1) naphthalene (C10H8) is nonpolar (dispersion forces) so soluble in CCl4 (as is I2)
2) Urea (H2NCONH2) is polar, so soluble in water
D) Metallic Crystals – not soluble in anything (delocalized electrons make a strong bond)
* however, some reactive metals will chemically react with some solvents
+1
-1
____ Na(s) + ____ H2O(l) Na + OH + H2
UNIT VIII – SOLUTIONS
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IX. Concentration Units
A) Percent by Mass
% by mass =
g solute
x100
g solution
A sample of 0.892g of KCl is dissolved in 54.6g of water. What is the percent by mass?
mass solution = 0.892g + 54.6 g = 55.492g
%=
0.892g
x100% = 1.61%
55.492g
B) Mole Fraction (X)
X=
mol A
mol of mixture
A chemist prepared a solution by adding 200.4g of pure ethanol (C2H5OH) to 143.9g of water. Calculate
the mole fractions of these two components.
200.4 g C 2 H 5OH ∗
143.9 g H 2O ∗
1 C 2 H 5OH
= 4.36 mol C 2 H 5OH
46 g C 2 H 5OH
1 H 2O
= 7.99 mol H 2O → 4.36 + 7.99 = 12.35 mol total
18 g H 2 O
4.36
= 0.353
12.35
7.99
=
= 0.647
12.35
X C 2 H 5OH =
X H 2O
C) Molarity (M) & molality (m)
Molarity =
moles solute
liters solution
molality =
moles solute
kg solvent
Calculate the molality of a sulfuric acid solution containing 24.4g of sulfuric acid in 198g of water.
24.4 g H 2SO 4 ∗
1 H 2SO 4
0.249 mol H 2SO 4
=
= 1.26m
0.198 kg
98.1 g H 2SO 4
How many grams of NaOH are required to make 50θ mL of a 2.00 M solution?
40.0 g NaOH
n = MV = (2.00 mol )(0.500 L) = 1.00mol ∗
= 40.0 g NaOH
L
1 mol NaOH
UNIT VIII – SOLUTIONS
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X. A comparison of concentration units
1) Calculate the molarity of an aqueous 0.396m glucose (C6H12O6) solution. The density of the solution is
1.16 g/mL.
0.396m → For every 1000g H 2 O you have 0.396 mol C 6 H12 O 6
0.396mol C 6 H12 O 6 ∗
18ΘgC 6 H12 O 6
= 71gC 6 H12 O 6 + 1000 g H 2 O = 1071g sol' n
1molC 6 H12O 6
1 mL
= 923mL sol' n = 0.923 L sol' n
1.16 g
0.396 mol
M=
= 0.429 M
0.923L
1071g sol' n ∗
2) The density of a 2.45 M aqueous methanol (CH3OH) solution is 0.976 g/mL. What is the molality of
the solution?
1L sol' n = 1000mL sol' n ∗
0.976 g
= 976 g sol' n
1mL
32 g CH 3OH
n = MV = (2.45 mol )(1L) = 2.45mol CH 3OH ∗
= 78.4 g CH 3OH
L
1mol CH OH
3
976g sol' n - 78g CH 3OH = 898g H 2 O = 898 kg H 2 O
m=
2.45mol CH 3OH
= 2.73m
0.898kg
3) Calculate the molality of a 35.4% aqueous solution of phosphoric acid (H3PO4).
35.4 g H 3 PO 4
→ 100 − 35.4 = 64.6 g H 2 O = 0.0646kg H 2 O
100 g sol' n
1 mol H 3PO 4
35.4g H 3PO 4 ∗
= 0.361mol H 3 PO 4
98 g H 3PO 4
m=
0.361mol H 3PO 4
= 5.59m
0.0646kg H 2O
UNIT VIII – SOLUTIONS
XI. The Effect of Temperature on Solubility
A. saturated solution – a solution which
is holding as much solute as it is able at that
temperature
B. unsaturated solution – a solution
which is able to hold more solute at that
temperature
C. supersaturated solution – a solution
which is holding more solute than it is able at that
temperature
* How is this possible? If a saturated
solution is allowed to cool, and no crystals are
able to form often a seed crystal is needed
D. fractional crystallization – a means of
separating a mixture of solutes by understanding
their solubilities
*example? A mixture of KNO3 and NaCl
can be separated due to the drastic difference in
their solubilities at different temperatures
E. Gas solubility – note that the solubility
decreases as T increases
* thermal pollution
* When hot water is introduced to a lake or stream, the solubility of gases, such as O2,
decreases. Lower dissolved oxygen (DO) counts can support less fish
1) How many grams of KCl could be dissolved in 100g of water at 50oC? in 250g of water?
43 g KCl
x
(43 g KCl)(250g H 2 O)
=
→x=
100 g H 2 O 250 g H 2 O
100 g H 2 O
x = 107.5 g KCl
2) If you had 200 mL (assume density of 1.00 g/mL) of a saturated KNO3 solution at 50oC, and you
cooled it to 20oC, how many grams of KNO3 crystals could you expect?
50 o →
88 g KNO3
x
=
→ x = 176 g KNO3
100 g H 2O 200 g H 2O
20 o →
34 g KNO3
x
=
→ x = 68 g KNO3
100 g H 2 O 200 g H 2 O
176 g KNO3 − 68 g KNO3 = 108g KNO3 crystals
XII. The Effect of Pressure on Gas Solubility
* Henry’s Law – the solubility of a gas is directly proportional to the pressure applied
* Henry’s Law is not followed when the gas reacts with water:
NH3(g) + H2O(l)
NH4+1(aq) + OH-1(aq)
CO2(g) + H2O(l)
H2CO3(aq)
* Although increasing the pressure will cause the reaction to proceed to the right more, the
changes that occur are chemical in nature. Releasing the pressure will cause the
reactions to proceed more to the left
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UNIT VIII – SOLUTIONS
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XIII. Colligative Properties of Nonelectrolyte Solutions
* Colligative properties – properties of solutions that vary from the original solvent, and do not
depend on the type of solute dissolved
A) Vapor Pressure Lowering: Raoult’s Law
* vp is lowered by the addition of any solute
∆P
= XPsolvent
If you dissolve 120g of urea (H2NCONH2) in 500 mL of water at 25oC, where the vapor pressure is
23.76 mmHg, what will the vapor pressure of the solution be?
120 g H 2 NCONH 2 ∗
500 g H 2O ∗
1 mol H 2 NCONH 2
= 2mol H 2 NCONH 2
60 g H 2 NCONH 2
1 mol H 2O
= 27.7mol H 2O + 2 mol H 2 NCONH 2 = 29.7mol total
18 g H 2 O
 27.7mol 
(23.76mmHg ) = 22.16mmHg
∆P = X solvent Ptotal = 

29
.
7
mol


B) Boiling Point Elevation and Freezing Point Depression
∆Tf = Kfm
∆Tb = Kbm
* for water: Kf = 1.86 oC/m, and Kb = 0.52 oC/m
Using the urea example above, calculate the boiling and freezing points of the solution prepared.
molality =
2mol solute
= 4m
0.500kg H 2 O
o
∆T f = (1.86 C )(4m) = 7.44 o C → T f = 0 o C − 7.44 o C = −7.44 o C
m
o
∆Tb = (0.52 C )(4m) = 2.08 o C → Tb = 100 o C + 2.08 o C = 102.08 o C
m
C) Osmotic Pressure – caused by a separation of pure solvent and solution across a semipermeable membrane (allows passage of solvent molecules, but not solute)
π = MRT
The average osmotic pressure of seawater is about 30.0 atm at 25oC. What is the molarity of seawater?
M=
π
RT
=
30.0atm
= 1.23 mol = 1.23M
L
atm • L 

 0.0821
(298 K )
mol • K 

UNIT VIII – SOLUTIONS
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D) Using Colligative Properties to Determine Molar Mass
*freezing point depression good for smaller molecules
*osmotic pressure good for larger molecules
1) A 7.85g sample of a compound with empirical formula C5H4 is dissolved in 301g of benzene
(Kf = 5.12 oC/m). The freezing point is 1.05oC below that of pure benzene. What are the molar
mass and molecular formula of the compound?
o
∆T f = K f m → 1.05o C = (5.12 C )m → m = 0.205molal
m
mol solute
x
molality =
→ 0.205m =
→ x = 0.0617 mol
kg solvent
0.301kg benzene
m
7.85 g
127
FM = =
= 127 g
→ FM C5 H 4 = 64 →
= 2 → mf : C10 H 8
mol
n 0.0617 mol
64
2) A solution is prepared by dissolving 35.0g of hemoglobin in enough water to make up 1.00L in volume.
If the osmotic pressure of the solution is found to be 10.0 mmHg at 25oC, calculate the molar mass
of hemoglobin.
1atm
= 0.01316atm
760mmHg
0.01316atm
π
M=
=
= 5.378 x10 − 4 mol ∗1.00 L = 5.378 x10 − 4 mol
L
RT (0.0821)(298 K )
m
35.0 g
FM = =
= 65,000 g
−4
mol
n 5.378 x10 mol
10.0mmHg ∗
XIV. Colligative Properties of Electrolyte Solutions
* need to add the van’t Hoff factor (i) which roughly corresponds to the number of ions
dissolved in solution per formula unit e.g. NaCl (i~2), MgCl2 (i~3)
∆Tf = iKfm
∆Tb = iKbm
π = iMRT
Calculate the freezing and boiling points of water if you dissolved 35.6g of NaCl in 155mL of water.
(i = 1.9)
35.6 g NaCl ∗
1 mol NaCl
= 0.609mol ÷ 0.155kg H 2O = 3.93molal
58.5 g NaCl
o
∆T f = (1.9)(1.86 C )(3.93m) = 13.9 o C → T f = −13.9 o C
m
o
∆Tb = (1.9)(0.52 C )(3.93m) = 3.9 o C → T f = 103.9 o C
m
UNIT VIII – SOLUTIONS
Solvent
water (H2O)
carbon tetrachloride
(CCl4)
chloroform (CHCl3)
benzene (C6H6)
carbon disulfide (CS2)
ethyl ether (C4H10O)
camphor (C10H16O)
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Boiling Point
(oC)
Kb (oC –
kg/mol)
Freezing Point
(oC)
Kf (oC –
kg/mol)
100.0
76.5
0.51
5.03
0
-22.99
1.86
3θ
61.2
80.1
46.2
34.5
208.0
3.63
2.53
2.34
2.02
5.95
-63.5
5.5
-111.5
-116.2
179.8
4.70
5.12
3.83
1.79
4θ