Chemistry
Unit 10:
Solutions
Solution Definitions
solution: a homogeneous mixture
-- evenly mixed at the particle level
-- e.g., salt water
alloy: a solid solution of metals
-- e.g., bronze = Cu + Sn
brass = Cu + Zn
solvent: the substance that dissolves the solute
e.g., water
e.g., salt
soluble: “will dissolve in”
miscible: refers to two liquids that mix
evenly in all proportions
-- e.g., food coloring and water
Factors Affecting the Rate of Dissolution
1. temperature
2. particle size
3. mixing
4. nature of solvent or solute
As temp.
, rate
As size
rate
,
With more mixing,
rate
We can’t control
this factor
Classes of Solutions
aqueous solution: solvent = water
water = “the universal solvent”
amalgam: solvent = mercury
e.g., dental amalgam for cavities
tincture: solvent = alcohol
e.g., tincture of iodine (for cuts)
organic solution: solvent
carbon
contains ________
Organic solvents include
benzene, toluene, hexane, etc.
Non-Solution Definitions
insoluble: “will NOT dissolve in”
e.g., sand and water
immiscible: refers to two liquids that
will NOT form a solution
e.g., water and oil
suspension: appears uniform while
being stirred, but
settles over time
e.g., liquid medications,
Italian dressing
H
H–C–H
H–C–H
H–C–H
H–C–H
H
Molecular Polarity
nonpolar molecules:
-- e– are shared equally
-- tend to be symmetric
e.g., fats and oils
polar molecules:
-- e– NOT shared equally
e.g., water
H
H
O
“Like dissolves like.”
polar + polar = solution
nonpolar + nonpolar = solution
polar + nonpolar = suspension (won’t mix evenly)
Using Solubility Principles
Chemicals used by body obey solubility principles.
-- water-soluble vitamins:
e.g., vitamin C
-- fat-soluble vitamins:
e.g., vitamins A & D
Anabolic steroids and HGH are
fat-soluble, synthetic hormones.
Using Solubility Principles (cont.)
Dry cleaning employs nonpolar liquids
-- polar liquids damage wool, silk
-- also, dry clean for stubborn stains
(ink, rust, grease)
-- tetrachloroethylene was in longtime use
Cl
Cl
C=C
Cl
Cl
emulsifying agent (emulsifier):
-- molecules w/both a polar AND a nonpolar end
-- allows polar and nonpolar substances to mix
e.g., soap
eggs
lecithin
MODEL OF A SOAP MOLECULE
Na+
POLAR
HEAD
NONPOLAR
HYDROCARBON
TAIL
detergent
soap
vs.
-- made from animal
and vegetable fats
detergent
-- made from petroleum
-- works better in hard
water
Na+
POLAR HEAD
NONPOLAR
HYDROCARBON
TAIL
Hard water contains minerals w/ions like Ca2+, Mg2+,
and Fe3+ that replace Na+ at polar end of soap
molecule. Soap is changed into an insoluble
H2O
precipitate (i.e., soap scum).
HO
2
micelle: a liquid droplet covered
w/soap or detergent molecules
oil
H2O
H2O
Solubility
sudden stress
causes this
much ppt
how much solute
dissolves in a given
amt. of solvent at a
given temp.
SOLUBILITY
CURVE
KNO3 (s)
KCl (s)
Solubility
(g/100 g
H2O)
HCl (g)
Temp. (oC)
unsaturated: sol’n could dissolve more
solute; below the line
saturated: sol’n dissolves “just right” amt.
of solute; on the line
supersaturated: sol’n has “too much” solute
dissolved in it; above the line
Solids dissolved
in liquids
Gases dissolved
in liquids
[O
Sol.
2]
Sol.
To
As To ,
solubility ___
To
As To ,
solubility ___
Using an available solubility curve, classify as
unsaturated, saturated,or supersaturated.
150
KI
140
80 g NaNO3 @ 30oC
unsaturated
45 g KCl @ 60oC
saturated
50 g NH3 @ 10oC
unsaturated
70 g NH4Cl @ 70oC
supersaturated
Solubility (grams of solute/100 g H2O)
per 100 g H2O
130
120
gases
NaNO3
110
solids
100
KNO3
90
80
HCl
NH4Cl
70
KCl
NH3
60
50
40
30
NaCl
20
10
SO2
0
10
20
30
40
50
KClO3
60
70 80
90 100
(Unsaturated, saturated, or supersaturated?)
Solubility vs. Temperature for Solids
Per 500 g H2O,
120 g KNO3 @ 40oC
So saturation pt.
@ 40oC for 500 g H2O
= 5 x 70 g
= 350 g
120 g < 350 g
unsaturated
KI
140
130
Solubility (grams of solute/100 g H2O)
saturation point
@ 40oC for 100 g H2O
= 70 g KNO3
150
120
gases
NaNO3
110
solids
100
KNO3
90
80
HCl
NH4Cl
70
KCl
NH3
60
50
40
30
NaCl
20
10
SO2
0
10
20
30
40
50
KClO3
60
70 80
90 100
Describe each situation below.
Solubility (grams of solute/100 g H2O)
150
(A) Per 100 g H2O,
140 KI
o
100 g NaNO3 @ 50 C. 130
120
unsaturated;
NaNO
110
all solute dissolves;
100
KNO
clear sol’n.
90
80
(B) Cool sol’n (A) very
HCl
NH Cl
70
slowly to 10oC.
NH
60
50
supersaturated;
40
extra solute remains
30
20
in sol’n; still clear
KClO
10
SO
(C) Quench sol’n (A) in
an ice bath to 10oC.
saturated; extra solute (20 g)
can’t remain in sol’n and becomes visible
3
gases
solids
3
4
3
KCl
NaCl
3
2
0
10
20
30
40
50
60
70 80
90 100
Solubility Curve
150
KI
140
130
Solubility (grams of solute/100 g H2O)
120
gases
NaNO3
110
solids
100
KNO3
90
80
HCl
NH4Cl
70
KCl
NH3
60
50
40
30
NaCl
20
10
SO2
0
10
20
30
40
50
KClO3
60
70
80
90
100
Glassware – Precision and Cost
beaker
vs.
1000 mL + 5%
volumetric flask
1000 mL + 0.30 mL
When filled to
1000 mL line,
how much liquid
is present?
WE DON’T KNOW.
min: 950 mL
max: 1050 mL
imprecise; cheap
Range:
Range:
5% of 1000 mL = 50 mL
min: 999.70 mL
max: 1000.30 mL
precise; expensive
water in
grad. cyl.
mercury in
grad. cyl.
measure to
bottom
measure to
top
** Measure to part of meniscus w/zero slope.
Concentration…a measure of solute-to-solvent ratio
concentrated
dilute
“lots of solute”
“not much solute”
“not much solvent”
“watery”
Add water to dilute a sol’n;
boil water off to concentrate it.
Selected units of concentration
A. mass % = mass of solute x 100
mass of sol’n
B. parts per million (ppm) = mass of solute x 106
mass of sol’n
 also, ppb and ppt
(Use 109 or 1012 here)
mol
-- commonly used for minerals or
contaminants in water supplies
M
L
C. molarity (M) = moles of solute
mol
L of sol’n
M 
L
-- used most often in this class
How many mol solute are req’d to make
1.35 L of 2.50 M sol’n?
mol = M L = 2.50 M (1.35 L )
= 3.38 mol
mol
M
L
A. What mass sodium hydroxide is this?
Na+
OH–
NaOH
40.0 g 

3.38 mol 
 = 135 g NaOH
 1 mol 
B. What mass magnesium phosphate is this?
Mg2+
PO43–
Mg3(PO4)2
262.9 g 

3.38 mol 
 = 889 g Mg3(PO4)2
 1 mol 
Find molarity if 58.6 g barium hydroxide are
Dissolved in 5.65 L sol’n. Ba2+ OH–
mol
Ba(OH)2
 1 mol 
M
L
58.6 g 
 = 0.342 mol
 171.3 g 
mol 0.342 mol
M 

= 0.061 M Ba(OH)2
L
5.65 L
You have 10.8 g potassium nitrate. How many mL
of sol’n will make this a 0.14 M sol’n? K+ NO3–
KNO3
 1 mol 
10.8 g 
 = 0.1068 mol
(convert to mL)
 101.1 g 
mol 0.1068 mol
 1000 mL 


L 

= 0.763 L
 1L 
M
0.14 M
= 763 mL
mass
vol.
Molarity and
Stoichiometry
mol
M
L
mass
vol.
mol
M
L
part.
part.
PbI2
V of sol’ns
Pb(NO3)2(aq) + 2 KI (aq) 
KI
V of gases
at STP
PbI2(s) + 2 KNO3(aq)
What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2?
Strategy:
(1) Find mol KI needed to yield 89 g PbI2.
(2) Based on (1), find volume of 4.0 M KI sol’n.
Pb(NO3)2(aq) + 2 KI (aq) 
PbI2(s) + 2 KNO3(aq)
What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2?
Strategy: (1) Find mol KI needed
to yield 89 g PbI2.
(2) Based on (1), find volume
of 4.0 M KI sol’n.
PbI2
KI
 1 mol PbI2   2 mol KI 
89 g PbI2 

 = 0.39 mol KI
 461 g PbI2   1 mol PbI2 
mol
0.39 mol KI
L 

M
4.0 M KI
= 0.098 L of 4.0 M KI
How many mL of a 0.500 M
CuSO4 sol’n will react w/excess
Al to produce 11.0 g Cu?
Cu
CuSO4
3 CuSO4(aq) + 2 Al(s)  3 Cu(s) + Al2(SO4)3(aq)
 1 mol Cu   3 mol CuSO4 
11.0 g Cu 


 63.5 g Cu   3 mol Cu 
= 0.173 mol CuSO4
0.173 mol CuSO4
mol

L 
0.500 M CuSO4
M
= 0.346 L
 1000 mL 


1
L


= 346 mL of 0.500 M CuSO4
Dilutions of Solutions
Acids (and sometimes bases) are purchased in
concentrated form (“concentrate”) and are easily
diluted to any desired
concentration.
**Safety Tip:
When diluting, add acid
(or base) to water.
Dilution Equation: MC VC = MD VD
C = conc.
D = dilute
Conc. H3PO4 is 14.8 M. What volume of concentrate
is req’d to make 25.00 L of 0.500 M H3PO4?
MC VC = MD VD
14.8 (VC) = 0.500 (25)
14.8
14.8
VC = 0.845 L
How would you mix the above sol’n?
0.845 L of conc. H3PO4.
1. Measure out _____
~20 L of cold H2O.
2. In separate container, obtain ____
3. In fume hood, slowly pour H3PO4 into cold H2O.
4. Add enough H2O until 25.00 L of sol’n is obtained.
Cost Analysis with Dilutions
2.5 L of 12 M HCl
(i.e., “concentrate”)
0.500 L of
0.15 M HCl
Cost: $25.71
Cost: $6.35
How many 0.500 L-samples of 0.15 m HCl can be
made from the bottle of concentrate?
MC VC = MD VD
(Expensive water!)
Moral:
Buy the concentrate
and mix it yourself to
any desired concentration.
12 M (2.5 L) = 0.150 M (VD)
VD = 200 L
400 samples @ $6.35 ea. =
$2,540
You have 75 mL of conc. HF (28.9 M); you need
15.0 L of 0.100 M HF. Do you have enough to do
the experiment?
MC VC = MD VD
28.9 M (0.075 L) = 0.100 M (15 L)
Calc. how much conc. you need…
28.9 M (VC) = 0.100 M (15 L)
VC = 0.052 L = 52 mL needed
Yes;
we’re OK.
Dissociation occurs when neutral combinations of
particles separate into ions while in aqueous solution.
sodium chloride NaCl  Na+ + Cl–
sodium hydroxide NaOH  Na+ + OH–
hydrochloric acid HCl  H+ + Cl–
sulfuric acid H2SO4  2 H+ + SO42–
acetic acid CH3COOH  CH3COO– + H+
acids yield hydrogen (H+) ions in aqueIn general, _____
ous solution; bases
_____ yield hydroxide (OH–) ions.
Strong electrolytes exhibit nearly 100% dissociation.
NOT in water:
in aq. sol’n:
Cl–
NaCl
Na+
1000
0
0
1
999
999
+
Weak electrolytes exhibit little dissociation.
CH3COOH
CH3COO–
NOT in water:
1000
0
0
in aq. sol’n:
980
20
20
+
H+
“Strong” or “weak” is a property of the substance.
We can’t change one into the other.
electrolytes: solutes that dissociate in sol’n
-- conduct elec. current because
of free-moving ions
-- e.g., acids,
bases,
most ionic compounds
-- are crucial for many
cellular processes
-- obtained in a healthy diet
-- For sustained exercise or
a bout of the flu, sports
drinks ensure adequate
electrolytes.
nonelectrolytes: solutes that DO NOT dissociate
-- DO NOT conduct elec.
current (not enough ions)
-- e.g., any type of sugar
Colligative Properties
 properties that depend on the conc. of a sol’n
Compared to
solvent’s…
a sol’n w/that
solvent has a…
…normal freezing
…lower FP
point (NFP)
(freezing point depression)
…normal boiling
point (NBP)
…higher BP
(boiling point elevation)
Applications of Colligative Properties
(NOTE: Data is fictitious.)
1. salting roads
in winter
FP
BP
water
0oC (NFP)
100oC (NBP)
water + a little salt
–11oC
103oC
water + more salt
–18oC
105oC
Applications of Colligative Properties (cont.)
2. Antifreeze (AF)
(a.k.a., “coolant”)
FP
BP
water
0oC (NFP)
100oC (NBP)
water + a little AF
–10oC
110oC
50% water + 50% AF
–35oC
130oC
Applications of Colligative Properties (cont.)
3. law enforcement
white powder
starts
melting
at…
finishes
melting
at…
penalty, if
convicted
A
120oC
150oC
comm.
service
B
130oC
140oC
2 yrs.
C
134oC
136oC
20 yrs.
Calculations with
Colligative Properties
Adding a nonvolatile solute to a solvent decreases
the solution’s freezing point (FP) and increases its
boiling point (BP).
The freezing point depression and
boiling point elevation are given by:
DTx = Kx m i
DTx = FP depression or BP elevation
Kx = Kf (molal FP depression constant) or
Kb (molal BP elevation constant)
-- they depend on the solvent
-- for water: Kf = 1.86oC/m, Kb = 0.52oC/m
m = molality of solute
mol of solute
m 
kg of solvent
i = van’t Hoff factor
(accounts for # of
particles in solution)
In aq. soln., assume that…
1 for nonelectrolytes
-- i = __
2 for KBr, NaCl, etc.
-- i = __
3 for CaCl2, etc.
-- i = __
Jacobus Henricus van’t Hoff
(1852 – 1911)
In reality, the van’t Hoff factor isn’t always
an integer. Use the guidelines unless
given information to the contrary.
Find the FP and BP of a soln. containing 360 g barium
chloride and 2.50 kg of water.
Kf = 1.86oC/m
DTx = Kx m i
Kb = 0.52oC/m
 1 mol 
 = 1.725 mol BaCl2
360 g BaCl2 
 208.7 g 
 2.50 kg
0.690 m
i=3
DTf = Kf m i = 1.86(0.690)(3) = 3.85oC
FP = –3.85oC
DTb = Kb m i = 0.52(0.690)(3) = 1.08oC
BP = 101.08oC