u10_hc

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Unit 10:
Solutions
Chemistry
Solution Definitions
solution: a homogeneous mixture
-- e.g., 
-- 
alloy: a solid solution of metals 
-- e.g., 
solvent: the substance that dissolves the solute
evenly mixed at the particle level 
salt water 
bronze = Cu + Sn
brass = Cu + Zn 
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
With more mixing,
	rate
As temp. , rate 
As size ,
 rate 
We can’t control
this factor
Classes of Solutions
aqueous solution: 
solvent = water 
amalgam:
e.g., 
tincture:
e.g., 
organic solution: 
solvent = alcohol 
dental amalgam for cavities
tincture of iodine (for cuts) 
solvent = mercury 
water = “the universal solvent” 
		solvent
contains ________ 
Organic solvents include
benzene, toluene, hexane, etc.
carbon
Non-Solution Definitions
insoluble: “will NOT dissolve in”
	e.g., 
immiscible: refers to two liquids that
		will NOT form a solution
	e.g., 
suspension: appears uniform while
		 being stirred, but
		 settles over time
	e.g., 
sand and water 
water and oil
liquid medications, 
Italian dressing 
Molecular Polarity
nonpolar molecules:	
	-- e– are shared equally 
-- tend to be symmetric 
e.g., 
e.g., 
polar molecules:	
	-- e– NOT shared equally 
“Like dissolves like.”	 
fats and oils 
H–C–H 
 H
H–C–H 
H–C–H 
H–C–H 
 H
water 
H
H
 O
polar + polar = solution 
nonpolar + nonpolar = solution 
polar + nonpolar = suspension (won’t mix evenly)
Anabolic steroids and HGH are
fat-soluble, synthetic hormones.
Using Solubility Principles
Chemicals used by body obey solubility principles. 
-- water-soluble vitamins: 	e.g., 
-- fat-soluble vitamins:		e.g., 
vitamin C 
vitamins A & D 
Dry cleaning employs nonpolar liquids 
Using Solubility Principles (cont.)
-- polar liquids damage wool, silk 
-- also, dry clean for stubborn stains
	(ink, rust, grease) 
-- tetrachloroethylene was in longtime use 
C=C
Cl
Cl
Cl
Cl
emulsifying agent (emulsifier): 
molecules w/both a polar AND a nonpolar end 
-- 
-- 
allows polar and nonpolar substances to mix 
e.g., soap 
lecithin 
eggs 
MODEL OF A SOAP MOLECULE
NONPOLAR
HYDROCARBON
TAIL
POLAR
HEAD
Na+
detergent 
soap		 vs.	 	 detergent
--				 --
made from animal
and vegetable fats 
made from petroleum 
--
works better in hard
water 
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
precipitate (i.e., soap scum). 
NONPOLAR
HYDROCARBON
TAIL
POLAR HEAD
Na+
micelle: a liquid droplet covered
 w/soap or detergent molecules
oil
H2O
H2O
H2O
H2O
Solubility 
Temp. (oC)
Solubility
(g/100 g H2O)
KNO3 (s)
KCl (s)
HCl (g)
SOLUBILITY
CURVE
unsaturated: sol’n could dissolve more
		 solute; 
saturated: sol’n dissolves “just right” amt.
	 of solute; 
supersaturated: sol’n has “too much” solute
		 dissolved in it; 
how much solute
dissolves in a given
amt. of solvent at a
given temp. 
below the line 
on the line 
above the line 
sudden stress
causes this
much ppt
Solids dissolved 		Gases dissolved
 in liquids		 	 in liquids
To
Sol.
To
 Sol.
As To ,
solubility ___ 
As To ,
solubility ___ 
 [O2]
supersaturated 
NaCl
0 10 20 30 40 50 60 70 80 90 100
Solubility (grams of solute/100 g H2O)
KI
KCl
20
10
30
40
50
60
70
80
90
110
120
130
140
100
NaNO3
KNO3
HCl
NH4Cl
NH3
KClO3
SO2
gases
solids
150
Using an available solubility curve, classify as unsaturated, saturated,or supersaturated. 
80 g NaNO3 @ 30oC 
45 g KCl @ 60oC 
50 g NH3 @ 10oC 
70 g NH4Cl @ 70oC 
per 100 g H2O
unsaturated 
saturated 
unsaturated
NaCl
0 10 20 30 40 50 60 70 80 90 100
Solubility vs. Temperature for Solids
Solubility (grams of solute/100 g H2O)
KI
KCl
20
10
30
40
50
60
70
80
90
110
120
130
140
100
NaNO3
KNO3
HCl
NH4Cl
NH3
KClO3
SO2
gases
solids
150
Per 500 g H2O,
120 g KNO3 @ 40oC 
saturation point
@ 40oC for 100 g H2O
	= 70 g KNO3 
So saturation pt.
@ 40oC for 500 g H2O
	= 5 x 70 g
	= 350 g
120 g < 350 g 
unsaturated 
(Unsaturated, saturated, or supersaturated?)
NaCl
0 10 20 30 40 50 60 70 80 90 100
Solubility (grams of solute/100 g H2O)
KI
KCl
20
10
30
40
50
60
70
80
90
110
120
130
140
100
NaNO3
KNO3
HCl
NH4Cl
NH3
KClO3
SO2
gases
solids
150
Describe each situation below.
(A) Per 100 g H2O,
 100 g NaNO3 @ 50oC. 
(B) Cool sol’n (A) very
 slowly to 10oC. 
(C) Quench sol’n (A) in
 an ice bath to 10oC. 
unsaturated;
all solute dissolves;
clear sol’n. 
supersaturated;
extra solute remains
in sol’n; still clear
saturated; extra solute (20 g)
can’t remain in sol’n and becomes visible 
Solubility Curve
NaCl
0 10 20 30 40 50 60 70 80 90 100
Solubility (grams of solute/100 g H2O)
KI
KCl
20
10
30
40
50
60
70
80
90
110
120
130
140
100
NaNO3
KNO3
HCl
NH4Cl
NH3
KClO3
SO2
gases
solids
150
Glassware – Precision and Cost 
beaker	 vs.	volumetric flask 
1000 mL + 5%		 1000 mL + 0.30 mL 
When filled to
1000 mL line,
how much liquid
is present? 
5% of 1000 mL 
= 50 mL 
min:
max:
Range:
WE DON’T KNOW.
min:
max:
Range:
imprecise; cheap 
precise; expensive 
950 mL 
1050 mL 
999.70 mL 
1000.30 mL 
** Measure to part of meniscus w/zero slope. 
water in
grad. cyl.
mercury in
grad. cyl.
measure to
bottom 
measure to
top 
Concentration…a measure of solute-to-solvent ratio
concentrated 
dilute 
Add water to dilute a sol’n;
boil water off to concentrate it. 
“lots of solute”
“not much solute”
“watery”
“not much solvent”
Selected units of concentration
A. 
mass % = mass of solute x 100
		mass of sol’n
parts per million (ppm) = mass of solute x 106
				 mass of sol’n 
B. 
 also, ppb and ppt 
-- commonly used for minerals or
 contaminants in water supplies 
molarity (M) = moles of solute
			L of sol’n 
C. 
-- used most often in this class 
 mol
 L
 M
 (Use 109 or 1012 here)
Na+
How many mol solute are req’d to make
1.35 L of 2.50 M sol’n? 
A. What mass sodium hydroxide is this? 
B. What mass magnesium phosphate is this? 
 mol
 L
 M
mol = M L 
= 2.50 M (1.35 L )
= 3.38 mol
3.38 mol
= 135 g NaOH
OH–
NaOH
Mg2+
3.38 mol
= 889 g Mg3(PO4)2
PO43–
Mg3(PO4)2
Find molarity if 58.6 g barium hydroxide are
Dissolved in 5.65 L sol’n. 
You have 10.8 g potassium nitrate. How many mL
of sol’n will make this a 0.14 M sol’n? 
 mol
 L
 M
58.6 g
Ba2+
OH–
Ba(OH)2
= 0.342 mol
= 0.061 M Ba(OH)2
K+
NO3–
KNO3
10.8 g
= 0.1068 mol
= 0.763 L
(convert to mL)
= 763 mL
Molarity and
Stoichiometry
 Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq)
What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2?
Strategy: 	(1)
 
		(2) 
Find mol KI needed to yield 89 g PbI2. 
Based on (1), find volume of 4.0 M KI sol’n. 
2
2
PbI2
KI
V of gases
at STP
V of sol’ns
 L
 M
part.
vol.
mass
mass
vol.
part.
mol
 L
 M
mol
Strategy: (1)
 
	 (2) 
Find mol KI needed
to yield 89 g PbI2. 
Based on (1), find volume
of 4.0 M KI sol’n. 
89 g PbI2
= 0.39 mol KI
= 0.098 L of 4.0 M KI
 Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq)
What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2?
2
2
PbI2
KI
How many mL of a 0.500 M
CuSO4 sol’n will react w/excess
Al to produce 11.0 g Cu? 
 CuSO4(aq) + Al(s) 
 Cu(s)
11.0 g Cu
= 0.173 mol CuSO4
= 346 mL of 0.500 M CuSO4
+ Al2(SO4)3(aq) 
3 
3 
2 
= 0.346 L
Cu
CuSO4
Cu
CuSO4
MC VC = MD VD
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: 
C = conc.
D = dilute
14.8
Conc. H3PO4 is 14.8 M. What volumeof concentrate
is req’d to make 25.00 L of 0.500 M H3PO4?
MC VC = MD VD
(VC)
=
0.500
(25)
How would you mix the above sol’n? 
1. Measure out _____ L of conc. H3PO4. 
2. In separate container, obtain ____ L of cold H2O. 
3. In fume hood, slowly pour H3PO4 into cold H2O. 
4. Add enough H2O until 25.00 L of sol’n is obtained. 
0.845
14.8
14.8
VC = 0.845
L
~20
Cost Analysis with Dilutions
2.5 L of 12 M HCl
(i.e., “concentrate”)
Cost: $25.71
0.500 L of
0.15 M HCl
Cost: $6.35
How many 0.500 L-samples of 0.15 m HCl can be
made from the bottle of concentrate?
12 M
MC VC = MD VD
(2.5 L)
=
0.150 M
VD = 
200 L 
400 samples 
@ $6.35 ea. = 
 $2,540 
Moral:
Buy the concentrate
and mix it yourself to
any desired concentration. 
(Expensive water!) 
(VD)
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? 
28.9 M
MC VC = MD VD
(0.075 L)
=
0.100 M
(15 L)
Yes;
we’re OK.
Calc. how much conc. you need…
28.9 M
(VC)
=
0.100 M
(15 L)
VC = 0.052
L
= 52 mL needed
Dissociation occurs when neutral combinations of
particles separate into ions while in aqueous solution.
In general, _____ yield hydrogen (H+) ions in aque-
ous solution; _____ yield hydroxide (OH–) ions. 
acids
bases
sodium hydroxide NaOH 
hydrochloric acid	 HCl 
sodium chloride	NaCl 
Na+ + Cl– 
Na+ + OH–
H+ + Cl– 
sulfuric acid H2SO4 
2 H+ + SO42–
acetic acid CH3COOH 
CH3COO– + H+
NOT in water: 
in aq. sol’n: 
“Strong” or “weak” is a property of the substance.
We can’t change one into the other.
NOT in water: 
in aq. sol’n: 
1000 
0 
0 
1
999 
999 
NaCl Na+ + Cl– 
1000 
0 
0 
980
20 
20 
CH3COOH CH3COO– + H+ 
Weak electrolytes exhibit little dissociation. 
Strong electrolytes exhibit nearly 100% dissociation.
electrolytes: solutes that dissociate in sol’n
-- conduct elec. current because
 of free-moving ions 
-- e.g., 
-- are crucial for many
 cellular processes 
-- obtained in a healthy diet 
-- 
acids,
bases,
most ionic compounds 
For sustained exercise or
a bout of the flu, sports 
drinks ensure adequate
electrolytes. 
nonelectrolytes: solutes that DO NOT dissociate
-- 
-- e.g., any type of sugar 
DO NOT conduct elec. 
current (not enough ions) 
Colligative Properties 
 properties that depend on the conc. of a sol’n 
Compared to
solvent’s… 
…normal freezing
 point (NFP) 
…normal boiling
 point (NBP) 
a sol’n w/that
solvent has a…
…lower FP 
(freezing point depression) 
…higher BP 
(boiling point elevation) 
water + more salt
water + a little salt
water
BP
FP
0oC (NFP)
Applications of Colligative Properties 
(NOTE: Data is fictitious.) 
1. salting roads
 in winter 
100oC (NBP)
–11oC
103oC
–18oC
105oC
50% water + 50% AF
water + a little AF
water
BP
FP
0oC (NFP)
Applications of Colligative Properties (cont.) 
2. Antifreeze (AF) 
 (a.k.a., “coolant”) 
100oC (NBP)
–10oC
110oC
–35oC
130oC
Applications of Colligative Properties (cont.) 
3. law enforcement
C
B
A
finishes melting at…
starts melting at…
white powder
penalty, if convicted
120oC
150oC
comm.
service
130oC
140oC
134oC
136oC
2 yrs.
20 yrs.
Calculations with
Colligative Properties 
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 
Adding a nonvolatile solute to a solvent decreases
the solution’s freezing point (FP) and increases its
boiling point (BP).
m = molality of solute 
i = van’t Hoff factor
 (accounts for # of
 particles in solution) 
-- i = __ for nonelectrolytes 
In aq. soln., assume that…
-- i = __ for KBr, NaCl, etc.
-- i = __ for CaCl2, etc.
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.
1
2
3
360 g BaCl2
Find the FP and BP of a soln. containing 360 g barium
chloride and 2.50 kg of water.
= 1.725 mol BaCl2
0.690 m
DTx = Kx m i
Kf = 1.86oC/m
Kb = 0.52oC/m
DTf = Kf m i
= 1.86(0.690)(3)
i = 3
= 3.85oC
FP = –3.85oC
DTb = Kb m i
= 0.52(0.690)(3)
= 1.08oC
BP = 101.08oC
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mol
 
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40.0
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262.9
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1000
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171.3
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101.1
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PbI
 
mol
 
1
KI
 
mol
 
2
M
mol
 
 
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4.0
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mol
 
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2
2
PbI
 
g
 
461
PbI
 
mol
 
1
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PbI
 
mol
 
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KI
 
mol
 
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M
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2
2
PbI
 
g
 
461
PbI
 
mol
 
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1
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1000
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Cu
 
g
 
63.5
Cu
 
mol
 
1
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Cu
 
mol
 
3
CuSO
 
mol
 
3
4
M
mol
 
 
L
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CuSO
 
M
 
0.500
CuSO
 
mol
 
0.173
 
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L
 
1
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1000
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Cu
 
g
 
63.5
Cu
 
mol
 
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Cu
 
mol
 
3
CuSO
 
mol
 
3
4
M
mol
 
 
L
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4
CuSO
 
M
 
0.500
CuSO
 
mol
 
0.173
 
=
solvent
 
of
 
kg
solute
 
of
 
mol
 
 
m
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kg
solute
 
of
 
mol
 
 
m
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g
 
208.7
mol
 
1
kg
 
2.50
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208.7
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1
kg
 
2.50
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