Solutions
Chemical Stewardship
• Be responsible in
how you dispose of
and use chemicals.
• Chemical pollution
can travel far – and
harm organisms.
Frog with three legs – it has
mutated from chemical exposure.
Solutions
• What the solute and the solvent are
determines
–whether a substance will dissolve.
–how much will dissolve.
• A substance dissolves faster if it is stirred or
shaken.
–The particles are made smaller.
–The temperature is increased.
Why?
Solution = Solute + Solvent
• Solute - gets dissolved
• Solvent - does the dissolving
– Aqueous
– Tincture
– Amalgam
– Organic
• Polar
• Non-polar
(water)
(alcohol)
(mercury)
Dental filling
Nightmare on White Street
Chem Matters, December 1996
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
water
salt
soluble: “will dissolve in”
miscible: refers to two gases or two liquids that form
a solution; more specific than “soluble”
-- e.g., food coloring and water
Types of Solutions
Solute
Solvent
Solution
gas
gas
air (nitrogen, oxygen, argon gases)
humid air (water vapor in air)
liquid
liquid
liquid
carbonated drinks (CO2 in water)
vinegar (CH3COOH in water)
salt water (NaCl in water)
solid
solid
dental amalgam (Hg in Ag)
sterling silver (Cu in Ag)
Gaseous Solutions
gas
liquid
Liquid Solutions
gas
liquid
solid
Solid Solutions
liquid
solid
Charles H.Corwin, Introductory Chemistry 2005, page 369
Factors Affecting the Rate of Dissolution
1. temperature
2. particle size
3. mixing
4. nature of solvent or solute
As To , rate
As size
, rate
More mixing, rate
Classes of Solutions
aqueous solution: solvent = water
water = “the universal solvent”
amalgam: solvent = Hg
e.g., dental amalgam
tincture: solvent = alcohol
e.g., tincture of iodine (for cuts)
organic solution: solvent contains carbon
e.g., gasoline, benzene, toluene, hexane
Solubility
Experiment 1:
Add 1 drop of red food coloring
Before
AFTER
Miscible – “mixable”
two gases or two liquids
that mix evenly
Water
Water
Water
Water
COLD
HOT
COLD
HOT
B
A
B
A
Solubility
Experiment 2:
Add oil to water and shake
AFTER
Before
Immiscible – “does not mix”
two liquids or two gases
that DO NOT MIX
Oil
Water
Water
T0 sec T30 sec
Muddy Water: Dissolved Solids
Experiment 3:
Add soil to water, shake well, and allow to settle
AFTER
Before
Dissolved solids can be calculated
as a percentage:
v/v
(volume/volume)
w/v
(weight/volume)
w/w
(weight/weight)
Muddy
Water
Water
5% v/v soil in water
5 mL solid / 95 mL water
T1 min
T5 min
5 mL / 100 mL = 5%
Centrifugation
• Spin sample very rapidly:
denser materials go to
bottom (outside)
• Separate blood into serum
and plasma
– Serum (clear)
– Plasma (contains red blood
cells ‘RBCs’)
AFTER
Before
Serum
Blood
RBC’s
• Check for anemia (lack of iron)
A
B
C
Making solutions
• In order to dissolve - the solvent molecules
must come in contact with the solute.
• Stirring moves fresh solvent next to the solute.
• The solvent touches the surface of the solute.
• Smaller pieces increase the amount of
surface of the solute.
Water Molecule
Water is a POLAR molecule
d+
H2O
d-
d+
H+
H+
O2-
d-
Water molecules
“stick” together to
create surface tension
to support
light weight objects.
Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.
Water Molecule
• What is a polar
molecule?
dHydrogen
bond
d+
H
• How does the
polarity of water
effect this
molecule?
O
H
• Hydrogen bonds occur
between two polar
molecules, or between
different polar regions
of one large macromolecule.
• One “relatively”
negative region is
attracted to a second
“relatively” positive
region.
H
O
Electronegative
atoms
H
Hydrogen
bond
N
H
H
H
Interstitial Spaces
Oil
Oil
Oil
Oil
Oil
Oil
Oil
Non-polar
"immiscible"
Layer
dissolved
solid
Water
Water
Water
Water
Water
Water
Water
Water
Polar
red food
coloring
Dissolving of solid NaCl
Na+ ClCl-
salt
Na+
Cl-
Na+
NaCl solid
(aq)
= Na+
Animation by Raymond Chang
All rights reserved.
= Cl-
Dissolving of NaCl
H
H
O
Na+
+
+
+
+
+
-
-
hydrated ions
-
Cl-
Timberlake, Chemistry 7th Edition, page 287
+
-
Dissolving of Salt in Water
Na+
ions
Water molecules
Clions
NaCl(s) + H2O  Na+(aq) + Cl-(aq)
Solubility vs. Temperature
Solubility (g solute / 100 g H2O)
200
180
160
140
120
100
80
60
NaCl
40
20
0
20
40
60
Temperature (oC)
Timberlake, Chemistry 7th Edition, page 297
80
100
Gas Solubility
CH4
2.0
Solubility (mM)
O2
CO
1.0
He
0
10
20
30
Temperature (oC)
40
50
Electrolytes
Electrolytes - solutions that carry an electric current
strong electrolyte
NaCl(aq)
Na+ + Cl-
Timberlake, Chemistry 7th Edition, page 290
weak electrolyte
HF(aq)
H+ + F-
nonelectrolyte
Effect of Salinity on Cells
no change
isotonic solution
hemolysis
hypotonic solution
crenation
Timberlake, Chemistry 7th Edition, page 312
hypertonic solution
Isotonic
(a) Cells in dilute salt solution
Hypotonic
(b) Cells in distilled water
Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.
Hypertonic
(c) Cells in concentrated salt solution
Concentration =
#1offish
fish
volume
1 (L) (L)
Concentration = 1 “fishar”
V = 1000 mL
V = 1000 mL
V = 5000 mL
n = 2 fish
n = 4 fish
n = 20 fish
Concentration = 2 “fishar”
[ ] = 4 “fishar”
[ ] = 4 “fishar”
Concentration =
# of moles
volume (L)
V = 250 mL
n = 8 moles
[ ] = 32 molar
V = 1000 mL
V = 1000 mL
V = 5000 mL
n = 2 moles
n = 4 moles
n = 20 moles
Concentration = 2 molar
[ ] = 4 molar
[ ] = 4 molar
Making
Molar
Solutions
…from liquids
(More accurately, from stock solutions)
Concentration…a measure of solute-to-solvent ratio
concentrated
“lots of solute”
vs.
dilute
“not much solute”
“watery”
Add water to dilute a solution; boil water off to concentrate it.
remove
sample
moles of
solute
initial solution
Making a
Dilute
Solution
mix
same number of
moles of solute
in a larger volume
diluted solution
Timberlake, Chemistry 7th Edition, page 344
Concentration
“The amount of solute in a solution”
A. mass % = mass of solute
mass of sol’n
% by mass – medicated creams
% by volume – rubbing alcohol
B. parts per million (ppm)  also, ppb and ppt
– commonly used for minerals or
contaminants in water supplies
C. molarity (M) = moles of solute
L of sol’n
mol
– used most often in this class
M =
D.
mol
L
molality (m) = moles of solute
kg of solvent
M
L
Glassware – Precision and Cost
beaker
vs.
volumetric flask
When filled to 1000 mL line, how much liquid is present?
beaker
5% of 1000 mL = 50 mL
volumetric flask
1000 mL + 0.30 mL
Range: 950 mL – 1050 mL
Range: 999.70 mL– 1000.30 mL
imprecise; cheap
precise; expensive
Markings on Glassware
Beaker
500 mL + 5%
Range = 500 mL + 25 mL
475 – 525 mL
Graduated Cylinder 500 mL + 5 mL
Range = 500 mL + 5 mL
495 – 505 mL
Volumetric Flask 500 mL + 0.2 mL
Range = 499.8 – 500.2 mL
TC 20oC “to contain at a temperature of 20 oC”
TD “to deliver”
22
s
T
“time in seconds”
How to mix solid chemicals
Lets mix chemicals for the upcoming soap lab.
We will need 1000 mL of 3 M NaOH per class.
How much sodium hydroxide will I need, for five classes, for this lab?
mol
M =
L
? mol
3M =
1L
How much will this weigh?
? = 3 mol NaOH/class
x 5 classes
15 mol NaOH
1 Na @ 23g/mol + 1O @ 16g/mol + 1 H @ 1 g/mol
MMNaOH = 40g/mol
X g NaOH = 15.0 mol NaOH
40.0 g NaOH
= 600 g NaOH
1 mol NaOH
FOR EACH CLASS:
To mix this, add 120 g NaOH into 1L volumetric flask with
~750 mL cold H2O.
Mix, allow to return to room temperature – bring volume to 1 L.
How to mix a Standard
Solution
Wash bottle
Volume marker
(calibration mark)
Weighed
amount
of solute
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480
Process of Making a Standard
Solution from Liquids
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 483
Reading a pipette
Identify each volume to two decimal places
(values tell you how much you have expelled)
4.48 - 4.50 mL
4.86 - 4.87 mL
5.00 mL
www.chalkbored.com
Dilution of Solutions
Molarity
Reagent
Percent
To Prepare 1
Liter of one molar
Solution
1.05
17.45
99.8%
57.3 mL
35.05
0.90
14.53
56.6%
69.0 mL
Formic Acid (HCOOH)
46.03
1.20
23.6
90.5%
42.5 mL
Hydrochloric Acid (HCl)
36.46
1.19
12.1
37.2%
82.5 mL
Hydrofluoric Acid (HF)
20.0
1.18
28.9
49.0%
34.5 mL
Nitric Acid (HNO3)
63.01
1.42
15.9
70.0%
63.0 mL
Perchloric Acid 60% (HClO4)
100.47
1.54
9.1
60.0%
110 mL
Perchloric Acid 70% (HClO4)
100.47
1.67
11.7
70.5%
85.5 mL
Phosphoric Acid (H3PO4)
97.1
1.70
14.8
85.5%
67.5 mL
Potassium Hydroxide (KOH)
60.05
1.05
17.45
99.8%
57.3 mL
Sodium Hydroxide (NaOH)
40.0
1.54
19.4
45.0%
85.5 mL
Sulfuric Acid (H2SO4)
98.08
1.84
18.0
50.5%
51.5 mL
Solution Guide
Formula
Weight
Specific
Gravity
Acetic Acid Glacial (CH3COOH)
60.05
Ammonium Hydroxide (NH4OH)
MConc.VConc. = MDiluteVDilute
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 = concentrate
D = dilute
Concentrated H3PO4 is 14.8 M. What volume of concentrate
is required to make 25.00 L of 0.500 M H3PO4?
MC VC  MD VD  14.8 M (VC )  0.500 M (25.00 L)
VC = 0.845 L = 845 mL
How would you mix the above solution?
1. Measure out 0.845 L of concentrated H3PO4 .
2. In separate container, obtain ~20 L of cold H2O.
3. In fume hood, slowly pour [H3PO4] into cold H2O.
4. Add enough H2O until 25.00 L of solution is obtained.
Be sure to wear your safety glasses!
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?
MCVC = MDVD
28.9 M (0.075 L) = 0.100 M (15.0 L)
Yes;
we’re OK.
2.1675 mol HAVE > 1.50 mol NEED
Dilution
• Preparation of a desired solution by
adding water to a concentrate.
• Moles of solute remain the same.
M1V1  M 2V2
Dilution
• What volume of 15.8M HNO3 is required to
make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3
Preparing Solutions
How to prepare 500 mL
of 1.54 M NaCl solution
– mass 45.0 g of NaCl
– add water until total volume is
500 mL
– Shake to dissolve salt
500 mL
mark
45.0 g NaCl
solute
500 mL
volumetric
flask
Preparing Solutions
molality
molarity
1.54m NaCl in
0.500 kg of water
500 mL of 1.54M NaCl
– mass 45.0 g of NaCl
– add 0.500 kg of water
– mass 45.0 g of NaCl
– add water until total volume is
500 mL
500 mL
water
45.0 g
NaCl
500 mL
mark
500 mL
volumetric
flask
Spec-20 Instrument
Sample holder
cover
Amplifier
control knob
Light control
knob
Wavelength
control knob
Spec-20 Absorbance
Insert
Photograph
Spec-20
Meter
Wavelength control
100 % Absorbance
0 % Transmittance
Wavelength scale
Sample-holder
Light control
Zero adjust
and power control
Absorbance
0 % Absorbance
100 % Transmittance
l (wavelength)
Schematic representation of a spectrophotometer
I
Io
Monochromator
Light
Source
(1)
Detector
Meter
Sample
cell
(2)
(5)
(4)
(3)
A spectrophotometer is an instrument that measures the fraction (I/Io) of an incident
beam of light (Io) which is transmitted (I) by a sample at a particular wavelength.
Meter
For a given substance,
the amount of light
absorbed depends on:
a) the concentration;
b) the cell or path length;
c) the wavelength of light; and
d) the solvent.
Wavelength control
Wavelength scale
Sample-holder
Light control
Zero adjust
and power control
Absorbance of Chlorophyll
1.5
Frequency (Hz)
1026
cosmic
rays
1.2
10-8
1022
1024
gamma
rays
10-4
10-6
1020
1018
x-rays
10-2
1016
ultraviolet
1
102
1010
1014
106
108
infraradio
radar tele- radio
red (microwave)
vision
104
106
108
1010
1012
102
104
power
transmission
1014
1016
Absorbance
Wavelength (nm)
UV
0.9
Violet
400 nm
Blue
Green
Yellow
Orange Red
700 nm
600 nm
500 nm
0.6
0.3
0.0
300
400
500
600
Wavelength (nm)
700
800
1
Near
Infrared
Calibration Curve
(fixed wavelength)
100 % Absorbance
0 % Transmittance
?
Absorbance
0 % Absorbance
100 % Transmittance
x2
out of linear range
(too concentrated)
Concentration
Dilute sample with water 50:50. Run sample, read concentration.
What mass of CaF2 must be added to 1,000 L of water
so that fluoride atoms are present at a conc. of 1.5 ppm?
X m’cule H2O = 1000 L 1000 mL 1 g 1 mol
1L
1 mL 18 g
6.02 x1023 m’cule
= 3.34 x 1028 m’cules H2O
1 mol
1.5 atom F
X atoms F
=
1,000,000 m’cule H2O
3.34 x 1028 m’cule H2O
X = 5.01 x 1022 atoms F
times
X g CaF2 = 2.505 x 1022 molecules
1 molecule CaF2
2 atoms F
1 mol CaF2
6.02 x 1023 molecules
= 2.505 x 1022 molecules CaF2
78.1 g CaF2
= 3.25 g CaF2
1 mol CaF2
mol
M
L
How many moles solute are required to make 1.35 L of 2.50 M solution?
mol = M L = 2.50 M (1.35 L) = 3.38 mol
A. What mass sodium hydroxide is this?
X g NaOH = 3.38 mol NaOH
40.0 g NaOH
=
1 mol NaOH
135 g NaOH
B. What mass magnesium phosphate is this?
X g Mg3(PO4)2 = 3.38 mol Mg3(PO4)2
262.9 g Mg3(PO4)2
= 889 g Mg3(PO4)2
1 mol Mg3(PO4)2
Find molarity if 58.6 g barium hydroxide are in 5.65 L solution.
Step 1). How many moles barium hydroxide is this?
X mol Ba(OH)2 = 58.6 g Ba(OH)2
1 mol Ba(OH)2
171.3 g Ba(OH)2
= 0.342 mol Ba(OH)2
Step 2). What is the molarity of a 5.65 L solution containing 0.342 mol solute?
M =
mol
L
M =
0.342 mol
5.65 L
= 0.061 M Ba(OH)2
 1mol Ba(OH) 2 

58.6 g Ba(OH) 2 
 171.3 g Ba(OH) 2  
XM
5.65 L
0.061 M Ba(OH) 2
You have 10.8 g potassium nitrate.
How many mL of solution will make this a 0.14 M solution?
 1 mol 
10.8 g KNO 3 

101.1
g

  0.763 L  1000 mL  
XL 


0.140 M
1
L


76.3 mL
convert to mL
Dissociation occurs when neutral combinations of particles separate
into ions while in aqueous solution.
sodium chloride
NaCl  Na1+ + Cl1–
sodium hydroxide
NaOH  Na1+ + OH1–
hydrochloric acid
HCl  H1+ + Cl1–
sulfuric acid
H2SO4  2 H1+ + SO42–
acetic acid
CH3COOH  CH3COO1– + H1+
In general, acids yield hydrogen (H?1+) ions
? 1–) ions.
in aqueous solution; bases yield hydroxide (OH
Strong electrolytes exhibit nearly 100% dissociation.
NOT in water:
in aq. solution:
NaCl
1000
1
Na1+
0
999
+
Cl1–
0
999
Weak electrolytes exhibit little dissociation.
CH3COOH
NOT in water:
in aq. solution:
1000
980
CH3COO1–
0
20
+
H1+
0
20
“Strong” or “weak” is a property of the substance.
We can’t change one into the other.
electrolytes: solutes that dissociate in solution
-- conduct electric 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 electric current (not enough ions)
e.g., any type of sugar
Colligative Properties  depend on concentration of a solution
Compared to solvent’s…
a solution w/that solvent has a…
…normal freezing point (NFP)
…lower FP
…normal boiling point (NBP)
…higher BP
FREEZING PT.
DEPRESSION
BOILING PT.
ELEVATION
Applications (NOTE: Data are fictitious.)
1. salting roads in winter
FP
water
0oC
(NFP)
water + a little salt
–11oC
water + more salt
–18oC
BP
100oC
(NBP)
103oC
105oC
2. antifreeze (AF) /coolant
FP
water
0oC
(NFP)
BP
100oC
(NBP)
water + a little AF
–10oC
110oC
50% water + 50% AF
–35oC
130oC
3. law enforcement
starts
melting
at…
finishes
melting
at…
penalty, if
convicted
A
109oC
175oC
comm. service
B
150oC
180oC
2 years
C
194oC
196oC
20 years
white powder
Effect of Pressure on Boiling Point
Boiling Point of Water at Various Locations
Feet
Boiling
Patm
Location
above
Point
(kPa)
(C)
sea level
Top of Mt. Everest, Tibet
29,028
32
70
Top of Mt. Denali, Alaska
20,320
45.3
79
Top of Mt. Whitney, California
14,494
57.3
85
Leadville, Colorado
10,150
68
89
Top of Mt. Washington, N.H.
6,293
78.6
93
Boulder, Colorado
5,430
81.3
94
Madison, Wisconsin
900
97.3
99
New York City, New York
10
101.3
100
-282
102.6
100.3
Death Valley, California
Calculations for Colligative Properties
The change in FP or BP is found using…
DTx = Kx m i
DTx = change in To (below NFP or above NBP)
Kx = constant depending on… (A) solvent
(B) freezing or boiling
m = molality of solute = mol solute / kg solvent
i = integer that accounts for any solute dissociation
any sugar (all nonelectrolytes)……………...i = 1
table salt, NaCl  Na1+ + Cl1–………………i = 2
barium bromide, BaBr2  Ba2+ + 2 Br1–……i = 3
Freezing Point Depression
DTf = Kf m i
Boiling Point Elevation
DTb = Kb m i
Then use these in conjunction with the NFP and NBP to
find the FP and BP of the mixture.
(Kf = cryoscopic constant, which is 1.86 K kg/mol for the freezing point of water)
(Kb = ebullioscopic constant, which is 0.51 K kg/mol for the boiling point of water)
168 g glucose (C6H12O6) are mixed w/2.50 kg H2O.
Find BP and FP of mixture. For H2O, Kb = 0.512, Kf = –1.86.
i=1
(NONELECTROLYTE)
168 g
mol C 6H12 O 6
180 g
m

 0.373 m
kg H2O
2.50 kg
DTb = Kb m i = 0.512 (0.373) (1) = 0.19oC
BP = (100 + 0.19)oC = 100.19oC
DTf = Kf m i = –1.86 (0.373) (1) = –0.69oC
FP = (0 + –0.69)oC = –0.69oC
168 g cesium bromide are mixed w/2.50 kg H2O. Find BP and FP of mixture.
For H2O, Kb = 0.512, Kf = –1.86.
Cs1+ Br1–
i=2
CsBr  Cs1+ + Br1–
168 g
m
mol CsBr

kg H2O
212.8 g
 0.316 m
2.50 kg
DTb = Kb m i = 0.512 (0.316) (2) = 0.32oC
BP = (100 + 0.32)oC = 100.32oC
DTf = Kf m i = –1.86 (0.316) (2) = –1.18oC
FP = (0 + –1.18)oC = –1.18oC
Molarity and Stoichiometry
M=
M
mol
L
V
M
mol
mol
V
mol = M L
P
ML
ML
P
__
1 Pb(NO3)2(aq) + __
2 KI (aq)  __
1 PbI2(s) + __
2 KNO3(aq)
What volume of 4.0 M KI solution is required to yield 89 g PbI2?
Stoichiometry for Reactions in Solution
Step 1) Identify the species present in the combined solution,
and determine what reaction occurs.
Step 2) Write the balanced net ionic equation for the reaction.
Step 3) Calculate the moles of reactants.
Step 4) Determine which reactant is limiting.
Step 5) Calculate the moles of product or products, as required.
Step 6) Convert to grams or other units, as required.
Stoichiometry steps for reactions in solution
1 Pb(NO3)2(aq) + 2 KI (aq)  1 PbI2(s) + 2 KNO3(aq)
? L 4.0 M
89 g
What volume of 4.0 M KI solution is required 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 solution.
X mol KI = 89 g PbI2
M=
mol
L
L=
1 mol PbI2 2 mol KI
461 g PbI2 1 mol PbI2
= 0.39 mol KI
mol
0.39 mol KI
= 0.098 L of 4.0 M KI
=
M
4.0 M KI
How many mL of a 0.500 M CuSO4 solution will react
w/excess Al to produce 11.0 g Cu?
Al3+ SO42–
__CuSO4(aq) + __Al (s)  __Cu(s) + __Al2(SO4)3(aq)
3 CuSO4(aq) + 2 Al (s)  3Cu(s) + 1Al2(SO4)3(aq)
x mol
11 g
X mol CuSO4 = 11 g Cu
mol
M =
L
1 mol Cu
63.5 g Cu
mol
L =
M
3 mol CuSO4
= 0.173 mol CuSO4
3 mol Cu
0.173 mol CuSO4
0.500 M CuSO4
0.346 L 1000 mL = 346 mL
1L
= 0.346 L
Stoichiometry Problems
• How many grams of Cu are required to react
with 1.5 L of 0.10M AgNO3?
Cu + 2AgNO3  2Ag +
?g
1.5
L
Cu(NO3)2
1.5L
0.10M
.10 mol
AgNO3
1 mol
Cu
1L
2 mol
AgNO3
63.55
g Cu
= 4.8 g
1 mol
Cu
Cu
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Limiting Reactants
• 79.1 g of zinc react with 0.90 L of 2.5M HCl.
Identify the limiting and excess reactants.
How many liters of hydrogen are formed at
STP?
Zn + 2HCl 
79.1 g 0.90 L
2.5M
ZnCl2 + H2
?L
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Limiting Reactants
Zn + 2HCl 
79.1 g 0.90 L
2.5M
79.1
g Zn
ZnCl2 + H2
?L
1 mol
Zn
1 mol
H2
22.4 L
H2
65.39
g Zn
1 mol
Zn
1 mol
H2
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
= 27.1 L
H2
Limiting Reactants
Zn + 2HCl 
79.1 g 0.90 L
2.5M
0.90
L
2.5 mol
HCl
1 mol
H2
1L
2 mol
HCl
ZnCl2 + H2
?L
22.4
L H2
= 25 L
H2
1 mol
H2
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Limiting Reactants
Zn: 27.1 L H2
HCl: 25 L H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
A Hydrocarbon
• Typical petroleum product
• Non-polar
CH2
CH3
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
C18H38
CH2
CH2
CH2
CH2
CH3
CH2
Oil and Water Don’t Mix
• Oil is nonpolar
• Water is polar
“Like dissolves like”
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 470
Hydrogenation
tub (soft)
margarine
+ H2
heat, nickel
catalyst
vegetable oils
Timberlake, Chemistry 7th Edition, page 570
shortening
stick margarine
Molecular Polarity
-- e– are shared equally
-- tend to be symmetric
nonpolar molecules:
e.g., fats and oils
H
H
O
polar molecules:
-- e– NOT shared equally
e.g., water
“Like dissolves like.”
polar + polar = solution
nonpolar + nonpolar = solution
polar + nonpolar = suspension (won’t mix evenly)
H
H–C–H
H–C–H
H–C–H
H–C–H
H
Using Solubility Principles
Chemicals used by body obey solubility principles.
-- water-soluble vitamins: e.g., vit. C
-- fat-soluble vitamins:
e.g., vits. A, D
Dry cleaning employs nonpolar liquids.
-- polar liquids damage wool, silk
-- also, dry clean for stubborn stains (ink, rust, grease)
-- (tetra) perchloroethylene is in common 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
detergent
lecithin
MODEL OF A SOAP MOLECULE
Na1+
POLAR
HEAD
NONPOLAR
HYDROCARBON
TAIL
eggs
Saponification
Process of making soap from animal fat or vegetable oil using a base.
O
CH2 – O – C – (CH2)14CH3
CH2 – OH
O
O
CH – O – C – (CH2)14CH3 + 3 NaOH
CH – OH + 3 Na+ -OC – (CH2)14CH3
O
CH2 – O – C – (CH2)14CH3
glyceryl tripalmitate
(tripalmitin)
CH2 – OH
sodium
hydroxide
glycerol
3 sodium palmitate
(soap)
A Phospholipid
polar head
nonpolar tails
(a) chemical structure of a phospholipid
polar head
nonpolar tails
(b) simplified way to draw a phospholipid
Timberlake, Chemistry 7th Edition, page 576
A Model of a Cell Membrane
Polar
Nonpolar
Cholesterol
Timberlake, Chemistry 7th Edition, page 587
Proteins
Phospholipid
bilayer
Cleaning Action
of Soap
Micelle
Timberlake, Chemistry 7th Edition, page 573
SOAP
-- made from animal and
vegetable fats
vs.
DETERGENT
-- made from petroleum
-- works better in hard water
Hard water contains minerals w/ions like Ca2+, Mg2+, and Fe3+ that
replace Na1+ at polar end of soap molecule. Soap is changed into
an insoluble precipitate (i.e., soap scum).
micelle: a liquid droplet covered
w/soap or detergent molecules
Solvation
“Like Dissolves Like”
NONPOLAR
POLAR
NONPOLAR
POLAR
Solvation
 Soap
/ Detergent
• polar “head” with long nonpolar “tail”
• dissolves nonpolar grease in polar water
micelle
Lava Lamp
It is…
H
H
O
Polar mixture
Water
Polyethylene glycol
H H
Chlorinated paraffin
Paraffin from kerosene
H H
H H
H-O-C-C-O-C-C-O-C-C-O-C-C-O-H
H H
mixture
theNonpolar
moment
H H
H H
H H
H H
Cl H H H H Cl H H H H H Cl H H H H Cl H H H
H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-H
H H H Cl H H H H Cl H H H H H Cl H H H Cl H
Heat transfer coil
H H H H H H H H H H H H H H H H H H H H
Bulb gives heat and light
prehistoric
here to stay
H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-H
H H H H H H H H H H H H H H H H H H H H
Dialysis
A semi-permeable membrane
allows small particles to pass
through while blocking larger
particles.
Dialysis is used to clean blood
when people suffer kidney
failure.
Solution, Suspension, Colloid
Timberlake, Chemistry 7th Edition, page 309
Solubility
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form
increasing concentration
Solubility vs. Temperature for Solids
140
KI
130
Solubility
Table
shows the dependence
of solubility on temperature
Solubility (grams of solute/100 g H2O)
120
NaNO3
110
gases
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517
10 20 30 40 50 60 70 80 90 100
Solubility
Solubility
maximum grams of solute that will dissolve
in 100 g of solvent at a given temperature
varies with temperature
based on a saturated solution
H
2
3
4
5
6
7
1
Li
He
Be
B
C
N
O
F
2
Ne
3
4
Na Mg
5
Al
6
Si
7
P
8
S
9
Cl
10
Ar
13 14 15 16
Cu Zn Ga Ge As Se
17
Br
18
Kr
36
Xe
11
K
12
Ca Sc
Ti
V
Cr Mn Fe Co
Ni
19 20
Rb Sr
21
Y
22 23 24 25 26 27 28 29 30 31 32 33 34
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
35
I
37 38
Cs Ba
39
40
Hf
55
Fr
87
56
Ra
88
*
W
41
Ta
42
W
43 44 45
Re Os Ir
72 73 74 75 76 77
Rf Db Sg Bh Hs Mt
46 47 48 49 50
Pt Au Hg Tl Pb
51 52
Bi Po
53 54
At Rn
78
83
85
79
80
81
82
Hydrogen Bonding
84
86
H 2O
H2Te
0
-100
-200
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 443
Boiling Points
of Covalent Hydrides
100
104 105 106 107 108 109
Boiling point (oC)
1
H 2S
H2Se
2
3
4
Period of X (H2X)
5
Boiling Points of Covalent Hydrides
H 2O
Boiling point (oC)
100
0
H2Te
-100
H 2S
H2Se
SnH4
GeH4
SiH4
-200 CH4
50
100
Molecular mass
150
Group16 Hydrogen Compounds
Compound
Molar Mass
Melting
Point (oC)
Boiling
Point (oC)
H fusion
(cal/mol)
H vapor
(cal/mol)
H2O
18
0.0
100.0
1440
9720
H2S
34
-85.5
-60.7
568
4450
H2Se
81
-60.4
-41.5
899
4620
H2Te
130
-48.9
-2.2
1670
5570
Precipitation Reaction Between
AgNO3 and KCl
AgCl
AgNO3(aq) + KCl(aq)
Ag+ + NO3- + K+ + Cl-
Ag+
K+
Cl-
NO3-
+
NO3- Ag
Ag+
?
ClK+
Cl-
ClK+
NO3-
in silver
nitrate
solution
unknown white solid
unknown white solid
in potassium
chloride
solution
Ag+ + NO3- + K+ + Cl-
Ag+ + Cl- + K+ + NO3product
Ag+
AgNO3(aq) + KCl(aq)
product
AgCl(s) + KNO3(aq)
AgCl precipitate
AgNO3(aq) + KCl(aq)  KNO3 (aq) + AgCl(s)
Clogged Pipes – Hard Water
Step 1: Acid rain is formed
H2O + CO2
H2CO3
carbonic acid
Step 2: Acid rain dissolves limestone
H2CO3 + CaCO3
Ca(HCO3)2
H2CO3 + MgCO3
Mg(HCO3)2
limestone
‘hard’ water
Water softener
Pipes develop Scales
Step 3: Hard water is heated and deposits scales
Ca(HCO3)2
D
CaCO3(s) + H2O + CO2
scales on pipes
Mg(HCO3)2
D
MgCO3(s) + H2O + CO2
1 gram = 1/7000th pound
300 gallons / day with 10 grain water supply
No More Hard Water Scale
No More Ugly Hard Water Spotting
Protects Plumbing and Appliances
Saves Money on Cleaning Products
Water Purification
Cation
Exchanger
Hard
Water
H+
OH-
H+
(b)
Fe3+
OH-
Deionized
Water
OH-
H+
(a)
Ca2+
H+
Mg2+
Na+
Anion
Exchanger
H+
OH-
H+
OH-
(c)
OH-
H+
OH-
H+
OH-
Hard water is softened by exchanging Na+ for Ca2+, Mg2+, and Fe3+.
Corwin, Introductory Chemistry 2005, page 361
(a)
(b)
(c)
The cations in hard water are exchanged for H+.
The anions in hard water are exchanged for OH-.
The H+ and OH- combine to give H2O.
Which ions are removed from hard water
to produce soft water?
Na+(aq) + H (resin)
Cl-(aq) +
(resin) OH
Na (resin)
(resin) Cl
+
+
H+(aq)
OH-(aq)
Notice that the ion exchange resin
produces both hydrogen ions and hydroxide ions
which can readily combine to give water.
H+(aq) + OH- (aq)
H2O (aq)
The net result is that the resin removes all ions from water passing through the deionizing system.
Hard water is softened by exchanging Na+ for Ca2+, Mg2+, and Fe3+.
Corwin, Introductory Chemistry 2005, page 361
Reverse Osmosis
semipermeable
membrane
fresh
water
pressure
salt
water