Solutions
Solution Definitions
• Solution: homogeneous mixture, evenly mixed at the
particle level (like salt water).
Solute: substance
being dissolved
Solvent: present in
greater amount that
can dissolve the
solute
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
Solution Definitions
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 the term
“soluble”
Solubility
Solubility - maximum grams of solute
that will dissolve in 100 g of solvent at a
given temperature
varies with temperature
Solubility
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form
increasing concentration
Solubility  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:
saturated:
supersaturated:
above the line
solution could hold more solute; below line
solution has “just right” amt. of solute; on line
solution has “too much” solute dissolved in it;
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
Temperature (°C)
Classify as unsaturated, saturated, or supersaturated.
per
100 g
H2O
80 g NaNO3 @ 30oC
unsaturated
60 g KCl @ 60oC
saturated
50 g NH3 @ 10oC
unsaturated
70 g NH4Cl @ 70oC
supersaturated
Per 500 g H2O, 120 g KNO3 @ 40oC
saturation point @ 40oC for 100 g H2O = 66 g KNO3
So sat. pt. @ 40oC for 500 g H2O = 5 x 66 g = 330 g
120 g < 330 g
unsaturated
Solubility
 Solids
are more soluble at...
• high temperatures.

Gases are more soluble at...
low temperatures
Describe each situation below.
(A) Per 100 g H2O, 100 g
NaNO3 @ 50oC.
Unsaturated; all solute
dissolves; clear solution.
(B) Cool solution (A) very
slowly to 10oC.
Supersaturated; extra
solute remains in solution;
still clear.
(C) Quench solution (A) in
an ice bath to 10oC.
Saturated; extra solute
(20 g) can’t remain in
solution, becomes visible.
Non-Solution Definitions
insoluble: “will NOT dissolve in”
e.g., sand and water
immiscible: refers to two gases or two liquids that will NOT form a solution
e.g., water and oil
suspension: appears uniform while being stirred, but settles over time
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
Solutions
• What the solute and the solvent are
determines:
–whether a substance will dissolve.
–how much will dissolve.
Factors Affecting the Rate of Dissolution
1. temperature
2. particle size
3. mixing
4. nature of solvent or solute
As T , rate
As size
, rate
More mixing, rate
Solid Solubility
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
Particle Size and Mixing
• 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.
Molecular Polarity
• Nonpolar molecules: electrons are shared
equally
– Molecule has no charge
• e.g. Ethane
Polar Molecules
• Electrons are not
equally shared
• Molecule has
slight charge
“Like dissolves like”
(Polarity)
d+
H+
H+
O2-
d-
Solubility Rules
General Rules for Solubility of
Ionic Compounds (Salts) in Water at 25 oC
1. Most nitrate (NO31-) salts are soluble.
2. Most salts of Na+, K+, and NH4+ are soluble.
3. Most chloride salts are soluble. Notable exceptions
are AgCl, PbCl2, and Hg2Cl2.
4. Most sulfate salts are soluble. Notable exceptions
are BaSO4, PbSO4, and CaSO4.
5. Most hydroxide compounds are only slightly
soluble.* The important exceptions are NaOH and
KOH, Ba(OH)2 and Ca(OH)2 are only moderately
soluble.
6. Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-) salts are only slightly soluble.
*The terms insoluble and slightly soluble really mean the same
thing. Such a tiny amount dissolves that it is not possible to
detect it with the naked eye.
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 218
Oil and Water Don’t Mix
•
•
•
•
•
Oil is nonpolar
Water is polar
“Like dissolves like”
polar + polar = solution
nonpolar + nonpolar =
solution
• polar + nonpolar =
suspension (won’t mix
evenly)
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 470
Solvation
 Soap
/ Detergent
• polar “head” with long nonpolar “tail”
• dissolves nonpolar grease in polar water
micelle
Solubility of Common Compounds
Soluble compounds
NO3- salts
Na+, K+, NH4+ salts
Cl-, Br-, I- salts
Except for
Ag+, Hg22+, Pb2+
those containing
SO42- salts
Except for
2+, Pb2+, Ca2+
Ba
those containing
Insoluble compounds
S2-, CO32-, PO43- salts
OH1- salts
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 218
Except for
those containing
Na+, K+, Ca2+
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.
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
Molarity (M) =
# 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
Glassware
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
Range: 950 mL – 1050 mL
imprecise; cheap
volumetric flask
1000 mL + 0.30 mL
Range: 999.70 mL– 1000.30 mL
precise; expensive;
good for making solutions
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
How to mix solid chemicals
How many grams of sodium hydroxide would we need to make 100. mL of 3.0 M
NaOH?
mol
? mol
M =
3M =
L
1L
How much will this weigh?
1 Na @ 23g/mol + 1O @ 16g/mol + 1 H @ 1 g/mol
MMNaOH = 40g/mol
X g NaOH = 3.0 mol NaOH
40.0 g NaOH
= 120 g NaOH
1 mol NaOH
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.
mol
M
L
How many moles of 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 of sodium hydroxide is this?
X g NaOH = 3.38 mol NaOH
40.0 g NaOH
=
1 mol NaOH
135 g NaOH
B. What mass of 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 of 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
æ 1 mol Ba(OH)2 ö
58.6 g Ba(OH)2 ç
÷
è 171.3 g Ba(OH)2 ø
=
5.65 L
= 0.0605 M Ba(OH)2
0.0605 M Ba(OH)2
You have 10.8 g potassium nitrate.
How many mL of solution will make this a 0.14 M solution?
æ 1 mol ö
10.8 g KNO3 ç
÷
æ 1000 mL ö
101.1
g
è
ø
XL =
= 0.763 L ç
÷ =
è 1L ø
0.140 M
76 mL
convert to mL
Molality
moles of solute
molality (m) 
kg of solvent
0.25 mol
0.25m 
1 kg
mass of solvent only
1 kg water = 1 L water
Molality
 Find
the molality of a solution containing
75 g of MgCl2 in 250 mL of water.
75 g MgCl2
1 mol MgCl2
95.21 g MgCl2 0.25 kg water
mol
m
kg
= 3.2m MgCl2
Molality
 How
many grams of NaCl are required to
make a 1.54 m solution using 0.500 kg of
water?
0.500 kg water 1.54 mol NaCl
1 kg water
1.5 mol
1.5m 
1 kg
58.44 g NaCl
1 mol NaCl
= 45.0 g NaCl
Preparing Solutions
molality
1.54m NaCl in
0.500 kg of water
– mass 45.0 g of NaCl
– add 0.500 kg of water (volume
will be just over 500 total mL)
molarity
500 mL of 1.54M NaCl
– 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
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
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+
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
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, (a sugar, nonelectrolyte) are mixed w/2.50 kg H2O.
Find BP and FP of mixture. For water, 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) = –0.694oC
FP = (0 + –0.69)oC = –0.694oC
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?
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
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 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 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