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
water
soluble: “will dissolve in”
salt
miscible: refers to two gases or two liquids that form
a solution; more specific than “soluble”
-- e.g., food coloring and water
Factors Affecting the Rate of Dissolution
1. temperature
As To , rate
2. particle size
As size
3. mixing
More mixing, rate
4. nature of solvent or solute
, 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
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
H
Molecular Polarity
nonpolar molecules:
H–C–H
H–C–H
H–C–H
H–C–H
H
-- e– are shared equally
-- tend to be symmetric
e.g., fats and oils
-- e– NOT shared equally
polar molecules:
H
e.g., water
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., 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)
-- tetrachloroethylene 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
eggs
MODEL OF A SOAP MOLECULE
Na1+
POLAR
HEAD
soap
vs.
-- made from animal and
vegetable fats
NONPOLAR
HYDROCARBON
TAIL
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
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:
sol’n could hold more solute; below line
saturated:
sol’n has “just right” amt. of solute; on line
supersaturated: sol’n has “too much” solute dissolved in it;
above the line
Solids dissolved in liquids
Sol.
Gases dissolved in liquids
Sol.
To
As To , solubility
To
As To , solubility
Classify as unsaturated, saturated, or supersaturated.
per
100 g
H2O
80 g NaNO3 @ 30oC
unsaturated
45 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
Describe each situation below.
(A) Per 100 g H2O, 100 g
Unsaturated; all solute
NaNO3 @ 50oC.
dissolves; clear sol’n.
(B) Cool sol’n (A) very
Supersaturated; extra
slowly to 10oC.
solute remains in sol’n;
still clear.
(C) Quench sol’n (A) in an
ice bath to 10oC.
Saturated; extra solute
(20 g) can’t remain in
sol’n, becomes visible.
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?
beaker
volumetric flask
5% of 1000 mL = 50 mL
Range: 999.70 mL
Range: 950 mL – 1050 mL
imprecise; cheap
– 1000.30 mL
precise; expensive
water in
grad. cyl.
mercury in
grad. cyl.
~
~
~
~
~
~
~
~
Measure to part of meniscus w/zero slope.
Concentration…a measure of solute-to-solvent ratio
concentrated
dilute
“lots of solute”
“not much solute”
“watery”
Add water to dilute a sol’n; boil water off to concentrate it.
Selected
units:
A. mass % = mass of solute
mass of sol’n
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
-- used most often in this class
M 
mol
L
mol
M
L
D.
molality (m) = moles of solute
kg of solvent
7.85 kg KCl are dissolved in 2.38 L of sol’n. Find molality.
m
kg solute 7.85 kg


L sol' n
2.38 L
3.30 m KCl
24.8 g table sugar (i.e., sucrose, C12H22O11) are mixed into
450 g water. Find molality.
m
kg solute 0.0248 kg


L sol' n
0.450 L
0.055 m C12H22 O11
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?
23
 1000 mL  1 g  1 mol  6.02 x 10 m' cule 
X m' cule H2O  1000 L 




1
L
1
mL
18
g
1 mol





= 3.34 x 1028 m’cules H2O
1.5 atom F
X atoms F

1,000,000 m' cule H2O 3.34 x 1028 m' cule H2O
 1 m' c CaF2 
22
X  5.01 x 1022 at. F  times 
  2.505 x 10 m' c CaF2
 2 at. F 
1 mol

  78.1 g 
X g CaF2  2.505 x 1022 m' c 

  3.25 g CaF2
23
 6.02 x 10 m' c   1 mol 
1: 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
A. What mass sodium hydroxide is this?
 40.0 g 
X g NaOH  3.38 mol 
  135 g NaOH
1
mol


B. What mass magnesium phosphate is this?
 262.9 g 
X g Mg 3 (PO 4 )2  3.38 mol 
  889 g Mg 3 (PO 4 )2
1
mol


2: Find molarity if 58.6 g barium hydroxide are in
5.65 L sol’n.
 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
3: You have 10.8 g potassium nitrate. How many mL of
sol’n will make this a 0.14 M sol’n?
 1 mol 
10.8 g KNO 

101.1
g

  0.763 L  1000 mL  
XL 


0.140 M
 1L

3
763 mL
convert to mL
Molarity and Stoichiometry
M
V
P
M
mol
ML
mol
ML
V
P
__Pb(NO3)2(aq) + __KI (aq)  __PbI2(s) + __KNO3(aq)
1 Pb(NO3)2(aq) + 2 KI (aq)  1 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.
 1mol PbI2   2 mol KI 
 
  0.39 mol KI
X mol KI  89 g PbI2 
 461 g PbI2   1mol PbI2 
M 
mol
mol
0.39 mol KI
 L 

 0.098 L of 4.0 M KI
L
M
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?
Al3+ SO42–
__CuSO4(aq) + __Al (s)  __Cu(s) + __Al2(SO4)3(aq)
3 CuSO4(aq) + 2 Al (s)  3 Cu(s) + 1 Al2(SO4)3(aq)
 1 mol Cu   3 mol CuSO 4 
X mol CuSO 4  11 g Cu 


63.5
g
Cu
3
mol
Cu




= 0.173 mol CuSO4
M 
0.173 mol CuSO 4
mol
mol
 L 

 0.346 L
L
M
0.500 M CuSO 4
 1000 mL 
0.346 L 
 
1
L


346 mL
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 M (VC )  0.500 M (25.00 L)
VC = 0.845 L = 845 mL
How would you mix the above sol’n?
1. Measure out 0.845 L of conc. 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 sol’n is obtained.
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 (15.0 L)
Yes;
we’re OK.
2.1675 mol HAVE > 1.50 mol NEED
Dissociation occurs when neutral combinations of particles
separate into ions while in aqueous solution.
NaCl  Na1+ + Cl1–
sodium chloride
sodium hydroxide
NaOH  Na1+ + OH1–
hydrochloric acid
HCl  H1+ + Cl1–
H2SO4  2 H1+ + SO42–
sulfuric acid
CH3COOH  CH3COO1– + H1+
acetic acid
In general, acids yield hydrogen (H1+) ions
in aqueous solution; bases yield hydroxide (OH1–) ions.
Strong electrolytes exhibit nearly 100% dissociation.
NOT in water:
in aq. sol’n:
NaCl
Na1+
Cl1–
1000
0
0
1
999
999
+
Weak electrolytes exhibit little dissociation.
CH3COOH
CH3COO1–
+
H1+
NOT in water:
1000
0
0
in aq. sol’n:
980
20
20
“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  depend on conc. of a sol’n
Compared to solvent’s…
a sol’n w/that solvent has a…
…normal freezing point (NFP)
…lower FP
FREEZING PT.
DEPRESSION
…normal boiling point (NBP)
…higher BP
BOILING PT.
ELEVATION
Applications (NOTE: Data are fictitious.)
1. salting roads in winter
water
water + a little salt
water + more salt
FP
0oC (NFP)
–11oC
–18oC
BP
100oC (NBP)
103oC
105oC
FP
0oC (NFP)
–10oC
–35oC
BP
100oC (NBP)
110oC
130oC
2. antifreeze (AF) /coolant
water
water + a little AF
50% water + 50% AF
3. law enforcement
white powder
A
B
C
starts
finishes
melting at… melting at…
120oC
150oC
130oC
140oC
134oC
136oC
penalty, if
convicted
comm. service
2 years
20 years
Calculations for Colligative Properties
The change in FP or BP is found using…
Tx = Kx m i
Tx = 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
Tf = Kf m i
Boiling Point Elevation
Tb = Kb m i
Then use these in conjunction with the NFP and NBP to
find the FP and BP of the mixture.
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

Tb = Kb m i = 0.512 (0.373) (1) = 0.19oC
BP = (100 + 0.19)oC = 100.19oC
Tf = 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–
CsBr  Cs1+ + Br1–
i=2
168 g
m
mol CsBr

kg H2O
212.8 g
 0.316 m
2.50 kg
Tb = Kb m i = 0.512 (0.316) (2) = 0.32oC
BP = (100 + 0.32)oC = 100.32oC
Tf = Kf m i = –1.86 (0.316) (2) = –1.18oC
FP = (0 + –1.18)oC = –1.18oC