1. Chemistry
Unit 10:
Solutions

2. 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

3. soluble: “will dissolve in”
miscible: refers to two liquids that mix
evenly in all proportions
-- e.g.,
food coloring and water

4. Factors Affecting the Rate of Dissolution
1. temperature
As temp. , rate
As size ,
rate
2. particle size
3. mixing
With more mixing,
rate
We can’t control
this factor.
4. nature of solvent or solute

5. Classes of Solutions
aqueous solution:
solvent = water
water = “the universal solvent”
amalgam:
solvent = mercury
e.g.,
dental amalgam
tincture:
solvent = alcohol
e.g.,
tincture of iodine (for cuts)
organic solution:
solvent
contains ________
carbon
Organic solvents include
benzene, toluene, hexane, etc.

6. 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

7. -- e– are shared equally
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 H O
e.g.,
water
“Like dissolves like.”
polar + polar = solution
nonpolar + nonpolar = solution
polar + nonpolar = suspension (won’t mix evenly)

8. 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.

9. 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
C=C Cl

10. MODEL OF A SOAP MOLECULE
emulsifying agent (emulsifier): --
molecules w/both a polar AND a nonpolar end --
allows polar and nonpolar substances to mix
e.g., soap
eggs
lecithin
detergent
MODEL OF A SOAP MOLECULE
NONPOLAR
HYDROCARBON
TAIL
POLAR
HEAD
Na1+

11. Hard water contains minerals w/ions like Ca2+, Mg2+,
soap vs detergent
made from animal
and vegetable fats
made from petroleum --
works better in hard
water
NONPOLAR
HYDROCARBON
TAIL
POLAR HEAD
Na1+
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).
oil
H2O
micelle: a liquid droplet covered
w/soap or detergent molecules

13. unsaturated: sol’n could hold more solute; below the line
Temp. (oC)
Solubility
(g/100 g H2O)
KNO3 (s)
KCl (s)
HCl (g)
SOLUBILITY
CURVE
sudden stress
causes this
much ppt
Solubility
how much solute
dissolves in a given
amt. of solvent at a
given temp.
unsaturated: sol’n could hold more
solute;
below the line
saturated: sol’n has “just right” amt.
of solute;
on the line
supersaturated: sol’n has “too much” solute
dissolved in it;
above the line

14. Solids dissolved Gases dissolved
in liquids in liquids To
Sol. To
Sol.
[O2]
As To ,
solubility ___
As To ,
solubility ___

15. Using an available solubility
curve, classify as
unsaturated, saturated,
or supersaturated.
per 100 g H2O
80 g 30oC
unsaturated
45 g 60oC
saturated
30 g 30oC
supersaturated
70 g 60oC
unsaturated

16. (Unsaturated, saturated, or supersaturated?)
Per 500 g H2O,
100 g 40oC
saturation point
@ 40oC for 100 g H2O
= 63 g KNO3
So saturation pt.
@ 40oC for 500 g H2O
= 5 x 63 g
= 315 g
100 g < 315 g
unsaturated

17. Describe each situation below.
(A) Per 100 g H2O,
100 g 50oC.
unsaturated;
all solute dissolves;
clear sol’n.
(B) Cool sol’n (A) very
slowly to 10oC.
supersaturated;
extra 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 and becomes visible

18. Glassware – Precision and Cost
beaker vs. volumetric flask
1000 mL + 5% mL mL
When filled to
1000 mL line,
how much liquid
is present?
WE DON’T KNOW.
5% of 1000 mL
= 50 mL
min:
max:
Range:
min:
max:
Range:
950 mL mL
1050 mL mL
imprecise; cheap
precise; expensive

19. ** Measure to part of meniscus w/zero slope.
water in
grad. cyl.
mercury in
grad. cyl.
measure to
bottom
measure to
top
** Measure to part of meniscus w/zero slope.

21. 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.

22. Selected units of concentration
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 or here)
mol
-- commonly used for minerals or
contaminants in water supplies M L C.
molarity (M) = moles of solute
L of sol’n
-- used most often in this class

23. mol L M
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 )
= mol
A. What mass sodium hydroxide is this?
Na1+
OH1–
NaOH
3.38 mol
= g NaOH
B. What mass magnesium phosphate is this?
Mg2+
PO43–
Mg3(PO4)2
3.38 mol
= g Mg3(PO4)2

24. Find molarity if 58.6 g barium hydroxide are in 5.65 L sol’n. mol
OH1–
Ba(OH)2
58.6 g
= mol
= M Ba(OH)2
You have 10.8 g potassium nitrate. How many mL
of sol’n will make this a 0.14 M sol’n?
K1+
NO31–
KNO3
10.8 g
= mol
(convert to mL)
= L
= mL

25. Molarity and StoichiometryV P
mol
M L
V of gases
at STP
PbI2
V of sol’ns KI 1 2 1 2
__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.

26. 1 2 1 2 89 g PbI2 = 0.39 mol KI mol L M = 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?
Strategy: (1)
(2)
Find mol KI needed
to yield 89 g PbI2.
PbI2 KI
M L
Based on (1), find volume
of 4.0 M KI sol’n.
89 g PbI2
= 0.39 mol KI
mol L M
= L of 4.0 M KI

27. Cu
CuSO4
How many mL of a M
CuSO4 sol’n will react w/excess
Al to produce 11.0 g Cu?
M L
Al3+
SO42–
__CuSO4(aq) + __Al(s)  3 2
__Cu(s) 3
+ __Al2(SO4)3(aq) 1
11.0 g Cu
= mol CuSO4
mol L M
= L
= mL of M CuSO4

29. 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.
C = conc.
D = dilute
Dilution Equation:
MC VC = MD VD

30. Conc. H3PO4 is 14.8 M. What volume of concentrate
is req’d to make L of M H3PO4?
MC VC = MD VD
14.8
(VC) =
0.500
(25)
14.8
VC = 0.845 L
How would you mix the above sol’n?
1. Measure out _____ L of conc. H3PO4.
0.845
2. In separate container, obtain ____ L of cold H2O.
~20
3. In fume hood, slowly pour H3PO4 into cold H2O.
4. Add enough H2O until L of sol’n is obtained.

31. Calc. how much conc. you need…
You have 75 mL of conc. HF (28.9 M); you need
15.0 L of 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.

32. Dissociation occurs when neutral combinations of
particles separate into ions while in aqueous solution.
sodium chloride NaCl 
Na Cl1–
sodium hydroxide NaOH 
Na OH1–
hydrochloric acid HCl 
H Cl1–
sulfuric acid H2SO4 
2 H SO42–
acetic acid CH3COOH 
CH3COO1– H1+
In general, _____ yield hydrogen (H1+) ions in aque-
ous solution; _____ yield hydroxide (OH1–) ions.
acids
bases

33. Strong electrolytes exhibit nearly 100% dissociation.
NaCl Na Cl1–
NOT in water:
1000
in aq. sol’n: 1
999
999
Weak electrolytes exhibit little dissociation.
CH3COOH CH3COO1– H1+
NOT in water:
1000
in aq. sol’n:
980 20 20
“Strong” or “weak” is a property of the substance.
We can’t change one into the other.

34. 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.

35. nonelectrolytes: solutes that DO NOT dissociate--
DO NOT conduct elec.
current (not enough ions)
-- e.g., any type of sugar

36. 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
point (NFP)
…lower FP
(freezing point depression)
…normal boiling
point (NBP)
…higher BP
(boiling point elevation)

37. Applications of Colligative Properties (NOTE: Data are fictitious.)
1. salting roads
in winter
water + more salt
water + a little salt
water BP FP
0oC (NFP)
100oC (NBP)
–11oC
103oC
–18oC
105oC

38. Applications of Colligative Properties (cont.)
2. Antifreeze (AF)
(a.k.a., “coolant”)
50% water + 50% AF
water + a little AF
water BP FP
0oC (NFP)
100oC (NBP)
–10oC
110oC
–35oC
130oC

39. Applications of Colligative Properties (cont.)
3. law enforcement
starts melting at…
finishes melting at…
penalty, if convicted
white powder
comm.
service A
120oC
150oC B
130oC
140oC
2 yrs. C
134oC
136oC
20 yrs.