1. Chemistry
Chapter 14:
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. 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

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

10. 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

11. 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

12. 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.

13. 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

14. 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(mol/L) (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

15. 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

16. 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.

17. 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

18. 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

19. 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

20. 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.

21. Cost Analysis with Dilutions
2.5 L of 12 M HCl
(i.e., “concentrate”)
0.500 L of
0.15 M HCl
Cost: $25.71
Cost: $6.35
How many L-samples of 0.15 m HCl can be
made from the bottle of concentrate?
MC VC = MD VD
(Expensive water!)
12 M
(2.5 L) =
0.500 M
(VD)
Moral:
Buy the concentrate
and mix it yourself to
any desired concentration.
VD =
200 L
400 samples
@ $6.35 ea. =
$2,540

22. 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.