AP Chemistry – Solutions
solution: homogeneous mixture; solid, liquid, or gas
solvent: substance present in the greatest amount
solute: also present; there might be more than one
Intermolecular forces (IMFs) operate between
solvent and solute particles. Solutions form
when solute-solvent IMFs are comparable
to solute-solute IMFs.
--
+
d
H
O
d–
H
solvent particles
“pull” solute ions
from the crystal
d+H
H
O
+
O
+–+–+–+–+–+–
–+–+–+–+–+–+
H
H
-- then solvation:
solvent particles surround solute
particles (if solvent is H2O, we say “hydration”)
Energy Changes and Solution Formation
(1) solute particles are separated
from the crystal lattice;
DH1 > 0
(2) solvent particles separate to
make room for solute particles;
DH2 > 0
(3) solvent and solute particles are
attracted to each other;
DH3 < 0
S = DHsoln
--
(3) above isn’t very (–) for
oil and water and, with (1)
and (2) requiring energy
input, this doesn’t form a solution.
If DHsoln is too (+)
(i.e., too endo),
soln won’t form.
For polar/nonpolar, (3) isn’t (–) enough, and
(1) and (2) lead to a DHsoln that is too (+).
Spontaneous processes tend to occur...
exo
(A) when DHsoln is (–), (i.e., ____)
(B) when the disorder of the system increases
** NOTE:
(B) is MORE influential in
determining whether or not
soln forms
The enthalpy of solution formation (i.e., DHsoln) for
ammonium nitrate is (+), but the system’s disorder
increases from NH4NO3(s) to NH4+(aq) + NO3–(aq), and
so solution formation is still spontaneous.
Sometimes it is not readily apparent whether a solution has been formed or
whether a chemical reaction took place. To help you decide, consider
the following:
If the product is evaporated to dryness, a solution would give you what you started with.
e.g.,
NaCl(s) + H2O(l)  NaCl(aq)
Ni(s) + 2 HCl(aq)  NiCl2(aq) + H2(g)
(SOLN)
(RXN)
Solubility
Crystallization is the opposite of the solution process.
SOLUTE + SOLVENT
SOLUTION
crystallization
-- When the rates of solution and crystallization
are equal, __________
equilibrium is established.
A saturated solution of NaCl,
in which the rates of solution
and crystallization are equal.
solution
solubility: the amount of solute needed to form a
saturated solution in a given quantity of
solvent under given conditions of T and P
saturated: soln is in eq. w/undissolved solute
i.e., there is solid at the bottom
unsaturated: more solute could dissolve
i.e., soln is clear (MIGHT be colorless)
supersaturated: the amount of dissolved solute
exceeds the solubility
--
soln has a clear, water-like appearance, but
is VERY unstable
--
addition of a seed crystal causes excess
solute to crystallize, leaving a sat. soln.
(w/visible solid)
A supersaturated solution crystallizing
upon the addition of a seed crystal. The
resulting solution is then saturated.
http://www.youtube.com/watch?v=HnSg2cl09
PI
http://www.youtube.com/watch?feature=ends
creen&v=yxi6nxAqyew&NR=1
Sol. Curve for a Typical Substance
Dissolved in a Liquid
solid in liquid
supersaturated
Solubility
(g/100 g H2O)
saturation limit
(i.e., solubility)
gas in liquid
unsaturated
T (oC)
-- for solids, as T , sol. ___
-- for gases, as T , sol. ___
Factors Affecting Solubility
Solute-Solvent Interactions
-- As IMFs between solute and solvent increase,
increases
solubility _________.
miscible: describes pairs of liquids that mix in
all proportions (v. immiscible)
d+
d+
d+
d–
d–
Methanol, which is used to fuel race cars,
is miscible with water due to its highly
polar nature.
Low molar mass alcohols are completely miscible in
water, due to H-bonding of hydroxyl group (–OH);
as molar mass increases, the polarity of the alcohol
molecule...
decreases (it behaves more like a pure
hydrocarbon) and miscibility decreases.
e.g., CH3OH vs. CH3CH2CH2CH2OH
-- Substances with similar IMFs tend
to be soluble in one another;
“like dissolves like.” (pol/pol and np/np)
-- Some network solids aren’t
soluble in either polar or nonpolar solvents because of
strong forces within the solid.
Pressure Effects
-- Pressure has no effect on the solubility of solids in
liquids, but as P increases, gas solubility ___.
-- Henry’s law:
Sg = k Pg
Sg = solubility of the gas in
the solution (M)
k = Henry’s law constant; it
depends on solute, solvent,
and temp. (M/pres. unit)
Pg = partial pressure of the
gas over the solution
(pres. unit)
William Henry
(1775 – 1836)
A bottled soft drink at 25oC has CO2 gas at a pressure
of 5.0 atm over the liquid. If the partial pressure of CO2
in the atmosphere is 4.0 x 10–4 atm and the Henry’s
law constant for CO2 over water at 25oC is
3.1 x 10–2 M/atm, calculate the solubility of the CO2
both before and after the bottle is opened.
Sg = k Pg
BEFORE
Sg = 3.1 x 10–2 M/atm (5.0 atm)
= 0.16 M
(fresh)
AFTER
Sg = 3.1 x 10–2 M/atm (4.0 x 10–4 atm)
= 1.2 x 10–5 M
(flat)
Ways of Expressing Concentration
qualitative:
concentrated v. dilute
quantitative:
mass of component
mass % 
x 100
mass of solution
6
ppm =
x 10
ppb =
x 109
ppt =
x 1012
1 ppm  1 mg/L of soln
moles of component
mole fraction, X 
total moles
moles of solute
molarity, M 
liters of soln
moles of solute
molality, m 
kg of solvent
Unlike molarity, molality doesn’t
change with temp. because...
mass remains constant w/
changing T.
(V changes w/T.)
To go between molarity and
molality, you need…
the soln’s density.
A 5.5-g sample of well water contains 0.75 mg of
lead ions. In ppm, find the concentration of lead ions.
mass of comp.
ppm 
x 10 6
mass of soln.
0.75 x 106 g
ppm 
x 106
5.5 g
= 0.14 ppm
140 ppb
(!!!)
The federal limit for lead in
drinking water is 15 ppb.
If a commercial bleach is 4.35% sodium hypochlorite
by mass, calculate the bleach’s mole fraction and
molality of the sodium hypochlorite.
100 g
bleach
4.35 g NaClO
95.65 g H2O
 1mol 


 74.5 g 
 1 mol 


 18.0 g 
0.05839
X
0.05839  5.3139
0.05839 mol
m
0.09565 kg
= 0.05839 mol NaClO
= 5.3139 mol H2O
= 0.0109
= 0.610 m
Colligative Properties
Use this eq. once for every
substance in mix. that has
a measureable VP.
-- these depend on the concentration of particles in a
solution, but not... the kind of particles
(A) Adding a volatile/nonvolatile
solute to a solvent increases/
decreases the solution’s vapor
pressure (VP).
Raoult’s law:
PA, mix = XA PoA
PA, mix = VP of particle type “A” above the mixture
XA = mole fraction of particle type “A” in the mixture
PoA = VP of “pure A”
Francois
Marie Raoult
(1830 – 1901)
Ideal solutions obey Raoult’s law. Such solutions
have...
-- a low concentration
(i.e., are relatively dilute)
-- solute and solvent particles that are similar
in size and have similar IMFs
A solution of sodium chloride and water has a vapor
pressure of 0.854 atm at 100.oC. Find the mole
fraction of sodium chloride.
(Key #1: NaCl is nonvolatile, so VP above
the mixture is due entirely to H2O.)
 PH2O, mix = XH2O PoH2O
PA, mix = XA PoA
(Key #2: The VP of H2O @ 100.oC should be recognizable.)
0.854 atm = XH2O (1.00 atm)
 XH2O = 0.854
and so X is 0.146 for all other particles
If the solute were sucrose, Xsucrose = 0.146.
But here… XNaCl = 0.073
because then… XNa+ + XCl– = 0.146
A solution contains 89.7 g ethanol and 241.4 g water.
What is the vapor pressure above the mixture at 100.oC,
if ethanol’s vapor pressure at this temp. is 1694 torr?
(Find VP of each substance, then
use Dalton’s law to find Ptot.)
CH3CH2OH
 1 mol 


 46 g 
 1 mol 


 18 g 
89.7 g eth
241.4 g H2O
= 1.95 mol eth
= 13.41 mol H2O
= 215 torr
Ptot = 879 torr
PH2O, mix = XH2O PoH2O
= 0.873(760)
XH2O = 0.873
Ptot, mix = PH2O, mix + Peth, mix
Peth, mix = Xeth Poeth
= 0.127(1694)
Xeth = 0.127
= 664 torr
(B) Adding a nonvolatile solute to a solvent decreases
the solution’s freezing point (FP) and increases its
boiling point (BP).
Phase diagram
(not water’s)
P
pure solvent
solution
solid
liquid
Note that FP
and BP .
1 atm
gas
T
NFP
NBP
The freezing point depression and
boiling point elevation are given by:
DTx = Kx m i
DTx = FP depression or BP elevation
Kx = Kf (molal FP depression constant) or
Kb (molal BP elevation constant)
-- they depend on the solvent
-- for water: Kf = 1.86oC/m
Kb = 0.52oC/m
m = molality of solute
i = van’t Hoff factor
(accounts for # of
particles in solution)
In aq. soln., assume that…
1 for nonelectrolytes
-- i = __
2 for KBr, NaCl, etc.
-- i = __
3 for CaCl , etc.
-- i = __
2
In reality, the van’t Hoff factor isn’t always
an integer. Use the guidelines unless
given information to the contrary.
Jacobus Henricus van’t Hoff
(1852 – 1911)
Find the FP and BP of a soln. containing 360. g barium
chloride and 2.50 kg of water.
Kf = 1.86oC/m
DTx = Kx m i
Kb = 0.52oC/m
 1 mol 


 208.7 g 
360 g BaCl2
= 1.725 mol BaCl2
 2.50 kg
0.690 m
i=3
DTf = Kf m i
= 1.86(0.690)(3)
= 3.85oC
FP = –3.85oC
DTb = Kb m i
= 0.52(0.690)(3)
BP = 101.08oC
= 1.08oC
Camphor, C10H16O, has an NFP of 179.8oC and a
Kf of 40.0oC/m. When 0.186 g of a nonelectrolytic
substance is dissolved in 22.01 g of camphor,
the
mixture’s new freezing point is 176.7oC. Find the
unknown’s molar mass.
DTf = Kf m i
3.1oC
40.0
1 (organic = nonelec.)
(Multiply by
0.02201 kg
camphor to get…)
Solve for m = 0.0775 mol unknown
kg camphor
1.71 x 10–3 mol unk.
0.186 g unk.
=
1 mol unk.
X g unk.
X = 109 g/mol
(C) Adding a nonvolatile solute to a solvent increases
the solution’s osmotic pressure.
more conc. (in terms of solute) = higher osmotic pres.
Osmosis is the net movement of solvent away
from a soln. w/a lower solute [ ] toward a soln. w/a
higher solute [ ]. Another way
to say this is that osmosis is
the diffusion of a solvent
– i.e., from an area of higher
solvent [ ] to an area of lower
solvent [ ] – through a
semipermeable membrane.
Dialysis tubing used with
various sugar solns.
p
osmotic pressure, p:
the external pressure
req’d to prevent osmosis
“p”
Solvent A
(pure
solvent)
Solution B
(mixture)
solvent tends to flow
;
application of p... prevents flow
A soln’s osmotic pressure p can be thought of as a
negative pressure; that is, the greater a soln’s p, the
greater is the tendency for solvent to flow _____
INTO the
solution.
p
p
hypertonic solns: large p; “conc.” (in terms of solute)
hypotonic solns: small p; “dilute” (in terms of solute)
conc. in salt
dil. in salt
dil. in H2O
conc. in H2O
net H2O flow
semipermeable
membrane
hypertonic
hypotonic
Reverse osmosis is the process of applying a pressure (P > p) such
that solvent is forced to flow (“against its will”) from low solvent conc.
to high solvent conc. Reverse osmosis is sometimes used in the
desalination (or purification) of water.
osmotic pressure equation:
pV=nRTi
n = # of moles of particles
V = solution volume, in L
R = 8.314 L-kPa/mol-K = 0.08206 L-atm/mol-K
T = absolute temp. (i.e., in K)
i = van’t Hoff factor
An early form of the osmotic pressure
equation was proposed by van’t Hoff.
The equation was improved to its
current form by Harmon Northrop
Morse, an American chemist who
lived from 1848 to 1920.
1.5 mg of a certain protein are dissolved in water
to make 10.0 mL of soln. The soln’s osmotic pressure
was found to be 2.35 torr at 25oC. Calculate the
protein’s molar mass.
p V  nR T
pV
0.3133 (0.010)

n
8.314 (298)
RT
= 1.2645 x 10–6 mol
1.5 x 103 g
Xg

6
1.2645 x 10 mol 1 mol
X = 1200 g/mol