Use the following diagrams for questions concerning solution process.
1
)
H soln
> 0 (endothermic); product (solution) has a higher energy than reactants (solute & solvent).
Attractive forces between solute and solvent (unlike particles) are not as strong as solute-solute and solvent-solvent AF (like particles). More energy required to separate solute-solute and solvent-solvent particles than is released when form solute-solvent AF.
(
)
H soln
> 0)
A solution process with
)
H soln
> 0 (+) tends to not be spontaneous (but can be). It also depends on the entropy of solution,
)
S soln
. See page 3.

)
H soln
< 0 (exothermic); product (solution) has a lower energy than reactants (solute & solvent).
Attractive forces between solute and solvent (unlike particles) are stronger than solute-solute and solvent-solvent AF (like particles). Less energy required to separate solute-solute and solvent-solvent particles than is released when form solute-solvent AF.
(
)
H soln
< 0)
A solution process with
)
H soln
< 0 (-) tends to be spontaneous (but may not be). It also depends on the entropy of solution,
)
S soln
. See page 3.
2

3
)
H soln
= 0 (ideal); product (solution) has same energy as reactants (solute & solvent).
Attractive forces between solute and solvent (unlike particles) are similar to solute-solute and solvent-solvent AF (like particles). Energy required to separate solute-solute and solvent-solvent particles is about the same as that released when form solute-solvent AF. Generally, particles with only LF are the most likely to form ideal solutions.
(
)
H soln
= 0)
For ideal solutions (
)
H
must be positive (
)
S soln soln
= 0) and endothermic (
)
H soln
> 0) to form the change in entropy (disorder)
> 0, disorder must inc.). Generally,
)
S soln
> 0 for mixing.
Remember, as given in class:
)
H soln
=
)
H
1
+
)
H
2
+
)
H
3
)
H soln
=
)
H or sep solute
+
)
H sep solvent
+
)
H solvation
)
G =
)
H - T
C )
S and
)
G < 0 (negative) for a spontaneous process
)
G > 0 (positive) for a nonspontaneous process
If,
)
H > 0 (+) or
)
H = 0 then,
MUST have
)
S > 0 (+) (an increase in disorder) in order to have
)
G < 0 (-), at some temperature, in order for a solution to form. A
)
S > 0 (+) tends to make a process spontaneous.
If
)
H < 0 (-),
)
S can be negative (-) and still have
)
G < 0 (-), so a solution can form at some temp even with a decrease in disorder (inc. in order).

13.2) When an ionic compound dissolves in water the cation becomes surrounded by the
H
2
O molecules (see figure below).
4
Courtesy of Prentice Hall (BLB 10 th ed.)
This process is called hydration (specific term for solvation when the solute is an ion and the solvent is H
2
O). This is an Ion-Dipole AF between the ion and the
polar solvent (H
2
O). This AF is quite strong. The attractive forces we’ve discussed have the relative order in terms of increasing strength:
LF < DD < H-bonds < Ion-Dipole < Ionic & Covalent
In terms of the energy diagrams depicted on pages 1-3 above (Figure 13.4 in the text, p. 532) to which step does this interaction correspond? This corresponds to step 3 in the diagrams. This is
)
H
3
(
)
H solvation
). For this specific case of an ion in
H
2
O this is usually referred to as the heat of hydration,
)
H hyd
This step always
releases energy and is exothermic. Whenever attractive forces are formed energy is released.
Lattice energy is the main compoent of step 1,
)
H
1
, the enthalpy required to separate the solute particles (ions). If
)
H
1
is too large the dissolving process is prohibitively endothermic and the substance won’t be very soluble. In other words, not enough energy is released in step 3,
)
H
3
(
)
H hyd
), from the solutesolvent interactions (ion-dipole) to overcome the enthalpy required in the first two steps and give a small enough positive
)
H soln
so any positive change in
)
S soln
(inc in entropy) can overcome this and make the compound soluble.
If you remember I said in class that ionic compounds with high lattice energies and thus high melting points would also tend to not be very soluble due to the fact that the ions are harder to pull apart.

13.5) Which of the following pictures best represents a saturated solution?
5
Courtesy of Prentice Hall (BLB 10 th ed.)
Diagram (b) is the best representation of a saturate solution.
A saturated solution has some undissolved solid in equilibrium with dissolved solute in the solution. As much solute has dissolved as can dissolve, leaving some undissolved solid in contact with a saturated solution. Liken it to a sponge that you are using to absorb a larger puddle of liquid. After awhile the sponge can’t absorb any more liquid (it won’t hold any more) and the sponge is saturated. A saturated solution won’t “hold” any more solute (no more will dissolve).
An unsaturated solution is one in which the maximum amount of solute has not dissolved. More solute can dissolve. Again, this would be like using a sponge to absorb a drop of liquid. The sponge could still absorb more liquid and is not saturated (it is unsaturated).
A supersaturated solution is one in which there is actually more solute dissolved at a given temperature than there should be (more than in a saturated solution). This is a metastable state. You generally get such a thing by heating the solution and more solute dissolves (which is usually the case). Then you do not allow the solute to crystalize out as you cool the solution (some solutes have trouble crystallizing).
This “traps” the added heat in the system. When you do something to this metastable system (such as add a small crystal, scratch the side of the container, shake it, etc.) the excess solute (that doesn’t “belong” in the solution at the lower temperature) will crystalize out. As it does the system releases the “trapped” heat.
This is one of the processes used in reusable hot packs.

13.7) The structures of Vitamin B
6
and E are shown below.
6
Courtesy of Prentice Hall (BLB 10 th ed.)
Vitamin B
6
is a relatively small molecule consisting of a benzene-like 6-membered ring with a N atom in place of a C atom (5 C atoms and 1 N atom in a ring). It also has 3 polar -OH groups. These groups and the N atom make the molecule somewhat polar and the -OH groups can form H-bonds to water molecules in solution. The Vitamin B
6
molecule is small enough that the H
2
O molecules can effectively surround the molecule and interact with it and keep it in solution.
Vitamin E has one end that has a six-member and a 5-member ring fused together with only one -OH group (polar; DD & H-bonding) and one ether group, -O- (polar and H-bonding to water molecules). However, there is a long nonpolar hydrocarbon chain coming off the rings. The whole molecule has a slightly polar
“head” and a long nonpolar “tail”. This makes the molecule behave more like a nonpolar molecule, with large London Forces (LF). The same thing happens in fats, which are largely nonpolar. The H2O molecules can not effectively interact with the long nonpolar end of Vitamin E and surround the molecules to keep them from coming back together and keep them in solution.

7
13.11) There is a balloon made of a semipermeable membrane containing 0.25 L of a 0.2 M solution of some solute which is submerged in a 0.1 M solution of the same solute (as shown below). Assuming the volume of solution outside the balloon is large, what would you expect to happen to the solution volume inside the balloon once the system has come to equilibrium through osmosis?
Courtesy of Prentice Hall (BLB 10 th ed.)
Ideally, one would expect a volume in the balloon of 0.50 L (double what you started with). If the volume of solution outside the balloon, solvent will flow across the semipermeable membrane until the molarities of the solutions inside and outside the balloon are equal, 0.1 M. In order for the solution inside the balloon to decrease from 0.2 M to 0.1 M the volume inside the balloon must double (be twice as large as the initial volume, M
2
V
2
= M
1
V
1
), to 0.50 L. In actuality, osmosis across the membrane is not perfect (non-ideal). The solution concentration inside the balloon will be slightly greater than 0.1 M and the volume of the balloon will thus be slightly less than 0.50 L.

13.14) a)
)
H soln
=
)
H sep ions
+
)
H sep H2O
+
)
H hyd
(
)
H solvation
is
)
H hyd
when solvent is water)
You get the same general diagram for NaCl in benzene, C
6 the energy released in step 3 (
)
H solvation
H
6
, a nonpolar solvent. However,
) will be a lot less than for NaCl in H
2
O because you are trying to replace ionic AF with nonpolar LF between ions and the nonpolar C
6
H
6
solvent molecules.
)
H soln
=
)
H sep ions
+
)
H sep C6H6
+
)
H solvation
Remember,
)
H soln
> 0 when solute-solvent interactions (unlike forces) are less favorable than solute-solute & solvent-solvent interactions (like forces). This does NOT lead in general to a spontaneous dissolution (solution formation).
)
G =
)
H - T
C )
S and
)
G < 0 (negative) for a spontaneous process
If
)
H > 0 (+) to get
)
G < 0, must have
)
S > 0 (+) so an increase in randomness (disorder) is an important factor.
** See next page **
8

9
13.12) (cont.)
In these cases,
)
H
1
(
)
H sep ions
) is the same. In step 2, the energy required to separate the H
2
O molecules is greater than that required to separate C
6
H
6
molecules (
)
H sep H2O
>
)
H sep C6H6
). This is due to the fact that for H
2
O you have to break LF, DD and H-bonding to separate the molecules, while for C
6
H
6
you only break LF (dispersion forces). In step 3 (involving solutesolvent interactions), the energy released when H
2
O surrounds the Na + and Cl ions (forming strong ion-dipole AF) is much greater than that when the NaCl tries to dissolve in the nonpolar
C
6
H
6
(which can form only LF with the ions). This large difference in
)
H
3
in the two cases causes NaCl to be soluble in H
2
O but not C
6
(
)
H soln
H
6
. In both cases the solution process is endothermic
> 0) and an increase in disorder, entropy, (
)
S > 0) is necessary for a solution to form. In the case of NaCl in C
6
H
6
the increase in disorder is not enough to overcome the positive enthalpy change (
)
H soln
> 0) and make the solution form, whereas it is for NaCl in H
2
O).
Lattice energy is the main compoent of step 1,
)
H
1
, the enthalpy required to separate the solute particles (ions). If substance won’t be very soluble. In other words, not enough energy is released in step 3,
)
H
(
)
H hyd
)
H
1
is too large the dissolving process is prohibitively endothermic and the
3
), from the solute-solvent interactions (ion-dipole) to overcome the enthalpy required in the first two steps and give a small enough positive
)
H soln
so any positive change in
)
S soln
(inc in entropy) can overcome this and make the compound soluble.
If you remember I said in class that ionic compounds with high lattice energies and thus high melting points would also tend to not be very soluble due to the fact that the ions are harder to pull apart.
b)
A cation (or anion) is strongly hydrated (strong ion-dipole AF) when the ion is highly charged and small, i.e. a large charge density. The ion-dipole energy is given by:
E
I-D
Q
:
%
-------
d 2
Q: charge on ion
:
: dipole moment of solvent molecule d: distance between centers of ions & usually determined
by adding radii of the ion and the molecule.
From this one can see that this energy (AF) increases as the charge on the ion increases and the size of the ion decreases. It also depends on the size of the dipole moment of the solvent molecule, i.e. how polar the solvent molecule is. The more polar the solvent molecule the more soluble will be the ionic solute.

13.15)
10
Indicate the most important type of solute-solvent AF. Decide whether the solute and solvent are ionic, polar, covalent or nonpolar covalent (molecular).
a) CCl
4 in C
6
H
6
(benzene)
nonpolar nonpolar
LF LF
Both are nonpolar and have only LF so the only solute-solvent IAF will be
LF (dispersion).
b) CH
3
-OH in H
2
O
polar polar
LF
DD
LF
DD
H-bonding H-bonding
The solute-solvent IAF are LF, DD and H-bonding. Since H-bonding is the strongest of these it is the most important. In a solution of CH
3
OH in H the CH
3
OH can form H-bonds to H
2
2
O,
O through the lone pair e
&
on the O atom of CH
3
OH to H atoms of H
2
O. Also, the -OH H-atom can from H-bonds to lone pair e
&
on the O atom of H
2
O.
Note: Some molecules which can NOT form H-bonds in the pure liquid (or solid) CAN form H-bonds to the solvent molecules in solution
(solvent must have a H bonded to N, O, F in the solvent molecule). An example would be CH
3
- O - CH
3
. This can NOT form H-bonds as a pure substance. However, it can form H-bonds to H
2
O molecules through the lone pair e
&
on the O atom to the H atoms on the H
2
O solvent molecules.)

11
13.15) (cont.) c) KBr
ionic
ion-ion in H
2
O
polar
LF
DD
H-bonding
H
2
O
KBr (s) -----------> K + (aq) + Br
&
(aq)
Ionic solutes & polar solvents form Ion-Dipole IAF.
Ion-Dipole (weaker than ionic bonds, but stronger than LF, DD & H-bonds)
E
Q
C : ion-dipole
%
-------
d 2
Q
C :
(Att. Force, F
%
------- )
d 3
Q = chg on ion
:
= dipole moment of polar molecule d = distance between ion
& polar molecule
Note: Strength of solute-solvent IAF:
LF
.
DD < H-bonding < Ion-Dipole < Ionic & Covalent
Remember, for ionic compounds the lattice energy and AF are given by the following:
Q + Q
&
LE
%
----------
d
Q + Q
&
(Att. Force, F
%
---------- )
d 2
This is very similar to the Ion-Dipole energy and AF between ions and polar solvents.

12
13.15) (cont.) d) HCl
polar
LF
DD in CH
3
C
/
N:
polar
LF
DD
The solute-solvent IAF are LF and DD. In this case, since both molecules are about the same size and not that large and fairly polar the Dipole-Dipole
IAF are the most important.
13.16)

13.16) (cont.)
13

14
13.17) a) The energy diagrams for the solution process are given on pages 1-3. Equation
13.1 is:
)
H soln
=
)
H
1
+
)
H
2
+
)
H
3 or
)
H soln
=
)
H sep solute
+
)
H sep solvent
+
)
H solvation
If the solvent is water (
)
H solvation
=
)
H hyd
, the enthalpy of hydration)
)
H soln
=
)
H sep solute
+
)
H sep H2O
+
)
H hyd
When dissolving an ionic solute the lattice energy, LE,
Q + Q -
LE
%
-------
d
Q: charge on ion s d: distance between centers of ions & usually determined
by adding ionic radii.
is the amount of energy required to completely separate a mole of ions from the solid to its gaseous ions. This corresponds to
)
H
1
(solute-solute interactions).
b)
In eqn. 13.1, the 3 rd step,
)
H
3
, is always exothermic. Formation of attractive forces, no matter how weak, always release energy (lowers the energy of the system), relative to the energy of the isolated particles.
Step 1:
Step 2:
Step 3: separate solute particles separate solvent particles form solute-solvent AF
)
H
1
> 0 (+), endothermic
)
H
2
> 0 (+), endothermic
)
H
3
> 0 (-), exothermic

13.19)
15

13.20)
16

13.20) (cont.)
17