18
13.22)
19
13.24)
20
13.27)
21
13.27) (cont.)
22
13.27) (cont.)
23
13.29)
Remember,
“Like Dissolves Like”
Polar & ionic solutes dissolve in Polar solvents.
Ex:
NH3, CH3OH, CH3CH2OH, NaCl, etc. dissolve in H2O, NH3, CH3OH,
CH3-C-CH3, etc.
||
O
Nonpolar solutes dissolve in nonpolar solvents
Ex:
greases, oils, fats, hydrocarbons dissolve in nonpolar solvents
(hydrocarbons such as benzene, C6H6, cyclohexane, C6H12, hexane,
C6H14, toluene, C7H8, carbon tetrachloride, CCl4, etc.)
For a solute to be appreciably soluble in a solvent they both must possess similar
attractive forces. Then the energy required to separate solute particles and separate
solvent particles will be comparable to the energy released when the solute-solvent
attractions form.
ˆ For appreciable solubility solute and solvent should have similar AF
-
“Like Dissolves Like”
Hexane, C6H14, is a nonpolar solvent that has only LF. The more soluble solute
will have AF that are most like those in hexane. You are breaking LF between the
hexane molecules and if the solute and solvent both have essentially LF as the
major AF then the solute will dissolve in the nonpolar C6H14 since they will interact
effectively (replacing LF with LF when forming solute-solvent AF).
* continued on next page *
24
13.29) (cont.)
a)
CCl4
nonpolar
LF
CaCl2
ionic
Ionic AF
C6H14
nonpolar
LF
CCl4 is more soluble in C6H14. Both are nonpolar and have only LF. These
LF between solute-solute (CCl4) and solvent-solvent (C6H14) molecules must
be broken and then LF of similar size are formed between solute and solvent.
These would form a relatively ideal solution ()Hsoln . 0).
CaCl2 is ionic. In this case, the ionic (ion-ion) AF between the ions between
solute particles are broken. Only LF can form between the ions and the
nonpolar C6H14 solvent molecules in the solution. Trying to replace strong
ionic AF between solute particles with much weaker LF. This won’t happen.
In terms of the energy diagrams given on pages 1-3, this would be an
endothermic process ()Hsoln > 0). The )Hsoln would be so large and positive
that any increase in entropy is not large enough to over come this and have a
solution form.
b)
C6H6
nonpolar
LF
OH OH OH
|
|
|
CH2 - CH - CH2
polar
LF, DD
H-bonding
C6H14
nonpolar
LF
C6H6 is more soluble in C6H14. Both are nonpolar and have only LF. These
LF between solute-solute (C6H6) and solvent-solvent (C6H14) molecules are
broken and then LF of similar size are formed between solute and solvent.
These would form a relatively ideal solution ()Hsoln . 0).
CH2(OH)CH(OH)CH2(OH) is polar and has lots of H-bonding. In this case,
the LF, DD and H-bonds between solute molecules are broken. Only LF can
form between the polar solute molecules and the nonpolar C6H14 solvent
molecules in the solution. Trying to replace H-bonding between solute
particles with much weaker LF. This won’t happen. Again, this would be an
endothermic process ()Hsoln > 0) with a very large )Hsoln and any increase in
entropy is not large enough to over come this and have a solution form.
25
13.29) (cont.)
c)
CH3(CH2)6COOH
mostly nonpolar
LF (large)
DD (small)
CH3COOH
polar
LF, DD
H-bonding
C6H14
nonpolar
LF
The octanoic acid, CH3CH2CH2CH2CH2CH2CH2COOH, is more soluble in
C6H14. The octanoic acid has a small polar “head” (-COOH group) that can
form H-bonds. However, it has a larger nonpolar rod-like hydrocarbon “tail”
This large nonpolar part makes this molecule behave more like a nonpolar
molecule with fairly large LF. The nonpolar C6H14 molecules can form LF
with the mostly nonpolar octanoic acid molecules and effectively surround
the molecule. Even though the -COOH group can form H-bonds in the solute
the AF are mostly LF due to the large nonpolar part.
The acetic acid, CH3COOH, is a small molecule with the polar -COOH group
and a small nonpolar part (CH3 group). This molecule is still quite polar and
can from H-bonds. The nonpolar solvent, C6H14, won’t be able to interact
well with the fairly polar CH3COOH molecules and effectively surround
them to form a solution.
26
13.33)