1. The electrostatic Coulomb attraction between bare ions is very powerful. The melting
point of NaCl(s) for instance, the temperature required to separate the Na+ and Cl–
ions enough that they can flow past one another as opposed to remaining fixed in their
crystal packing, is 804 ◦ C.
As another example, formation of a plasma (a “gas” with positive ions and free electrons
that have been separated from one another) requires temperatures in excess of 7000 K
or extremely powerful electric discharges. Without this input of energy, the cations will
simply recombine with the electrons because of the Coulomb attraction.
Given this information, explain how a salt like NaCl can possibly dissolve in H2 O. As
part of your explanation, indicate the local structure of the water molecules surrounding
the Na+ and Cl– ions in an aqueous solution.
H2 O is an extremely polar molecule which is capable of stabilizing both positive and
negative ions by ion-dipole interactions, a feature that is not present in molten salts
or plasmas. The water molecules directly adjacent to the Na+ and Cl– ions in solution
do not behave like water molecules in the “bulk” (molecules far away from Na+ or
Cl– ions that only see other water molecules). In contrast to bulk H2 O molecules, the
H2 O molecules adjacent to the ions are highly ordered in a way that stabilizes the ion.
Around a cation like Na+ , the water molecules orient their oxygen atoms (which carry
a δ−) inward towards the ion to stabilize it. Around a anion like Cl– they orient their
hydrogens (δ+) inwards towards the ion which also stabilizes it.
Furthermore, the orienting behavior of the water molecules around the charges attenuates the Coulombic interaction between charged centers(you need to take a physics
electricity and magnetism class to really understand this, but it’s because to the high
dielectric constant of H2 O which is directly given by its high dipole moment).
Fundamentally, the fantastic stability of these ion dipole interactions is what makes the
dissolution of salts in H2 O and the overcoming of their Coulombic attraction possible
at normal temperatures.
2. 10.0 mL of 1.00 M HCl is added to 50.0 mL of 0.5 M NH3 (Kb = 1.8 × 10−5 ).
Calculate the pH of the resulting solution.
To model this process, we break it up into two parts. The first is a reaction that goes to
completion and sets up our “initial” amounts of species for the second reaction, which
is an equilibrium.
H + NH3 −→ NH+
4
+
−
−
*
NH3 + H2 O −
NH
)−−
4 + OH
If we take the first reaction to completion, we have a limiting reagent problem. The
number of moles of HCl added (which is really justH+ ions since it dissociates completely) is:
10.0 mL HCl × 1.00
mol
= 10.0 mmols HCl
L
and the number of moles of NH3 before the addition of HCl is
50.0 mL NH3 × 0.5
mol
= 25.0 mmols HCl
L
Because the reaction has 1:1 stoichiometry and there are fewer moles of HCl, HCl is the
limiting reagent. After they react (assumed to completion) we would therefore have 15
mmols of NH3 remaining and 10 mmols of NH+
4 . In order to enter these into our ICE
table, we need to express them as concentrations. Note that the total volume of the
solution is now 10.0 mL + 50.0 mL = 60.0 mL .
15.0 mmols
= 0.25 M
60.0 mL
10.0 mmols
= 0.167 M
=
60.0 mL
[NH3 ]initial =
[NH+
4 ]initial
These “initial” concentrations are somewhat fictitious, but they’re an important part of
the calculation. They represent what the concentration of the ammonia and ammonium
would be after the 1st reaction has totally gone to completion but the second reaction
hasn’t done anything.
NH3
H2 O
I
0.25
C
−x
E 0.25 − x
-
−
−
*
)
−
−
NH+
4
0.167
+x
0.167 + x
OH–
0
+x
x
Now using this ICE table, we can solve for x, the concentration of OH– at equilibrium.
We get
Kb = 1.8 × 10−5
x(0.167 + x)
=
0.25 − x
x = 2.7 × 10−5
Note: just use a graphing calculator
This gives us a pOH = − log(2.7 × 10−5 ) = 4.6, and a pH of 14 − 4.6 = 9.4.
3. For each pair, indicate which is more acidic. Why?
vs.
(a)
(b) H3 PO4 vs. H3 AsO4
(c) H2 ClCOOH vs. H2 BrCOOH
(d) 1M NH3 vs. 0.01 M NaOH
What we’re predominantly looking is whether the conjugate base for one species is more
stable than the other or not. The big difference in stability that controls the ∆G and
K for these reactions is how well the conjugate base can stabilize the excess charge.
(a) The key here is RESONANCE. Draw out the Lewis structure for 4-nitrophenol (the
first one). You should have a 1+ formal charge on the nitrogen and a 1- formal
charge on one of the two oxygens attached to the nitrogen (it’s really halfway on
both by resonance). Now, draw the conjugate base of 4-nitrophenol. There is an
important resonance structure that moves charge from the deprotonated oxygen
towards the nitrogen through the π system of the ring.
This resonance delocalizes the negative charge, thus stabilizing the conjugate base
and making 4-nitrophenol a strong acid.
(b) H3 AsO4 is a stronger acid than H3 PO4 . This can be predicted or rationalized based
on size and polarizability. Arsenic is bigger than phosphorous, so the H2 AsO–4 ion
delocalizes that negative charge over a bigger volume than does H2 PO–4 . Also, As
is more polarizable than P (again because it’s bigger), and thus is better able to
stabilize an extra negative charge.
(c) H2 ClOOH is a stronger acid than H2 BrCOOH because Cl is a more electronegative
substituent than Br. In general, putting these halogen substituents onto acetic
acid (or any acid) makes them more acidic because it stabilizes the build up of
charge on the conjugate base through inductive effects. This is all based on
electronegativity, so Cl is more effective than Br.
(d) √
The [OH– ] of 1 M NH3 is (if this isn’t obvious
√ to you, write out the ICE table),
Kb , which means the pH is 14 + log10 Kb = 11.6.
The pH of 0.01 M NaOH is 14 + log10 (0.01) = 12, so the NH3 solution is more
acidic.
4. For each of the following solutions, indicate whether the pH is less than 7, greater than
7, or equal to 7.
(a) Ca(HSO4 )2 (s) dissolved in H2 O(l)
(b) H2 O(l) which has had CO2 (g) bubbling up through it for about 10 minutes.
(c) (NH4 )2 SO4 (s) dissolved in H2 O(l)
(a) Ca(HSO4 )2 , calcium bisulfate, is going to be acidic because calcium does not significantly hydrolyze, and HSO–4 is an acid (it has no tendency to take a proton and
go to H2 SO4 since that is a strong acid, but will give up its proton to go to SO2–
4 ).
So it will have pH less than 7.
(b) When you bubble CO2 (g) through water, you get CO2 (aq), which reacts with H2 O
to form carbonic acid H2 CO3 , which is, well, an acid. So you have an acidic solution
with pH < 7.
(c) Since NH3 has a pKa of 1.8 × 10−5 , NH4 has a pKa of 1.0 × 10−14 /1.8 × 10−5 =
5.6 × 10−10 . Since HSO–4 has a pKa of 2.0, the sulfate anion SO2–
4 can act as a
base, with pKb = 1.0 × 10−14 /1.0 × 10−2 = 1.0 × 10−12 . Because this value is
+
significantly less than the pKa of NH+
4 , the acidic dissociation of NH4 is going to
be the dominant effect and thus the solution will be acidic.
5. Methanol, CH3 OH, is a commonly used organic solvent. Like water, methanol has a
characterization auto-ionization reaction,
+
−
−−
*
CH3 OH + CH3 OH )
−
− CH3 OH2 + CH3 O
K = 2.0 × 10−17
What is the pH of neat (pure) CH3 OH?
pH = −log10 ([H+ ]), represents the concentration of dissociated protons in solution, but
remember that even in water they’re not truely “free” H+ , they’re bound to one (or
more) water molecules. Thus we have species in H2 O like H3 O+ , or even possibly H5 O+
2
(the so-called Zundel cation) or even H9 O+
4 (so-called Eigen cation) where two or four
water neutral molecules all surround and bind the excess proton. But for the purpose
of pH, it doesn’t matter how many water molecules that extra proton is bound too, just
that it is an extra proton. You get the same answer if you use [H+ ] or [H3 O+ ]. So if we
switch solvents to methanol, the pH is simply − log10 [CH3 OH+
2]
−
2
−7
In neat methanol, we have [CH3 OH+
2 ] = [CH3 O ] = x and x = K = 2.0 × 10 , giving
+
−9
−9
x = [CH3 OH2 ] = 4.5 × 10 M . Thus the pH is − log10 4.5 × 10 = 8.3.
6. Describes what chemistry happens (if any) in the following situations, i.e. what would
you observe happen?
(a) Some powdered AgNO3 (s) and NaI(s) are put in an empty glass beaker.
(b) Some powdered AgNO3 (s) and NaI(s) are put in a beaker of H2 O(l).
(a) Nothing happens. You just mixed two solids together.
(b) Both dissolve in the water, and then the Ag+ and I– precipitate out, since they’re
insoluble together. Note that both AgNO3 (s) and NaI(s) are white powders but
AgI is a pale yellow precipitate, so you would definitely notice this.
7. Calculate the pH of a saturated solution of calcium hydroxide (Ksp = 4.68 × 10−6 ).
Ca(OH)2
I
C
E
-
−−
*
)
−
−
Ca2+
0
+x
x
OH–
0
+2x
2x
Ksp = (x)(2x)2
= 4x3
1
Ksp 3
x=
4
[OH] = 2x
pOH = − log10 (2x)
pH = 14 − pOH
= 14 + log10 (2x)
"
1 #
Ksp 3
= 14 + log10 2 ×
4
= 12.3
This is a basic pH (> 7), which is what we expect for a solution of Ca(OH)2 , so we’re
happy.
8. Ksp problem (Hard).
The solubility of AlF3 in deionized water is 6.7
g
L
.
(a) Calculate the Ksp of AlF3 .
(b) Calculate the solubility by mass of AlF3 in a solution buffered at pH 2.00. The
pKa of HF is 3.18.
3+
–
–
−−
*
−−
*
The possible reactions to consider are AlF3 (s) )
−
− Al + 3 F and F + H2 O )
−
− HF +
–
−14+3.18
OH . Because the equilibrium for the second, Kb = 10
<< 1, we can ignore it
for our neutral water situation.
Thus we have Ksp = x · (3x)3 = 27x4 , where x = 6.7
Ksp = 0.0011.
g
L
1 mol
× 83.98
= 0.080 M . This gives
g
In acid, we have to consider the F– acid-base behavior, because we known that since
F– is a base, in an acidic solution much of it is going to be converted to HF which will
reduce the F– concentration and shift the position of the solubility equilibrium to the
right.
If the solution is buffered at pH = 2.00, we can assume it’s pH doesn’t change over the
course of the dissolution. Therefore we can calculate the pOH = 14.00 - 2.00 = 12.00.
[OH− ] = 1 × 10−12 M .
E
E
AlF3
-
−−
*
)
−
−
F–
3x − y
H2 O
-
Al3+ (aq)
x
−−
*
)
−
−
3 F−(aq)
3x − y
HF
y
OH–
1 × 10−12
Ksp = x · (3x − y)3
Kb = 1 × 10−12 ·
y
3x − y
3xKb − yKb =1 × 10−12 y
3xKb =(Kb + 1 × 10−12 )y
3xKb
y=
Kb + 1 × 10−12
3
3xKb
Ksp = x · 3x −
Kb + 1 × 10−12
3
Kb
4
Ksp = 27x 1 −
Kb + 1 × 10−12
1
− 34
Ksp 4
Kb
x=
1−
27
Kb + 1 × 10−12
x = 0.64 M
So, the solubility by mass is 0.64
mol
L
× 83.98 g = 54
g
L
9. Setup the equations (no need to actually solve) that would yield the [H+ ] or pH of
a solution containing a certain number of moles – say α and β – of two independent
weak acids, HA1 and HA2 , together in a 1 L aqueous solution. There should be three
equations and three unknowns.
−
−
*
)
−
−
I
C
E
HA1
α
−a
α−a
H+
0
+a
a+b+c
A–1
0
+a
a
−−
*
)
−
−
I
C
E
HA2
β
−b
β−b
H+
0
+b
a+b+c
A–2
0
+b
b
I
C
E
H2 O
-
−
−
*
)
−
−
H+
0
+c
a+b+c
OH–
0
+c
c
a · (a + b + c)
α−a
b · (a + b + c)
Ka2 =
β−b
Kw = c · (a + b + c)
pH = − log(a + b + c)
Ka1 =
The thing to note with these ICE tables is that the equilibrium concentration of say
H+ includes a contribution from each reaction, and at equilibrium it must be uniform
throughout the solution and thus it must be the same E value in each equation.
This is a system of three equations with three unknowns (a, b, c), so it’s solvable. Sure
the algebra would be a little messy, its doable.