Midterm Exam III
CHEM 181: Introduction to Chemical Principles
Solutions
1. For reference, here are the pKa values for four weak acids (not all resonance
structures shown):
O
O
H 3C
H
N
C
C
O
H2
C
NH
C
H 3C
HC
N
H
OH
O
pKa = 9.7
pKa = 3.5
H 3C
OH
pKa = 4.8
pKa = 16
Now consider this list of compounds (not all resonance structures shown):
O
OH
OH
C
C
H
N
N
HC
CH
HC
CH
N
HC
CH
HC
CH
C
H
C
H
OH
O
HC
O
CH
H 3C
HC
CH
C
H
N
C
C
C
C
H2
C
OH
C
HC
CH
HC
C
N
1
CH
O
(a) An unknown acid has a pKa of 9.9 and the following IR spectrum:
i. Identify this compound from the list on page 2 (redraw the Lewis
structure below).
ii. Circle the most acidic proton of this compound.
iii. Draw any resonance structures that are important in determining the
acidity of this proton.
O
O
O
O
HC
CH
HC
CH
HC
CH
HC
HC
CH
HC
CH
HC
CH
HC
C
H
C
H
C
H
CH
CH
C
H
iv. Label any peaks on the IR spectrum that you can assign (there will
likely only be a few.)
Just as important is that there are not peaks for either C=O or C≡N
stretches.
2
(b) An unknown acid has a pKa of 4.6 and the following IR spectrum:
i. Identify this compound from the list on page 2 (redraw the Lewis
structure below).
ii. Circle the most acidic proton of this compound.
iii. Draw any resonance structures that are important in determining the
acidity of this proton.
N
N
N
Two minor structures (single/triple-bonded, with a −2 charge on one
nitrogen) could be included, too.
iv. Label any peaks on the IR spectrum that you can assign (try to label
most.)
The large peak at ∼2200 cm−1 can’t be a C=O. Given how close it is
to a C≡N, it’s a sensible guess that the N≡N stretch would be around
here, too. (Note that bond orders are probably a little under 3 and a
little over 1 for N–N bonds.)
3
2. For each of the following, rank the compounds by most acidic (basic) to least
acidic (basic).
(a) Write “strongest” next to the strongest acid and “weakest” next to the
weakest.
(b) Write “strongest” next to the strongest acid and “weakest” next to the
weakest.
(c) Write “strongest” next to the strongest BASE and “weakest” next to the
weakest BASE. (It is the NH2 group that acts as the base.)
(d) Write “strongest” next to the strongest BASE and “weakest” next to the
weakest BASE. (It is the N (or NH) that acts as the base.)
4
3. Match the compounds on the following page to the 1 H NMR spectra on this
page. (All peaks are shown. Ignore very small spikes in the baseline, as well as
the calibration peak at 0 ppm.)
The data here are plotted with an integration trace (red) which lets you measure
the area of each peak and makes it much easier to tell which peaks are real and
which are blips of noise. Because you didn’t have this information on the exam,
we accepted two correct answers for #4.
5
4. Succinic acid, C4 H6 O4 , is a diprotic acid:
Ka (C4 H6 O4 ) = 6.3 × 10−5
−6
Ka (C4 H5 O−
4 ) = 2.5 × 10
Aniline, C6 H5 NH2 , is a weak base:
Kb (C6 H5 NH2 ) = 7.4 × 10−10
(a) If you start with a 0.1 M solution of aniline in water, and add succinic acid
until 50% of the aniline is converted to its conjugate acid (C6 H5 NH+
3 ),
what is the pH of the resulting solution?
The important thing here is that there are equal quantities of aniline and
its conjugate base (anilineH+ ) at equilibrium. So
−
[C6 H5 NH+
3 ][OH ]
Kb (C6 H5 NH2 ) =
[C6 H5 NH2 ]
[C6 H5 NH+
3]
7.4 × 10−10 =
· [OH− ]
[C6 H5 NH2 ]
[OH− ] = 7.4 × 10−10
At equilibrium, it also must be true that
Kw = 10−14 = [H+ ][OH− ]
so
[H+ ][OH− ] = 10−14
10−14
[H+ ] =
[OH− ]
10−14
=
7.4 × 10−10
= 1.35 × 10−5 M
pH = − log10 [H+ ]
= 4.87
6
(b) At this point, what are the concentrations of the succinic acid and its two
2−
conjugate bases (C4 H6 O4 , C4 H5 O−
4 , and C4 H4 O4 )?
We know the H+ concentration at equilibrium, and we can use that to
figure out where the two acid equilibria lie for succinic acid:
[HSucc− ][H+ ]
Ka (C4 H6 O4 ) =
[H2 Succ]
[HSucc− ]
· 1.35 × 10−5
6.3 × 10−5 =
[H2 Succ]
−
[HSucc ]
= 4.67
[H2 Succ]
[Succ2− ][H+ ]
Ka (C4 H5 O−
)
=
4
[HSucc− ]
[Succ2− ]
2.5 × 10−6 =
· 1.35 × 10−5
[HSucc− ]
[Succ2− ]
= 0.19
[HSucc− ]
In other words, because we know the pH, we also know the proportion of
H2 Succ, HSucc− , and Succ2− in solution at equilibrium.
For every HSucc− we create, we must also create one anilineH+ , while we
create two anilineH+ with every Succ2− . So the stoichiometry is
[anilineH+ ] = [HSucc− ] + 2[Succ2− ]
and plugging in the above proportions lets us solve
0.05
0.05
[HSucc− ]
[Succ2− ]
[H2 Succ]
=
=
=
=
=
[HSucc− ] + 2[Succ2− ]
[HSucc− ] + 2 · 0.19 · [HSucc− ]
3.7 × 10−2 M
7.0 × 10−3 M
7.9 × 10−3 M
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5. Decanoic acid (C10 H20 O2 ) has a Ka of 1.26 × 10−5 . It is a white solid that is
only slightly soluble in water
Ksp (C10 H20 O2 ) = 8.7 × 10−4 M
If 85 g (0.5 mole) of decanoic acid is added to 1 L of water, what is the lowest
pH at which all of it will dissolve?
Assume that
• pH is changed by adding strong base (e.g. NaOH),
• the solution volume does not change, either when adding base or adding
decanoic acid, and
• C10 H19 O−
2 is completely soluble in water.
The reactions are
HDec(s) HDec(aq)
HDec(aq) Dec− (aq) + H+ (aq)
The equilibrium expressions are
HDec(aq) = 8.7 × 10−4 M
[Dec− (aq)][H+ (aq)]
1.26 × 10−5 =
[HDec(aq)]
Stoichiometry is
HDec(aq) + Dec− (aq) = 0.5 M
So
[Dec− (aq)][H+ (aq)]
[HDec(aq)]
0.5 − 8.7 × 10−4 +
−5
1.26 × 10
=
[H (aq)]
8.7 × 10−4
[H+ (aq)] = 2.2 × 10−8 M
pH = 7.66
1.26 × 10−5 =
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