Midterm Exam II
CHEM 181: Introduction to Chemical Principles
November 7, 2013
Answer Key
1. (a) An unknown acid has a pKa of 4.11 and the following IR spectrum:
i. Identify this compound from the list on page 2 (redraw the Lewis
structure below).
H
H
O
O
C
N
C
C
C
C
H
C
O
H
N
O
O
The C=O stretch peak is not visible, which rules out all but two of
the compounds. Of those, one will be more acidic then phenol (better
resonance structures) and one will be less acidic.
ii. Circle the most acidic proton of this compound.
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iii. Draw any resonance structures that are important in determining the
acidity of this proton.
O
H
O
C
C
N
C
C
H
O
H
C
O
C
C
C
H
H
O
H
N
O
major (2x)
major
O
C
C
N
O
N
C
C
H
O
C
C
O
C
C
N
O
H
N
C
C
H
O
O
O
O
minor (3x)
iv. Label any peaks on the IR spectrum that you can assign (there will
likely only be a few.)
The C–H and O–H peaks are similar to what’s in the phenol spectrum
(it is surprising and maybe confusing that the O–H looks less broad
here.) The NO2 groups should have characteristic stretching frequencies, but they are not on your list. Since the N–O bond order is 1.5,
a reasonable guess is that they should be higher in frequency than a
C–O stretch but lower than a C=O stretch.
2
(b) An unknown acid has a pKa of 2.5 and the following IR spectrum:
i. Identify this compound from the list on page 2 (redraw the Lewis
structure below).
H
H
O
O
C
C
C
O
H
H
This has both a C=O stretch and a hugely broad OH stretch, so it
must be one of the three carboxylic acids (COOH). Because it is substantially more acidic than hexanoic acid, there has to be some other
electronegative group—either the extra C=O or the C≡N—stabilizing
the conjugate base. If you gave the C≡N compound as your answer
(and it does have the right acidity), this is ok, provided you can point
to a bump somewhere that you assign to a C≡N stretch. (There is no
C≡N stretch in this spectrum, but the argument would be that it’s
there, camouflaged by the wide O–H.)
ii. Circle the most acidic proton of this compound.
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iii. Draw any resonance structures that are important in determining the
acidity of this proton.
H
H
O
O
C
C
C
O
H
H
H
O
O
C
C
C
O
H
equivalent major structures
iv. Label any peaks on the IR spectrum that you can assign (there will
likely only be a few.)
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2. The 1 H NMR spectra on this page are all of compounds with the molecular
formula C3 H7 NO:
5
3. (a) Calculate the pH of a solution where the concentration of Na2 CO3 is 0.1 M.
−11
(The Ka of HCO−
. As a sodium salt, this is completely soluble
3 is 4.7×10
(and completely dissociates) in water.)
This is a 0.1 M solution of CO2−
3 . The relevant reaction is that of a weak
base:
−
−
CO2−
3 + H2 O HCO3 + OH
We calculate the Kb from the Ka of the conjugate acid (HCO−
3 ):
−
Kb (CO2−
3 )Ka (HCO3 ) = Kw
Kw
Ka (HCO−
3)
−14
1 × 10
=
4.7 × 10−11
= 2.13 × 10−4
Kb (CO2−
3 ) =
So
−
[HCO−
3 ][OH ]
= 2.13 × 10−4
2−
[CO3 ]
Our ICE table:
[CO2−
[OH− ]
3 ]
I
0.1
∼0
C
−x
+x
x
E 0.1 − x
[HCO−
3]
0
+x
x
So
x2
0.1 − x
x2
x2 + 2.13 × 10−4 x − 2.13 × 10−5
x
= 2.13 × 10−4
= 2.13 × 10−4 (0.1 − x)
= 0
= 4.5 × 10−3
So
[OH− ] = 4.5 × 10−3
+
1 × 10−14
H
=
4.5 × 10−3
= 2.2 × 10−12
+
− log10 H
= 11.65
(Assuming 0.1 − x ≈ 0.1 gives you a pH of 11.66.)
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(b) The Kb of ammonia is 1.78 × 10−5 . What is the pH of a 0.1 M solution of (NH4 )2 CO3 (As an ammonium salt, this is completely soluble (and
completely dissociates) in water.)?
Dissociation is complete, so you are starting with a solution that is 0.2 M in
2−
NH+
4 and 0.1 M in CO3 . Acid/base equilibria for these species are:
[H+ ][NH3 ]
[NH+
4]
−
[OH ][HCO−
3]
2−
Kb (CO3 ) =
2−
[CO3 ]
Ka (NH+
4) =
We have the Kb from part (a), and we figure out Ka (NH+
4 ) from Kb of NH3 :
Ka (NH+
4) =
1 × 10−14
= 5.6 × 10−10
1.78 × 10−5
The acid-base equilibrium that’s going to be most important (by far) in determining concentrations is
2−
−
NH+
4 + CO3 NH3 + HCO3
We need to calculate K for this using what we have:
K =
2−
Ka (NH+
4 )Kb (CO3 ) =
=
=
So
K=
[NH3 ][HCO−
3]
+
2−
[NH4 ][CO3 ]
[H+ ][NH3 ] [OH− ][HCO−
3]
+
2−
[NH4 ]
[CO3 ]
−
[NH3 ][HCO3 ] +
−
+
2− [H ][OH ]
[NH4 ][CO3 ]
K · Kw
2−
Ka (NH+
5.6 × 10−10 · 2.13 × 10−4
4 )Kb (CO3 )
= 12
=
Kw
1 × 10−14
Now we ICE:
[NH+
[CO2−
[NH3 ]
4]
3 ]
I
0.2
0.1
0
−x
−x
+x
C
E 0.2 − x 0.1 − x
x
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[HCO−
3]
0
+x
x
[NH3 ][HCO−
3]
+
2−
[NH4 ][CO3 ]
x2
(0.2 − x)(0.1 − x)
x2
11x2 − 3.6x + 0.24
x
= K
= 12
= 12(x2 − 0.3 ∗ x + 0.02)
= 0
= 0.0932
So at equilibrium:
[NH+
4]
2−
[CO3 ]
[NH3 ]
[HCO−
3]
=
=
=
=
0.1068
0.0068
0.0932
0.0932
Get [H+ ] from
[H+ ][NH3 ]
[NH+
4]
0.0932 +
[H ]
5.6 × 10−10 =
0.1068
+
H
= 6.42 × 10−10
Ka (NH+
4) =
or from
[H+ ][CO2−
3 ]
−
[HCO3 ]
0.0068 +
4.7 × 10−11 =
[H ]
0.0932
+
H
= 6.44 × 10−10
Ka (HCO−
3) =
For a pH of 9.19 either way.
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4. The equilibrium constant (Henry’s Law constant) for solvation of chlorine in
water:
Cl2 (g) Cl2 (aq)
is kH = 9.0 × 10−2 mol L−1 atm−1 . The reaction
Cl2 (aq) + H2 O(`) HOCl(aq) + HCl(aq)
has an equilibrium constant of K = 3.94 × 10−4 M.
HCl is a strong acid, and
Ka (HOCl) = 2.95 × 10−8
(a) If the pressure of Cl2 (g) remains at 1 atm, what are the equilibrium concentrations of Cl2 (aq), HOCl(aq), OCl− (aq), and H+ (aq)?
[Cl2 (aq)] = kH PCl2 (g)
= 9.0 × 10−2 M atm−1 · 1 atm
= 9.0 × 10−2 M
Because the pressure of Cl2 remains at 1 atm, this necessarily also fixes
the concentration of Cl2 (aq). Now,
K=
[HOCl][HCl]
[Cl2 (aq)]
The equilibrium value of [Cl2 (aq)] is fixed, and [HOCl] = [HCl], so:
[HOCl][HCl]
[Cl2 (aq)]
x2
3.94 × 10−4 =
9.0 × 10−2
x = 5.92 × 10−3
K =
A 5.92 × 10−3 M solution of HCl(aq) is necessarily a 5.92 × 10−3 M solution
of H+ (aq). All that’s left is to consider the equilibrium of HOCl and OCl− :
[H+ ][OCl− ]
Ka (HOCl) =
[HOCl]
You could ICE this if you wanted to
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[HOCl]
I
5.92 × 10−3
C
−x
E 5.92 × 10−3 − x
[H+ ]
5.92 × 10−3
+x
5.92 × 10−3 + x
[OCl− ]
0
+x
x
but we assume that
5.92 × 10−3 − x ≈ 5.92 × 10−3 + x ≈ 5.92 × 10−3
so
[H+ ][OCl− ]
[HOCl]
5.92 × 10−3 · x
2.95 × 10−8 =
5.92 × 10−3
x = 2.95 × 10−8
Ka (HOCl) =
which confirms our assumptions very well. Final answers:
[Cl2 (aq)]
[HOCl]
OCl−
+
H
10
=
=
=
=
9.0 × 10−2 M
5.92 × 10−3 M
2.95 × 10−8 M
5.92 × 10−3 M
(b) A 1 L container is filled with 0.95 L of a 0.1 M solution of Cl2 (aq) in water.
There is no initial concentration of HOCl(aq) or OCl− (aq). If the system
is allowed to come to equilibrium at a pH of 7.5 (for example, by adding
as much strong base as is needed), what are the final concentrations of
Cl2 (aq), HOCl(aq), and OCl− (aq), and the final pressure of Cl2 (g)?
Now the Cl2 (aq) concentration is no longer fixed, but the H+ concentration—
and therefore the HCl concentration—is:
[H+ (aq)] = [HCl(aq)] = 3.16 × 10−8
The ICE table becomes:
I
C
E
[HOCl]
[H+ ]
0
?
+x
+x
x
3.16 × 10−8
[Cl2 ]
0.1
−x
0.1 − x
[HOCl][HCl]
[Cl2 (aq)]
3.16 × 10−8 x
=
0.1 − x
K =
3.94 × 10−4
x
= 12468
0.1 − x
Because this number is so large, the arithmetic is easier if you assume
x ≈ 0.1. Then
0.1
≈ 12468
0.1 − x
0.1 − x ≈ 8 × 10−6
which confirms the assumption. So far, then, we have
[Cl2 (aq)] = 8.0 × 10−6 M
[HOCl] = 0.1 M
Now the HOCl acid equilibrium must be considered to figure out [OCl− ].
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Once again, [H+ ] is fixed.
[H+ ][OCl− ]
[HOCl]
Ka (HOCl) =
−8
−8
2.95 × 10
= 3.16 × 10
[OCl− ]
[HOCl]
[OCl− ]
= 0.93
[HOCl]
so
x
0.1 − x
x
1.93x
x
= 0.93
= 0.093 − 0.93x
= 0.093
= 0.048
This gives new concentrations of
[Cl2 (aq)] = 8.0 × 10−6 M
[HOCl] = 0.052 M
OCl− = 0.048 M
Revisiting the first equilibrium:
K =
[HOCl][HCl]
[Cl2 (aq)]
we had solved this before with values:
3.94 × 10−4 =
3.16 × 10−8 · 0.1
8.0 × 10−6
Because the HOCl concentration has changed by a factor of 0.52, the
Cl2 (aq) concentration must change by the same factor (note this will have
essentially zero effect on the HOCl at this point.) So final concentrations
are:
[Cl2 (aq)] = 4.2 × 10−6 M
[HOCl] = 0.052 M
OCl− = 0.048 M
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The pressure of Cl2 comes from Henry’s law:
[Cl2 (aq)] = kH PCl2 (g)
4.2 × 10−6
PCl2 (g) =
9.0 × 10−2
= 4.7 × 10−5 atm
Ideal gas law
n =
PV
RT
4.7 × 10−5 atm · 0.05 L
(0.08206 L atm mol−1 K−1 ) · (298 K)
= 9.5 × 10−8 mol
=
This is not enough to affect the concentration of Cl2 (aq).
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