CHEMISTRY 1AA3
PROBLEM SET 2 – WEEK OF JANUARY 21, 2002
1.
What is the conjugate acid of each of the following species?
(a)
(b)
(c)
(d)
2.
Base
S2−
CH3COO−
ClO2−
SO32−
Conjugate acid
HS−
CH3COOH
HClO2
HSO3−
What is the conjugate base of each of the following species?
(a)
(b)
(c)
(d)
3.
SOLUTIONS
Acid
H2CO3
HBr
H2PO4−
[(CH3)2NH2]+ (dimethylammonium ion)
Conjugate base
HCO3−
Br−
HPO42−
(CH3)2NH2 (dimethylamine)
Write balanced chemical reactions and Ka or Kb expressions for the first ionization of the
following acids and bases (i.e. assume in each case that only one H+ is ionized or added).
(a) HF
(b) H3PO4
(c) CH3NH2
(d) CH3COOH
F− + H3O+
(e) HClO3
Ka = [F−][ H3O+]
[HF]
(a)
HF + H2O
(b)
H3PO4 + H2O
H2PO4− + H3O+
Ka = [H2PO4−][ H3O+]
[H3PO4]
(c)
CH3NH2 + H2O
CH3NH3+ + OH−
Kb = [CH3NH3+] [OH−]
[CH3NH2]
(d)
CH3COOH + H2O
(e)
HClO3 is a strong acid and is fully dissociated at all concentrations in water. There
is no equilibrium expression to describe the behaviour of this acid.
H3O+ + CH3COO−
Ka = [H3O+] [CH3COO−]
[CH3COOH]
2
4.
(a)
(b)
(c)
(d)
What is the pH of a 0.035 M solution of Ba(OH)2 (aq)?
What is the pH of a 0.0042 M solution of HClO4 (aq)?
What is the pOH of the solution in part (b)?
A 30.0 mL sample of HClO4 was diluted to 500 mL, and the pH of the final
solution was 4.56. What was the concentration of HClO4 in the original sample?
SOLUTION:
(a)
Ba(OH)2 (aq) is a soluble hydroxide, and a strong base. It dissociates fully in
water, so Ba(OH)2 (aq) produces Ba2+ (aq) and 2 −OH (aq).
The original 0.035 M sample of Ba(OH)2(aq) therefore produces
0.035 M Ba2+ ions, and
2 × (0.035M) −OH ions (according to the formula of our hydroxide).
∴ [−OH]
=
2(0.035M)
From this we can find pOH: pOH
=
0.070 M
=
=
=
-log10[−OH]
-log10[0.070]
1.15
From pH + pOH = 14, rearrange to get
pH = 14 – pOH
(b)
=
14 – 1.15
=
12.85 = 12.9
HClO4 is a strong acid, and dissociates fully in water to give:
HClO4(aq) + H2O(l) → ClO4−(aq) + H3O+(aq)
The original 0.0042 M sample of HClO4(aq) therefore produces
0.0042 M ClO4− ions,
0.0042 M H3O+ ions
and
pH = -log10[H3O+]
=
-log10(0.0042) =
2.38
(c)
pOH = 14 – pOH
=
14 – 2.38
11.62 = 11.6
(d)
Since HClO4 is a strong acid, it dissociates fully in solution. In the final solution,
pH = 4.56, and we can find [H3O+] directly from pH:
=
3
Since pH = −log10[H3O+], rearrange to give
−pH = log10[H3O+] and take the antilog to give
Thus, [H3O+] = 10−4.56
=
10−pH = [H3O+]
2.75 × 10−5 mol/L
To find the initial concentration of HClO4, we need the number of moles of
HClO4 present in solution:
mol HClO4
=
=
(2.75 × 10−5 mol/L)(0.500 L)
1.38 × 10−5 mol
Original concentration of HClO4 =
= 1.38 × 10−5 mol
0.0300 L
5.
(a)
(a)
mol HClO4
original volume
= 4.60 × 10−4 M
What is the pH of a 10.0 M solution of HCl?
What is the pH of a 1.00 × 10−8 M solution of HCl?
SOLUTION:
(a)
pH
=
−log10[H3O+] =
−log10(10.0) =
−1
Very acidic! The point of this question is just to show that the pH scale doesn't
just go from 0 to 14, it's just that most of the substances we talk about lie within
that range.
(b)
Again,
pH
=
−log10[H3O+]
BUT in this case, what is [H3O+]?
Remember that from the autoionization of water, there is a concentration of
[H3O+] = 10−7 M (which we usually ignore in our calculations). In this case,
however, the concentration is very significant.
Thus, we have two sources of H3O+ in our solution: the H2O and the HCl:
∴[H3O+]
=
=
Now we can use
1.00 ×10−8 M (from HCl) + 1.00 × 10−7 M (from water)
1.10 × 10−7 M
pH = −log10[H3O+] = −log10(1.10 × 10−7) = 6.96
(If you ignore the [H3O+] from water in this case, the pH will work out to be 8,
which doesn't make sense, since we're talking about a dilute solution of acid!).
4
6.
Dimethylamine, (CH3)2NH, is a weak base (ionization constant Kb = 7.40 × 10−4).
(a)
What is the equilibrium concentration of dimethylammonium ion, [(CH3)2NH2]+
in a 0.400 M aqueous solution of (CH3)2NH?
(b)
What is the pH of a 0.400 M aqueous solution of (CH3)2NH?
SOLUTION:
(a)
This is a “weak base” problem. Start by writing down the chemistry:
(CH3)2NH
(CH3)2NH2+
+ H2O
+
−
OH
Set up an “ICE” table (Initial, Change, Equilibrium concentrations) in order to
obtain mathematical expressions that describe the reaction.
(CH3)2NH
Initial conc.
Change
Equilibrium conc.
−
(CH3)2NH2+
0.400 M
-x
0.400-x
OH
(10−7 M from water)*
+x
x
0
+x
x
* We ignore the amount of −OH present from the ionization of water, if the value (10−7 M) is
very small, as compared to x.
Since (CH3)2NH functions as a base in our equilibrium, set up a Kb expression:
Kb
=
[(CH3)2NH2+][ −OH] =
[(CH3)2NH]
7.4 × 10−4
Substitute the equilibrium concentration values from the “ICE” table into Kb:
(x)(x)
(0.40-x)
=
7.4 × 10−4
Assume x << 0.400, and thus (0.400-x) ≈ 0.400
x2
x
Check:
=
=
(7.4 × 10−4)(0.400)
0.0172 M
0.0172 × 100% = 4.30%
0.400
∴ Equilibrium [(CH3)2NH2+] = x = 0.0172 M
<5%, assumption is okay
5
(b)
Recall that pH + pOH = 14, which can be rearranged to give: pH = 14 – pOH
We need the value for pOH: pOH = -log10[−OH] = -log10[0.0172] = 1.76
∴ pH = 14 – 1.76
7.
=
12.24 = 12.2
Hydrofluoric acid (HF) is a weak acid with Ka = 7.2 × 10−4 at 25 °C. For a 0.45 M
solution of HF at 25 °C, determine:
(a)
(b)
(c)
(d)
[H3O+]
pOH
percent dissociation of HF
Kb for F−
SOLUTION:
This is a “weak acid” problem. Let’s start with the chemistry:
HF
Initial conc.
Change
Equilibrium conc.
Ka
=
+
H2O
0.45 M
-x
0.45-x
[F−][ H3O+]
[HF]
(x)(x)
(0.45-x)
=
F−
0
+x
x
=
+
H3O+
(10−7 M)
+x
x
7.2 × 10−4
7.2 × 10−4
Assume x << 0.45, and thus (0.45-x) ≈ 0.45
Check:
x2
=
(7.2 × 10−4)(0.45)
x
=
0.018 M
0.018 × 100% =
0.045
4.0%
Now we can answer the parts of the question:
(a)
[H3O+] =
(b)
pOH we can determine from pH:
x
=
0.018 M
<5%, assumption is okay
6
pH = -log10[H3O+]
=
-log10(0.018) =
1.7
since pH + pOH = 14, rearrange to get pOH = 14 – pH
∴ pOH = 14.0 – 1.7 =
(c)
12.3
percent dissociation of HF
This is the same calculation we do when we check our assumption. In other
words, compare the concentration of what you get at equilibrium (H3O+, in this
case,) to the concentration of what you started with (HF).
[H3O+]equilibrium × 100%
[HF]initial
=
0.018 × 100% =
0.045
4.0% dissociated
Kb for F−
(d)
Recall the equation that relates Ka and Kb for a conjugate acid-base pair:
8.
Ka × Kb = Kw
which can be rearranged to
Thus, Kb
1.0 × 10−14
7.2 × 10−4
=
=
Kb = Kw
Ka
1.4 × 10−11
A student dissolves 2.15 g of hydrazoic acid, HN3, to form one litre of solution. The pH
of the solution is measured and found to be 3.01. What is the ionization constant of
hydrazoic acid?
SOLUTION:
This is a “weak acid” problem. Start with the chemistry:
HN3 + H2O
H3O+ + N3−
Ka = [H3O+] [N3−]
[HN3]
Ka is unknown, but the pH of the solution is 3.01, and so we can determine the [H3O+]:
[H3O+] = antilog(-pH) = 10−pH
= 10−3.01
= 9.77 × 10−4M
7
Also, we can determine the initial concentration of the solution:
2.15 g = 0.0500 mol
43.0 g/mol
and this is in one litre of solution, so, initially [HN3] = 0.0500 M
Initial conc.
Change
Equilibrium conc.
HN3
0.0500 M
-x
0.0500-x
H3O+
0
+x
x
But we calculated [H3O+], and so
x = 9.77 × 10−4M = [H3O+] = [N3−]
and [HN3] = 0.0500 - x = 0.0490 M
Ka = (9.77 × 10−4)( 9.77 × 10−4) = 1.95 × 10-5
0.0490
N3−
(10−7 M)
+x
x