Name: ______[KEY]_____________________________ Date: ___________
Period: _____
WS Acids & Bases 2 [KEY]
1. Calculate the pH of the following.
(a) HNO3 with a concentration of 0.200 M. (1)
0.200 M HNO3
= 0.200 M H+
(b/c strong acid is 100% ionized)
pH = –log (0.200) = 0.699
(b) Potassium hydroxide with a concentration of 0.0100 M. (1)
0.0100 M KOH = 0.0100 M OH– = [OH–]
pOH = –log (0.0100) = 2.000
(b/c strong base is 100% ionized)
pH = 14 – pOH = 14 – 2.000 = 12.000
(c) Benzoic acid with a concentration of 0.10 M. (pKa for benzoic acid = 4.20,
assume the degree of ionization of benzoic acid is negligible). (2)
Ka = 10–4.20 = 6.3 x 10–5
I
C
E
6.3 x 10–5 = [H+][A]
[HA]
HA + H2O ↔ H3O+ + A–
0.10 M
0
0
–x
+x
+x
0.10–x
x
x
≈ 0.10
b/c Ka <<< 1
6.3 x 10–5 =
x2
(0.10)
x = 0.0025 = [H3O+]
pH = –log (0.0025) = 2.60
2. A 0.10 M solution of a weak acid has a pH of 2.80. Calculate Ka and pKa.
(Assume the degree of ionization of the acid is negligible). (3)
[H+] = 10–2.80 = 0.0016 M (at equilibrium)
HA + H2O ↔ H3O+
+
A–
I
0.10 M
0
0
C
–0.0016
+0.0016
+0.0016
E
0.10 M
0.0016 M
0.0016 M
(0.984 w/ 2 dec places kept)
Ka = [H+][A–]
[HA]
Ka = (0.0016)2 = 2.6 x 10–5
(0.10)
pKa = –log (2.6 x 10–5) = 4.60
© Adrian Dingle’s Chemistry Pages 2004, 2005, 2006. All rights reserved.
3.
(a) Given the Kw of water at 50oC is 5.47 x 10–14 , calculate the pH of water at this temperature. (3)
H2O
+ H 2O
↔
H3O+ + OH–
Kw = [H3O+][OH–]
[H3O+] = [OH–] (mol ratio is still 1:1)
5.47 x 10–14 = x2
x = 2.34 x 10–7 = [H3O+]
pH = –log (2.34 x 10–7) = 6.631
(b) From your answer to (a) and your knowledge of Le Chatelier’s Principle, suggest and explain
whether the ionization of water is exothermic or endothermic.
(Kw at 25oC = 1 x 10–14). (2)
Endothermic. At the higher temperature (323 K as opposed to 273 K), the pH goes down, the degree of
ionization has therefore increased (more H+ ions). This means heat is essentially a reactant that shifts the
equilibrium to the product side at the higher temperature.
4. Consider the following reaction
C6H5NH2 + H2O ↔ C6H5NH3+ + OH–
(a) Explain the role of C6H5NH2 in the forward (left to right) reaction. (2)
It acts as a base since it accepts H+.
(b) Explain the significance of the ↔ sign and what it means in relation to your answer to (a). (2)
The reaction is reversible, meaning the base in (a) is weak (not completely dissociated).
(c) Comment on the role of water in the forward (left to right) reaction. (2)
Water is an acid, it donates H+.
(d) Identify one acid/base conjugate pair. (1)
H2O and OH–
or
C6H5NH2 and C6H5NH3+.
© Adrian Dingle’s Chemistry Pages 2004, 2005, 2006. All rights reserved.