- Joined
- Feb 10, 2011
- Messages
- 895
- Reaction score
- 126
12) What would be the pH of a 1.0 M solution of the unknown salt hydroxide given that the metal is monovalent?
11
8.0
7.5
13.0
INCORRECT:
Your Answer: D I did pOH = -log [1] =0 so picked 13. This is wrong b.c 1 M is the molarity, I need to get the mL or L to know the [OH-] concentration correct?
Correct Answer: A
Using Kb = ([X+][OH-])/[XOH]; X is most probably a Group I metal since no decomposition occurred (P3, L3-4), implying that the anion (OH-) and cation (X+) are of comparable size (P2, L1-4).
Assuming that [X+] = [OH-] approximately
Kb = [OH-]2/[XOH], where XOH approximates 1.0 M at equilibrium, thus:
1.0 x 10-6 = [OH-]2/1
[OH-]2 = 1.0 x 10-6
[OH-] = 1.0 x 10-3 mol dm-3
pOH = -log[OH-] = -log(1.0 x 10-3) = -(-3) = 3
Using pH + pOH = 14, we get:
pH = 14 - pOH = 14 - 3 = 11
11
8.0
7.5
13.0
INCORRECT:
Your Answer: D I did pOH = -log [1] =0 so picked 13. This is wrong b.c 1 M is the molarity, I need to get the mL or L to know the [OH-] concentration correct?
Correct Answer: A
Using Kb = ([X+][OH-])/[XOH]; X is most probably a Group I metal since no decomposition occurred (P3, L3-4), implying that the anion (OH-) and cation (X+) are of comparable size (P2, L1-4).
Assuming that [X+] = [OH-] approximately
Kb = [OH-]2/[XOH], where XOH approximates 1.0 M at equilibrium, thus:
1.0 x 10-6 = [OH-]2/1
[OH-]2 = 1.0 x 10-6
[OH-] = 1.0 x 10-3 mol dm-3
pOH = -log[OH-] = -log(1.0 x 10-3) = -(-3) = 3
Using pH + pOH = 14, we get:
pH = 14 - pOH = 14 - 3 = 11