3. The pH of a solution at 25°C in which [OH – ] = 3.3 ´ 10 –5 M is: a) 4.48 b)
ID: 984724 • Letter: 3
Question
3. The pH of a solution at 25°C in which [OH–] = 3.3 ´ 10–5M is:
a) 4.48
b) 3.30
c) 9.52
d) 4.76
e) none of these
5. A monoprotic weak acid when dissolved in water is 0.36% dissociated and produces a solution with a pH of 3.10. Calculate the Ka of the acid.
a) 3.6 ´ 10–3
b) 2.2 ´ 10–1
c) 2.9 ´ 10–6
d) Need to know the initial concentration of the acid.
e) None of these.
14. Fill in the missing compounds in the following table of conjugate acid/base pairs.
Conjugate acid
Conjugate Base
H2CO3
NH2-
H3O+
Conjugate acid
Conjugate Base
H2CO3
NH2-
H3O+
Explanation / Answer
3)
pH of [OH-] = 3.3*10^-5
pOH = -log(OH) = -log(3.3*10^-5) = 4.48
pH = 14- pOH = 14-4.48 = 9.52
b)
Ka = [H+][A-]/[HA]
[H+] = 10^-pH = 10^-3.1 = 0.00079432823
% ion = [H+]/ M *100
M = [H+] / % *100 = 0.00079432823 / 0.36*100 = 0.22064
then
Ka = [H+][A-]/[HA]
Ka = (0.00079432823 )(0.00079432823 /(0.22064-0.00079432823)
Ka = 0.00000287
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