Calculate the pH for 100. mL of a 0.500 M solution of ammonia, (NH3 Kb=1.8x10-5) being titrated with 0.500 M HCl at the following positions in the titration.
i) The initial pH (before any HCl has been added).
A. |
9.44 |
|
B. |
10.81 |
|
C. |
11.48 |
|
D. |
12.00 |
|
E. |
11.75 |
Ammonia is weak base dissociate as
NH3 + H2O NH4+ + OH-
Kb = [NH4+] [OH-] / [NH3]
but [NH4+] = [OH-]
consider [NH4+] = [OH-] = X
then Kb = [X] [X] / [NH3]
Kb = X2 / [NH3]
X2 = Kb X [NH3]
substitute the value
X2 = (1.8 X 10-5) X (0.500) = 9 X 10-6
X = 0.003 M
[NH4+] = [OH-] = X = 0.003 M
[OH-] = 0.003 M
pOH = -log [OH-] = -log 0.003 = 2.52
pH = 14 - pH = 14 - 2.52 = 11.48
pH of 0.500 M NH3 = 11.48
Ans = C) 11.48
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