Strong Bases
There are six strong acids.
 HCl, HNO3, HClO4, HI, HBr, and H2SO4
 When given any of the above acids
always assume 100% ionization.
 100% ionization is shown by using a
single arrow in the ionization equation.
 The ionization equation for H2SO4 is not
intuitively obvious to the most casual
observer.
8-1
Strong Base Problem
What is the pH of a solution prepared by
dissolving 20.0 g Ba(OH)2/L?
m = 20.0 g Ba(OH)2/L
Ba(OH)2(aq) → Ba2+(aq) + 2OH-(aq)
[Ba(OH)2] = n/V
8-2
20.0 g Ba(OH)2 x
.
[Ba(OH)2] =
1 mol Ba(OH)2
171.32 g Ba(OH)2
1.00 L
[Ba(OH)2] = 0.117 M
[OH-] =
0.117 mol Ba(OH)2
1.00 L
2 mol OHx
1 mol Ba(OH)2
[OH-] = 0.234 M
8-3
pOH = -log[OH-] = -log(0.234) = 0.631
pH + pOH = 14.00
pH = 14.00 – pOH = 14.00 – 0.631 = 13.37
8-4
A Weak Base Problem
(a) Calculate the pH of an ammonia solution
with a molarity of 0.25 M.
(b) Determine the pKb of the solution.
(a) [NH3] = 0.25 M
Kb = 1.8 x 10-5
Take special note of the equilibria equation.
Unlike weak acid problems where H2O as a
reactant is optional, it is not the case with a
weak base problem.
8-5
NH3(aq) + H2O(l)
[ ]I
[ ]c
[ ]e
0.25
-x
0.25-x
+]
0
+x
x
[NH4
Ka =
[NH3]
1.8 x
10-5
=
NH4+(aq) + OH-(aq)
[OH-]
0
+x
x
Remember that
there no entries for
liquids and solids.
x× x
0.25 - x
8-6
Assume 0.25 – x ≈ 0.25 and check later.
1.8 x 10-5 =
x2
0.25
x = [NH4+] = [OH-] = 2.1 x 10-3 M
%ion =
[OH-]
[NH3]
× 100%
2.1 x 10-3 M
× 100% = 0.84 %
%ion =
-1
2.5 x 10 M
8-7
The %ion = 0.84% is well within the 5%,
therefore the assumption is valid.
pOH = -log[OH-] = -log(2.1 x 10-3) = 2.68
pH + pOH = 14.00
pH = 14.00 – 2.68 = 11.32
(b) pKb = -logKb = -log(1.8 x 10-5) = 4.74
8-8