1. 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.

2. 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

3. 1 mol Ba(OH)2 20.0 g Ba(OH)2 x 171.32 g Ba(OH)2 [Ba(OH)2] = 1.00 L
2 mol OH-
[OH-] = x
1 mol Ba(OH)2
1.00 L
[OH-] =
0.234 M

4. pOH = -log[OH-] = -log(0.234) = 0.631
pH + pOH = 14.00
pH = – pOH = – = 13.37

5. A Weak Base Problem Calculate the pH of an ammonia solution
with a molarity of 0.25 M.
(b) Determine the pKb of the solution.
[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.

6. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
[ ]I
[ ]c x x x
[ ]e 0.25-x x x
Remember that
there no entries for
liquids and solids.
[NH4+]
[OH-] Ka =
[NH3] x x x ×
1.8 x 10-5 =

7. Assume 0.25 – x ≈ 0.25 and check later.=
x = [NH4+] = [OH-] = 2.1 x 10-3 M
[OH-]
[NH3] ×
100%
%ion =
2.1 x 10-3 M
%ion = ×
100% =
0.84 %
2.5 x 10-1 M

8. 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 = – 2.68 = 11.32
(b) pKb = -logKb = -log(1.8 x 10-5) = 4.74