Finding the pOH of a strong base:
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Define a base
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A strong base is one which breaks up fully into its ions when it is added to water
Write an equation to show this for each of the following strong bases
NaOH
LiOH
Ca(OH)2
Use your equations above to work out the following:
a)
b)
c)
d)
e)
f)
g)
h)
How many OH- ions would be produced from 10 NaOH splitting up?
How many OH- ions would be produced from 10 LiOH splitting up?
How many OH- ions would be produced from 10 Ca(OH)2 splitting up?
How many OH- ions would there be in 1 litre of water if there were 15 NaOH added to a litre of
water?
How many mols of OH- ions would there be per litre if 15 mols of LiOH were added?
How many mols of OH- ions would there be per litre if 15 mols of Ca(OH)2 were added?
What would be the concentration of OH- ions, [OH-], in LiOH with a concentration of 0.01 mol dm-3?
What would be [OH-] in 1.50 mol dm-3 Ca(OH)2
For strong bases the [OH-] is directly related to the concentration of the original acid. It has all split up to
make hydroxide ions.
pOH = -log10 [OH-]
1) Calculate the pOHs of the following strong bases?
a) 0.0300 mol dm-3 NaOH
b) 0.00500 mol dm-3 LiOH
c) 0.120 mol dm-3 Ca(OH)2
2) Calculate the concentration of the following strong bases from their pOHs:
a) LiOH with a pOH of 0.8
b) Calcium hydroxide with a pOH of 2.5
c) NaOH with a pOH of 2.0
Calculating the pH of a strong base
pH + pOH = 14
From your earlier answers calculate the pH of the following strong bases:
a) 0.0300 mol dm-3 NaOH
b) 0.00500 mol dm-3 LiOH
c) 0.120 mol dm-3 Ca(OH)2
and the [H+] of the following strong bases:
a) LiOH with a pOH of 0.8
b) Calcium hydroxide with a pOH of 2.5
c) NaOH with a pOH of 2.0
[H+].[OH-] = 1 x 10-14 @ 25°C
Calculate the [OH-] in the following solutions:
a) HCl with a pH of 2.3
b) NaOH with a pH of 12.6
c) H2SO4 with a pH of 1
d) 0.25 mol dm-3 HNO3
e) 1.2 mol dm-3 Ca(OH)2
[OH-]
- log [OH-] = pOH
pH
[OH-] = 10-pOH
[H+].[OH-] = 1 x 10-14
- log [H+] = pH
[H+] = 10-pH
[H+]
pOH
pH + pOH = 14