2. Calculate [H3O+]total, [OH-]total, [H3O+]water, [OH-]water, the pH and the pO
ID: 945778 • Letter: 2
Question
2.
Calculate [H3O+]total, [OH-]total, [H3O+]water, [OH-]water, the pH and the pOH for a 3.2 x
10-2 M aqueous solution of perchloric acid, HClO4.
Compare your value for [H3O+]water in the perchloric acid solution with the value for [H3O+]water in pure water. Are they the same or different? If they are different, explain why they are different.
3. Calculate [H3O+]total, [OH-]total, [H3O+]water, [OH-]water, [Ca2+]total, the pH and the pOH for a 0.00025 M aqueous solution of calcium hydroxide, Ca(OH)2.
Explanation / Answer
3.2 x10-2 M aqueous solution of perchloric acid, HClO4.
[H3O+]total = 3.2 x 10-2M [H3O+]HClO4 + 10-7M [H3O+]H2O (contribution from water)
= 3.2 x 10-2M [H3O+] ([H3O+]H2O is negligible compared to [H3O+]HClO4)
[H3O+]water [OH-]water = 10-14
[OH-]water = 10-7M
[H3O+]total [OH-]total= 10-14
[OH-]total = 10-14/[H3O+]total
= 10-14/3.2 x 10-2
= 0.3125 x 10-12
pH = -log [H3O+]
= - log 3.2 x 10-2
= 1.495
pOH = 14 - pH
= 14 - 1.495
= 12.505
[H3O+] is same in both HClO4 and water
0.00025 M aqueous solution of calcium hydroxide, Ca(OH)2.
[OH-]Ca(OH)2 = 2 x 0.00025 M
=0.00050 M
= 5 x 10-4 M
[OH-]water = 10-7 M
[OH-]total = [OH-]Ca(OH)2 + [OH-]water
= 5 x 10-4 + 10-7 (negligible compared with [OH-]Ca(OH)2)
= 5 x 10-4
[H3O+]Ca(OH)2 = 0
[H3O+]water = 10-7
[H3O+]total[OH-]total = 10-14
[H3O+]total = 10-14/5 x 10-4
= 0.2 x 10-10
[Ca2+]total = 2.5 x 10-4
pOH = -log[OH-]total
= -log(5 x 10-4)
= 3.3
pH = 14 - pOH
= 14-3.3
=10.7
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