Solution.
i) [OH- ] = 0.100 moldm-3
[H+] = Kw/[OH- ] = 10-14/0.100 = 10-13 pH = -log[10-13] = 13.00
(ii) [OH- ] = 0.200 moldm-3
[H+] = Kw/[OH- ] = 10-14/0.100 = 5 x 10-14 pH = -log[5 x 10-14] = 13.30
iii) [OH- ] = 0.050 moldm-3
[H+] = Kw/[OH- ] = 10-14/0.100 = 2 x10-13 pH = -log[2 x 10-13] = 12.69(Ans.)
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