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Zakładam, że V = 1dm3
Ca(OH)2 --> Ca2+ + 2OH-
0,005mola 0,01mola
pOH = -log0,01 = 2
pH + pOH = 14 => pH = 14 - pOH
pH = 14 - 2 = 12
pH = -log[H3O+] => [H3O+] = 10^-pH
[H3O+] = 10^-12 [mol/dm3]