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[OH-] = √Kb x M
[OH-] = √10⁻⁵ x 0,1
[OH-] = √10⁻⁶ = 10⁻³
pOH = - log 10⁻³ = 3
pH = 14 - pOH = 14 - 3 = 11
HCl 0,1 M
[H+] = 0,1 M
pH = - log 0,1
pH = 1