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[OH]=M.nb
=0,0025.2
=0,005
POH=3-log 5
PH=11+log 5
[OH-] = b x Mb
[OH-] = 2 x 0,0025 = 0,005 = 5 x 10⁻³
pOH = 3 - log 5
Kw = 5 x 10⁻¹³
maka pKw = 13 - log 5
pH = pKw - pOH
pH = 13 - log 5 - 3 - log 5
pH = 13 - 3
pH = 10
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