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mula-mula : 10 0,1 - -
bereaksi : o,1 0,1 0,1 0,1
sisa : 9,9 - 0,1 0,1
(OH^-) = Kb x mol basa
mol garam
= 10^-5 x 9,9
0,1
= 10^-5 x99=99x10^-5
= 99 x10^-5
pOH = - log (OH^-)
= - log 99x10^-5
=5 - log 99
pH = 14 - pOH
= 14 - (5 - log 99)
= 9 + log 99
jadi,pH larutan penyangga nya 9 + log99