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jumah mol NH4OH = 50 x 0,1 = 5 mmol
2NH4OH (aq) + H2SO4 (aq) --> (NH4)2SO4 (aq) + 2H2O (l)
Awal 5 mmol 2,5 mmol - -
Bereaksi -2,5 mmol -2,5 mmol +2,5 mmol -
Akhir 2,5 mmol - 2,5 mmol -
[OH^-] = 1x10^-5 x 2,5/2,5
= 1 x 10^-5
pOH = - log 1 x 10^-5
= 5 - log 1
= 5
pH = 14 - 5 = 9