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*n NH4OH = 0,1 mol
*n NH4Cl = 0,05 mol
*Kb = 10^-5
*pH ?
[OH-] = Kb x n NH4OH/n NH4Cl
= 10^-5 x 0,1/0,05
= 2 x 10^-5
pOH = 5 - log 2
pH = 9 + log 2
= 10^-5 x 2 = 2 x 10^-5
pOH = 5 - log 2
pH = 9 + log 2