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pOH = 3 - log 1,41
pH = 11 + log 1,41
b.) 2NH4OH + H2SO4 --> (NH4)2SO4 + 2H2O
m 10 mmol 5 mmol
b 10 mmol 5 mmol 5 mmol 10 mmol
s - - 5 mmol 10 mmol
Hidrolisis
M = 2 x 5 mmol / 150 mL = 0,06 M
H+ = √Kw/Kb x M
H+ = √10^-14/2x10^-5 x 0,06
H+ = 0,54 x 10^-5
H+ = 5,4 x 10^-6
pH = 6 - log 5,4
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