Jawabanmu

2014-08-02T21:31:19+07:00
Mol  NH_{3} = 60mL X 0,2M = 12mmol
mol H_{2}SO_{4}  20mL X 0,5M = 10mmol

[ OH^{-} ] = Kb. \frac{mol basa}{mol garam}
[ OH^{-} ] = 2. 10^{-5} . \frac{12}{10}
[ OH^{-} ] = 2,4 x  10^{-5}

pOH = -log [ OH^{-} ]
pOH = -log 2,4 x  10^{-5}
pOH = 5 - log 2,4

pH = 14 - pOH
pH = 14 - (5 - log 2,4)
pH = 14 - 5 + log 2,4
pH = 9 + log 2,4