jika 66mg (NH4)2SO4 Mr= 132 dilarutkan dalam air hingga volume 500 ml, kb NH3 = 2×10^-5 PH larutannya adalah.. minta bantuannya ya..

n = m/mr
n = 66/132 = 0.5 mol
M = n/V
M = 0.5 mol/0.5 Liter = 1
[OH] = √Kb.M
[OH] = √2x10^-5.1
[OH] = √2 x 10^-2,5

pOH = -log [OH]
pOH = -log [√2 x 10^-2,5]
pOH = 2,5 - log √2

pH = 14 - (2,5 - log√2)
pH = 11,5 + log√2