15 gram alumunium hidroksida dilarutkan dalam 300 ml air berapakah ph alumunium hidrokaida kb=1×10-5

Diketahui
m = 15 gram
Kb = 1 x 10^-5
V = 300 ml = 0.3 L
Alumunium Hidroksida = Al(OH)3 = Ar Al = 27 maka :
27 + (16x3) + (1x3) = 78

n = gr/mr  = 15 / 78 =0.19
M = n / V(dalam liter) = 0.19/ 0.3 = 0.6
=
Rumus :

OH = √Kb. M
OH = √1x 10^-5 . 0.6
OH = √ 6 x10^-6
OH = 2.44 x 10 ^-3

pOH = 3 - log 2.44
pH = 14-pOH
pH = 14 - (3-log 2.44)
pH = 11 + log 2.44
pH = 11,38