pH larutan 500 mL Ca(OH)2 0,005 mol adalah....tolong bantu jawab ya

Dik : v = 500ml
n = 0,005 mol -> 5 mmol
M = n/v
M = 5 mmol/ 500ml
M = 0,01 M

Caoh2 -> Ca2+ + 2Oh-

[Oh-] = b x MCaOH2
[oh-] = 2 x 0,01
[oh-] = 2 x 10 pangkat -2
Poh = -log [oh-]
Poh = -log 2 x 10 pgkt -2
Poh = 2-log2
Ph = pkw - poh
Ph = 14 - (2-log2)
Ph = 12+log 2

Smoga ngebantu ya