Whats the ph of a 0.25 mol/L solution of magnesium hydroxide

help please :(

Mg(OH)2 = Mg2+ + 2OH-
Hence 0.25M solution of Mg(OH)2 contains 2 x 0.25 = 0.5M OH-
pOH = -log[OH-] = -log0.50 = 0.3010
pH = 14 - pOH = 14 - 0.3010 = 13.699 = 13.70

Mg(OH)2 0.25m/l is Mg+2 + 2OH- 2* 0.25 or 0.50m/l OH-

pOH= -log[OH-] =0.3 pH = 14-pOH = 14-0.3 = 13.7