Mg(OH)2 <------>Mg2+ + 2 OH¯
Ksp = [Mg2+] [OH¯]2
5.6 x 10¯12 = (x) (2x)2 = 4x3
After dividing by 4 and then taking the cube root:
x = 1.12 x 10¯4 M
This is an important point: what we have calculated is 'x' and it is NOT the [OH¯]. That value is '2x.'
[OH¯] = 2.24 x 10¯4 M
pOH = - log 2.24 x 10¯4 = 3.650
pH = 14 - 3.650 = 10.350
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