What concentration of magnesium hydroxide will make a solution of pH=10.87

Answer 1

....................Mg(OH)2 ==> Mg^2+ + 2OH^-
I....................solid.................0..............0
C...................solid.................x..............2x
E....................solid.................x.............2x
Ksp = 1.8|E-11 = (Mg^2+)(OH^-)^2

You want pH = 10.87 so pOH = 14-10.87 = 3.13
pOH = -log(OH^-) = -log(3.13) so (OH^-) = 7.4E-4 M
Plug in 7.4E-4 for OH^- in the Ksp expression and solve for x = [Mg(OH)2]

Answer 2

Wait I'm sorry what does the e-4 stand for?

Answer 3

Wait nvm that last question