pH = - log(H^+) = 8 so (H^+) = 1E-8 M
pH + pOH = 14 so pOH = 6 and (OH^-) = 1E-6 M
[Ca(OH)2] = 1/2*(OH^-) = 5E-4 M
I did not try to correct for the H3O^+ and/or OH^- in solution from the water. The numbers shown above are 10 times away from normal (10^-7 for each) so there will be only a very slight change if that were included. I've ignored the difference.
A Ca(OH)2 solution has a pH of 8.0. Determine the values of [H3O+], [OH– ], and [Ca(OH)2] for the solution
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