Can You Calculate [OH-] For 1.6*10^-3 M Sr(OH)2?

Calculate the [OH-] = 1.6 x 10 -3 M Sr(OH)2

(Ksp= 3.2×10^-4 at 25 °C) A moderately strong hydroxide.

If I use the reaction: Sn(OH)2 --> Sn+2 + 2OH- then in an 1.6 x 10 -3 M solution of Strontium hydroxide the concentration of [OH-] = 2 x 1.6 x 10-3M = 3.2 x 10-3M. Sr(OH)2 pOH = 2.49, so
we have pH + pOH = 14 or 14 - pOH = pH = 14 - 2.49 = 11.51

If I calculate using the Ksp or solubility constant of Sr(OH)2 = 3.2 x 10-4 @ 25oC
= [Sr+2][OH-]2/Sr[OH]2 = [x][x]2/[1.6x10-3M] =

= Rewritten: [1.6 x 10-3M][3.2 x 10-4] = [x]2 = 5.12 x 10-7
                                                            [x]  = sqrt [5.12 x 10-7]
                                                            [x]  = 7.16 x 10-4M
                                 -log[pOH] or -log[7.16 x 10-4M]= 3.14
pH + pOH = 14,    pH = 14 - 3.14 = 10.86

So, without using the solubility constant for a moderately strong hydroxide you can get a fairly good answer or an educated guess. Utilizing the  solubility constant does however give results that are statistically different.

With Ksp, pH = 10.86

Without the constant, assuming full dissociation pH = 11.51
Both give significantly strong basic values of 10.68 and 11.51. There can be other corrections; I think this will do...
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