Kw (water) = 1.011 * 10-14
Kw = [H3O+] * [OH-] (product of Hydronium and Hydroxide ion concentrations.
Given [OH-] = 8.9 * 10-3M
Therefore, [H3O+] = Kw / [OH-] = 1.011 * 10-14 / 8.9 * 10-3 = 0.1136 * 10-11 M = 1.136 * 10-12 M
Now pH = -log10[H3O+] = -log101.136 * 10-12 = 11.9
pOH = 14 - pH =14 - 11.9 = 2.1
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