Chapter 15 - Acids and Bases - Questions and Problems - Page 659: 15.77

Please see the work below.

We know that $log[H_3O^+]=-PH=-11.63$ $log[H_3O^{+}]=-11.63$ $[H_3O^+]=anti log(-11.63)$ $[H_3O^+]=10^{-11.63}=2.34\times 10^{-12}M$ We also know that $[OH]^-=\frac{K_w}{[H_3O]^+}$ $[OH]^-=\frac{1.0\times 10^{-14}}{2.34\times 10^{-12}}=4.27\times 10^{-3}$