Chapter 15 - Acids and Bases - Questions and Problems - Page 659: 15.78

Please see the work below.

We know that $log[H_3O^+]=-PH=-9.61$ $log[H_3O^{+}]=-9.61$ $[H_3O^+]=anti log(-9.61)$ $[H_3O^+]=10^{-9.61}=2.45\times 10^{-10}M$ We also know that $[OH]^-=\frac{K_w}{[H_3O]^+}$ $[OH]^-=\frac{1.0\times 10^{-14}}{2.34\times 10^{-10}}=4.08\times 10^{-5}M$