## General Chemistry 10th Edition

$1.19\times 10^{-10}$
We know that $H_3O^+=\frac{K_w}{OH^-}$ We plug in the known values to obtain: $H_3O^+=\frac{1.00\times 10^{-14}}{8.4\times 10^{-5}}=1.19\times 10^{-10}$