Answer
$[H_3O^+] = 5.0 \times 10^{- 6}M$.
Work Step by Step
1. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[OH^-]$, and solve for "$[H_3O^+]$".
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.0 \times 10^{- 9} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.0 \times 10^{- 9}}$
$[H_3O^+] = 5.0 \times 10^{- 6}M$
