Answer
In this $NaOH$ solution: $[H_3O^+] = 4.0 \times 10^{- 13}M$
Work Step by Step
1. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[OH^-]$, and solve for "$[H_3O^+]$".
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.5 \times 10^{- 2} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.5 \times 10^{- 2}}$
$[H_3O^+] = 4.0 \times 10^{- 13}M$
