Answer
In bleach: $[H_3O^+] = 4.8 \times 10^{- 12}M$
Work Step by Step
1. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[OH^-]$, and solve for $[H_3O^+]$:
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.1 \times 10^{- 3} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.1 \times 10^{- 3}}$
$[H_3O^+] = 4.8 \times 10^{- 12}M$
