Answer
In vinegar: $[H_3O^+] = 1.0 \times 10^{- 3}M$.
Work Step by Step
1. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[OH^-]$, and solve for "$[H_3O^+]$".
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1.0 \times 10^{- 11} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1.0 \times 10^{- 11}}$
$[H_3O^+] = 1.0 \times 10^{- 3}M$
