Answer
In coffee: $[OH^-] = 1.0 \times 10^{-9}M$
Work Step by Step
1. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[H_3O^+]$, and solve for $[OH^-]$:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$1.0 \times 10^{-5} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{1.0 \times 10^{-5}}$
$[OH^-] = 1.0 \times 10^{-9}M$
