Answer
$1.2\times10^{-10}$, Acidic
Work Step by Step
For this we need to use the equations $K_{w}=[OH^{-}][H_{3}O^{+}]$ and $K_{w}=1\times10^{-14}$. To calculate the concentration of $OH^{-}$ set the equations equal to each other and plug in the concentration of $H_{3}O^{+}$.
$1\times10^{-14}=(8.5\times10^{-5})[H_{3}O^{+}]$
Rearranging, we get:
$\frac{1\times10^{-14}}{8.5\times10^{-5}}=[H_{3}O^{+}]$ so $[H_{3}O^{+}]=1.2\times10^{-10}$
Since this value for $[OH^{-}]$ is less than the value for $[H_{3}O^{+}]$, the solution is acidic.
