Answer
$9.1\times10^{-6}$ , 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 $H_{3}O^{+}$ set the equations equal to each other and plug in the concentration of $OH^{-}$.
$1\times10^{-14}=(1.1\times10^{-9})[H_{3}O^{+}]$
Rearranging, we get:
$\frac{1\times10^{-14}}{1.1\times10^{-9}}=[H_{3}O^{+}]$ so $[H_{3}O^{+}]=9.1\times10^{-6}$
Since this value for $[H_{3}O^{+}]$ is greater than the value for $[OH^{-}]$, the solution is acidic.
