Calculate the {h+}, {oh-}, oh, poh, and % dissociation of a 1.0x10^-3 m solution of hcl

Since HCl is a strong acid, it readily dissociates, so we can assume that there are 1.0x10^-3 M of H+ ions. We can use the formula:

pH=-log[H+]
pH=-log(1.0x10^-3)
pH=3

We can find the pOH by subtracting 14-pH:

14-3=11
pOH=11

The [OH-] can be found by using the formula:

pOH=-log[OH-]
11=-log[OH-]
10^-11=[OH-]
[OH-]=1.0x10^-10

Where is the graph ?