Finding pH after dilution of acetic acid
$$\ce{CH_3COO- + H_2O&lt;=&gt;CH3COOH + OH-}$$
Approximating that all the $\ce{OH-}$ is from this equilibrium, $[\ce{CH3COOH}] = [\ce{OH-}]$.
\begin{align}
  [\ce{OH-}]_\text{initial} &amp;= [\ce{CH3COOH}]_\text{initial} \\
    &amp;= \pu{10^{-4}M}\\
  \frac{[\ce{OH-}]_\text{initial}[\ce{CH3COOH}]_\text{initial}}{[\ce{CH3COO-}]_\text{initial}[\ce{H2O}]_\text{initial}} 
  &amp;= \frac{[\ce{OH-}]_\text{final}[\ce{CH3COOH}]_\text{final}}{[\ce{CH3COO-}]_\text{final}[\ce{H2O}]_\text{final}}
\end{align}
Then you can choose to approximate $[\ce{H2O}]$ as a constant value and omit it from both sides (or not if you need great accuracy).
\begin{align}
  [\ce{CH3COO-}]_\text{initial} &amp;= \pu{1 M}\\
  [\ce{CH3COO-}]_\text{final} &amp;= \pu{\frac{10}{100} M}\\
  x &amp;= [\ce{OH-}]_\text{final} = [\ce{CH3COOH}]_\text{final}
\end{align}
Solve for $x$.