Finding pH after dilution of acetic acid We have $\pu{10 L}$ of solution of $\ce{CH3COONa}$ with concentration $C = \pu{1 M}$ and $\mathrm{pH} = 10$. Then we add $\pu{90 L}$ $\ce{H2O}$. What will be the pH of the new solution?

My work:
\begin{align}
  \ce{CH3COONa &amp;&lt;=&gt; CH3COO- + Na+}\\
  \ce{CH3COO- + H2O &amp;-&gt; CH3COOH + OH-}
\end{align}
We have $\pu{1 M } \ce{CH3COO-}$ and let $x$ denote the concentration of $\ce{OH-}$. If the $\mathrm{pH}$ of the solution is $10$, it's implied that
$$\mathrm{pOH}=4\iff-\log_{10} x=4 \iff x=\pu{10^{-4} M}.$$
I know I have to use that the amounts of substances didn't change, so $$C_1V_1 = C_2V_2,$$
but I don't know what values I have to put in there. Is the $x$ I found my $C_1$?
$$\ce{CH_3COO- + H_2O&lt;=&gt;CH3COOH + OH-}$$
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 $x$.