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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 &<=> CH3COO- + Na+}\\ \ce{CH3COO- + H2O &-> 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 $C_1$?

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$$\ce{CH_3COO- + H_2O<=>CH3COOH + OH-}$$

Approximating that all the $\ce{OH-}$ is from this equilibrium, $[\ce{CH3COOH}] = [\ce{OH-}]$.

\begin{align} [\ce{OH-}]_\text{initial} &= [\ce{CH3COOH}]_\text{initial} \\ &= \pu{10^{-4}M}\\ \frac{[\ce{OH-}]_\text{initial}[\ce{CH3COOH}]_\text{initial}}{[\ce{CH3COO-}]_\text{initial}[\ce{H2O}]_\text{initial}} &= \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} &= \pu{1 M}\\ [\ce{CH3COO-}]_\text{final} &= \pu{\frac{10}{100} M}\\ x &= [\ce{OH-}]_\text{final} = [\ce{CH3COOH}]_\text{final} \end{align}

Solve for $x$.

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