# 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 &<=> 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 $$C_1$$?

## 1 Answer

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

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 $$x$$.