What is the pH of a mixture of $ 1.0 \; L \; NaOH \ 0.50 \; \frac{mol}{L} $ with $ 1.0 \; L \; CH_3COOH \; 1.0 \; \frac{mol}{L} $? Write the (nonionic) equation of the reaction!
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$NaOH+CH_3COOH \rightarrow CH_3COONa+H_2O$
Complete the following table:
| $NaOH$ | + | $CH_3COOH$ |  |
$CH_3COONa$ | + | $H_2O$ | |
| initial nb. of moles | 0.50 | 1.0 | 0 | ||||
| variation of mole nbs | ........ | ........ | ........ | ||||
| final nb. of moles | ........ | ........ | ........ |
| $NaOH$ | + | $CH_3COOH$ | |
$CH_3COONa$ | + | $H_2O$ | |
| initial nb. of moles | 0.50 | 1.0 | 0 | ||||
| variation of mole nbs | -0.50 | -0.50 | +0.50 | ||||
| final nb. of moles | 0 | 0.50 | 0.50 |
Write the solutes present at the end with their molarities!
$[CH_3COONa]=\frac{0.5}{2.0}=0.25\frac{mol}{L}$ $[CH_3COOH]=\frac{0.5}{2.0}=0.25\frac{mol}{L}$
Analyse the final mixture !
$CH_3COONa$ dissociates to $CH_3COO^-$ and $Na^+$ $Na^+$ has no influence. $CH_3COO^-$ is a weak base. $CH_3COOH$ is the corresponding acid. We are in the presence of a buffer mixture!
Calculate its pH!
$pH$ $=$ $pK_a+log\frac{n_{CH_3COO^-}}{n_{CH_3COOH}}$ $=$ $4.75+log\frac{0.5}{0.5}$ $=$ $4.75$ .
