What is the pH of a mixture of $ 1.0 \; L \; NaOH \; 1.0 \; \frac{mol}{L} $ with $ 1.0 \; L \; HCl \; 0.5 \; \frac{mol}{L} $? Write the (nonionic) equation of the reaction !
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$NaOH+HCl \rightarrow NaCl+H_2O$
Complete the following table:
| $NaOH$ | + | $HCl$ |  |
$NaCl$ | + | $H_2O$ | |
| initial nb. of moles | 1.0 | 0.5 | 0 | ||||
| variation of mole nbs | ........ | ........ | ........ | ||||
| final nb. of moles | ........ | ........ | ........ |
| $NaOH$ | + | $HCl$ | |
$NaCl$ | + | $H_2O$ | |
| initial nb. of moles | 1.0 | 0.5 | 0 | ||||
| variation of mole nbs | -0.5 | -0.5 | +0.5 | ||||
| final nb. of moles | 0.5 | 0.0 | 0.5 |
Write the solutes present at the end with their molarities !
$[NaCl]=\frac{0.5}{2.0}=0.25\frac{mol}{L}$ $[NaOH]=\frac{0.5}{2.0}=0.25\frac{mol}{L}$
What is the final pH ?
$NaCl$ has no influence. $NaOH$ dissociates to form $OH^-$: $pOH$ $=$ $-log\;0.25$ $=$ $0.60$ $pH=14-0.60=13.40$
