What is the pH of a mixture of $ 1.0 \; L \; KOH \; 1,0 \; \frac{mol}{L} $ with $ 1.0 \; L \; HF \; 1.0 \; \frac{mol}{L} $? Write the (nonionic) equation of the reaction!
For answers, use (possibly several times) the arrows ↑ Down! and ↓ Up! Complete please this question before moving on to the next one!
$KOH+HF \rightarrow KF+H_2O$
Complete the following table:
| $KOH$ | + | $HF$ |  |
$KF$ | + | $H_2O$ | |
| initial nb. of moles | 1.0 | 1.0 | 0 | ||||
| variation of mole nbs | ........ | ........ | ........ | ||||
| final nb. of moles | ........ | ........ | ........ |
| $KOH$ | + | $HF$ | |
$KF$ | + | $H_2O$ | |
| initial nb. of moles | 1.0 | 1.0 | 0 | ||||
| variation of mole nbs | -1.0 | -1.0 | +1.0 | ||||
| final nb. of moles | 0.0 | 0.0 | 1.0 |
Write the solutes present at the end with their molarities !
$[KF]=\frac{1.0}{2.0}=0.50\frac{mol}{L}$
Analyse the final mixture!
$KF$ has dissociated to $K^+$ and $F^-$. $K^+$ has no influence. $F^-$ is a weak base $(pK_b=10,83)$.
Calculate the pH!
Let $y=[OH^-]$: $y^2$ $+$ $10^{-10.83}y$ $-$ $10^{-10.83}0.50$ $=$ $0$ donne: $y=5.4\cdot 10^{-6}$. $pOH=5.26$ $pH$ $=$ $14-5.26$ $=$ $8.74$ .
