What is the pH of a mixture of $ 1.0 \; L \; KOH \; 1.0 \; \frac{mol}{L} $ with $ 1.0 \; L \; HCN \; 0.50 \; \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!
$KOH+HCN \rightarrow KCN+H_2O$
Complete the following table:
| $KOH$ | + | $HCN$ |  |
$KCN$ | + | $H_2O$ | |
| initial nb. of moles | 1.0 | 0.5 | 0 | ||||
| variation of mole nbs | ........ | ........ | ........ | ||||
| final nb. of moles | ........ | ........ | ........ |
| $KOH$ | + | $HCN$ | |
$KCN$ | + | $H_2O$ | |
| initial nb. of moles | 1,0 | 0,5 | 0 | ||||
| variation of mole nbs | -0.50 | -0.50 | +0.50 | ||||
| final nb. of moles | 0.50 | 0.0 | 0.50 |
Write the solutes present at the end with their molarities!
$[KCN]=\frac{0.5}{2.0}=0.25\frac{mol}{L}$ $[KOH]=\frac{0.5}{2.0}=0.25\frac{mol}{L}$
Analyse the final mixture!
$KCN$ has dissociated to $K^+$ and $CN^-$. $K^+$ has no influence. $CN^-$ is a weak base $KOH$ has dissociated to $K^+$ and $OH^-$. $K^+$ has no influence. $OH^-$ is treated as a strong base .
Calculate the pH!
$pOH=-log[OH^-]=-log0.25=0.60$ $pH$ $=$ $14-0.60$ $=$ $13.4$ .
