I have a mixture of 100 ml HCN 0.148 mol/l and 50 ml KOH 0.296 mol/l.
I think HCN is a weak acid and KOH is a strong base. For the calculation of the PH I therefore look at the strong base.
I calculate the number of moles of both, this is 0.0148 moles for both.
If I write this in my reaction schedule now
HCN + KOH w KCN + H2O
0.0148 0.0148
-0.0148 -0.01488
Then there will be nothing left of my KOH.
So I thought PH would be 7. however according to my solution it is 11.15.
Where does my reasoning go wrong?
Answer
Hi Els,
That’s an interesting problem! You correctly note that HCN is a weak acid and KOH is a strong base. So both will react with each other. For this we calculate the concentrations. When 100 mL HCN with concentration 0.148 mol.L-1 is diluted with 50 mL of KOH then the final volume is equal to 150 mL. The final concentration after dilution is then equal to:
CHCN = 0.148*100/150 = 0.09867 mol.L-1
Analogous is the concentration of KOH after dilution
CKOH = 0.296*50/150 = 0.09867 mol.L-1
Both concentrations are equal. The following response is now on:
HCN + KOH –> KCN + H2O
A salt of a weak acid (KCN in this case) obviously also has acid-base properties, and that’s the mistake you made. A solution of cyanide, CN–, is a weak base (and in this case a fairly strong weak base!). The pH of such a solution is given by the formula:
pH = -log{[(Ka*Kw)/C]^0.5}
(This is an approximate formula, but the approximation is good enough for this case)
Filling in the constants: Ka = 10^-9.21 ; kw = 10^-14 and C = 0.09867 returns:
pH = 11.10
(The Ka-value is of course the Ka from HCN)
Answered by
Dr Etienne Jooken
Chemistry
Old Market 13 3000 Leuven
https://www.kuleuven.be/
.