A solution is prepared by dissolving 0.050 mol of HCN and 0.025 mol NaCN in 100 ml of water, what is the pH?
Okay, I know to redo molarities so HCN--> .5 and NaCN--> .25
pH=-log (H)^(1/2)
but NaCN is a salt right.. a neutral salt so pH= 7.0
The answer is 8.91 but I cant seem to get the process help!!!
2006-12-11 14:12:53 · 1 answers · asked by Myra G 5 in Science & Mathematics ➔ Chemistry
NaCN is the salt of weak acid HCN and a strong base NaOH. The salt NaCN will be alkaline and not neutral.
Noting Ka = [H+] [CN-]/[HCN] yields
pH = pKa - log([HCN]/[CN-])
Assuming the value of pKa for HCN is 9.3 then
pH =9.3 -log(0.05/0.025) = 9.3 - log 2 = 9.0
Note If you maintain the pH = 8.9, your pKa value must have been near 9.2. Such disagreement in pKa values is not uncommon.
2006-12-11 18:27:40 · answer #1 · answered by Anonymous · 0⤊ 0⤋