posted by Tim's Titration Lab .
If 10.0 mL of 0.250 mol/L NaOH(aq) is added to 30.0 mL of 0.17 mol/L HOCN(aq), what is the pH of the resulting solution?
NaOH + HOCN ==> NaOCN + HOH
mols NaOH to begin = M x L = 0.250 x 0.010 = ??
mols HOCN to begin = M x L = 0.17 x 0.030 = ??
There is more HOCN than NaOH. The difference in mols is how much HOCN is left unreacted. The NaOCN formed is the amount of the lesser chemical (in this case NaOH). So you have a buffer formed consisting of a weak acid (HOCN) and its salt (NaOCN).
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
Post your work if you get stuck.