[Nathaniel]

If 10.0 mL of 0.250 mol/L NaOH(aq) is added to 30.0 mL of 0.17 mol/L HOCN(aq0, what is the resulting solution?

My work:
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 HOCN HOCN is left unreacted. The NaOCN formed is the amount of the lesser chemical (in this case NaOH). So I have a buffer formed consisting of a weak acid(HOCN) and its salt (NaOCN)

Now I use the Henderson-Hasselback equation

pH = pKa + log(base)/(acid)

can someone please finish off the rest of the answer?

[Rabn]

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