[samsanof]

What is the pH of the solution resulting from 25 mL .25 M NaOH being added to 50 mL .25 M HCN

Ka HCN = 4.9 x[10][/-10]

I have tried this question countless times, using the henderson-hasselbalch equation. Since it is the midpoint of the titration, the HCN and CN will be equal, thus cancelling, so the pH of the solution should equal the pKa , which is 9.3. However, my instructor says that the answer is 5.2. Can someone explain why I am wrong?

[Borek]

Your instructor is wrong. 5.2 is more or less pH of a 0.125 M solution of HCN. Someone forgot about CN^- from the neutralization.

[mjc123]

5.2 is the pH of a 0.084M solution of HCN. So they remembered to correct for the change in volume, but...