(multiple choice)Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH.?

Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka= 4.9 X 10^-10) solution.
a. 5.15
b. 8.71
c. 5.85
d. 9.91
e. 13.0

I did this problem about 3 times and keep getting 9.13. PLZZZ HELP

That's a tricky one!

Part of the HCN has been neutralized so the new conc. of HCN is 0.0667 M. There has also been CN- ions put into solution as the salt NaCN the conc of CN- is 0.01667 M.
So putting these concs into the equation for Ka and solving for [H+] then taking logs I got a pH of 8.706.
Better check this!
I get 12.22 :S. If (for some strange reason, though my uni chem textbook has done this a couple of times) you assume that the addition of the NaOH doesnt affect the volume (once again, i can not see why, but i have seen it done) then i get 11.3. the only way to get an answer off that list is if u use 10mL as the volume :S. and that is for e) 13.0

does anybody else have any clue?

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