Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (Ka(HCN) = 4.9 × 10^–10)

Answer 1

millimols KOH initially = mL x M = 75.0 x 0.15M = 11.25
millimols HCN initially = 7.0
..........KOH + HCN ==> KCN + H2O
I...........0.........7.0.............0.........0
add......11.25..................................................
C........-7.0......-7.0..........7.0...........7.0
E..........4.25......0............7.0............7.0
From the E line you can see you have an excess of a very strong base (i.e., KOH) so the pH will be determined by the KOH. This is NOT a buffered solution.
pOH = -log(KOH)
Then pH + pOH = pKw = 14. You know pKw and pOH, solve for pH.