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)

Question

My answer is 9.52, but im not sure.

Answer 1

I don't buy that. I'm interested in how you obtained 9.52.
Initially HCN millimols = mL x M = 35.0 x 0.20 = 7
You add KOH millimols = mL x M = 75.0 x 0.15 = 11.25
..........KOH + HCN ==> KCN + H2O
I...........0..........7.0............0..........0
add...11.25..................................
C.........-7.0.......-7.0............+7.0................
E..........4.25..........0.............7.0................
So you end up with a solution of 4.15 millimols KOH and 7.0 millimols of the salt, KCN, in a total volume of 76.0+35.0 = 110 mL.
KOH is a strong base. KCN is a salt. Although KCN will hydrolyze somewhat, the pH determining material is KOH. (KOH) = 4.25 mmols/110 mL = ?
pOH = -log (KOH), then
pH + pOH = pKw = 14. You know pKw and pOH, solve for pH.
I think something like 12 or so.

I

Answer 2

Thank you. :D