[Violet89]

Calculate the pH of a solution that results from mixing 38.0 mL of 0.150 M HCN(aq) with 36.1 mL of 0.129 M NaCN. The Ka value for HCN is 4.9 x 10^-10.

pH = pKa + log([A-]/[HA]) (Henderson-Hasselbalch Eqn.)

Volume after mixing = 74.1 mL

[A-] = 0.129 M * 36.1 mL / 74.1 mL = 0.062846154 mL

[AH] = 0.150 M * 38.0 mL / 74.1 mL = 0.076923077 mL

pH = -log(4.9 x 10^-10) + log(0.062846154 / 0.076923077 ) = 9.22

Is my work correct?

[Borek]

Looks OK to me, just a typo:

[A-] = 0.129 M * 36.1 mL / 74.1 mL = 0.062846154 mL

These are not mL. Same about [HA].