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?