Question

What is the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 M HCN with 50.00 mL of 0.030 M NaCN? Assume that the volumes of the solutions are additive and that Ka = 4.9 x 10^ -10 for HCN.

Answer 1

H+ =

V = 50 ml of HCN

M = 0.1 M

V = 50 ml of NACN

M = 0.03

Ka = 4.9 x 10^ -10

Apply buffer equation

pH = pKa + log(A-/HA)

calculate A-

[A-] = M2*V2 /VT = 0.03*50 / 100 = 0.015

[HA] = M1*V1 /VT = 0.10*50 / 100 = 0.05

Substitue

pH = pKa + log(A-/HA)

pH = -log(4.9 x 10^ -10) + log( 0.015/0.05) = 8.786

pH = 8.79