Question

A 150.0 mL buffer solution containing 0.100 M hydrocyanic acid (HCN, Ka = 6.2 x 10^-10) and 0.100 M potassium cyanide (KCN) has 30.0 mL of 0.120 M KOH added to it. What is the pH of the solution after the KOH has been added

a. 8.73 b. 9.00 c. 9.21 d. 9.29 e. 9.42

Answer 1

we know that

moles = molarity x volume (L)

so

moles of HCN = 0.1 x 150 x 10-3 = 15 x 10-3

moles of KCN = 0.1 x 150 x 10-3 = 15 x 10-3

now

moles of KOH added = 0.12 x 30 x 10-3 = 3.6 x 10-3

now

the reaction is

HCN + OH- ---> CN- + H20

so

moles of HCN reacted = moles of KOH added = 3.6 x 10-3

so

moles of HCN left = 15 x 10-3 - 3.6 x 10-3 = 11.4 x 10-3

now

moles of CN- formed = moles of KOH added = 3.6 x 10-3

new moles of CN- = 15 x 10-3 + 3.6 x 10-3 = 18.6 x 10-3

now

for buffers

pH = pKa + log [conjugate base / acid ]

also

pKa = -log Ka

so

pH = -log Ka + log [CN-/HCN]

pH = -log 6.2 x 10-10 + log [ 18.6 x 10-3 / 11.4 x 10-3]

pH = 9.42

so

the answer is e) 9.42