Question

# A 150.0 mL buffer solution containing 0.100 M hydrocyanic acid (HCN, Ka = 6.2 x 10-10)...

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

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

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