Question

Part A. Calculate the pH of a buffer solution that is 0.247 M in HCN and 0.171 M in KCN. For HCN, Ka = 4.9×10⁻¹⁰ (pKa = 9.31).

Part B.

Calculate the pH of a buffer solution that is 0.230 M in HC2H3O2 and 0.170 M in NaC2H3O2. (Ka for HC2H3O2 is 1.8×10−5.)

Part C. Calculate the pH of 1.0 L of the original buffer, upon addition of 0.110 mol of solid NaOH.

Answer 1

Part-A

HCN = 0.247M

KCN = 0.171M

PKa = 9.31

for acidic buffer

PH = PKa + log[salt]/[acid]

PH = Pka + log[KCN]/[HCN]

PH = 9.31+log(0.171/0.247)

PH = 9.15

Part-B

concentration of HC2H3O2 = 0.230M

Concentration of NaC2H3O2 = 0.170M

Ka = 1.8x10^-5

-log(Ka) = -log(1.8x10^-5)

PKa = 4.74

PH = 4.74 + log(0.170/0.230)

PH = 4.61

part-C

a) In a original buffer

HCN = 1.0L of 0.247 M

number of moles of HCN = 0.247M x 1.0L = 0.247 moles

KCN = 1.0L of 0.171M

number of moles of KCN = 0.171M x 1.0L = 0.171 mole

number of moles of NaOH = 0.110M

after addition of NaOH

number of moles of HCN = 0.247 - 0.110 = 0.137 moles

number of moles of KCN = 0.171 + 0.110 = 0.281 moles

PH = 9.31 + log(0.281/0.137)

PH = 9.62

b)

HC2H3O2 = 1.0L of 0.230M

number of moles of HC2H3O2 = 0.230M x 1.0L = 0.230 moles

NaC2H3O2 = 1.0L of 0.170M

number of moles of NaC2H3O2 = 0.170M x 1.0L = 0.170 moles

number of moles of NaOH = 0.110 moles

after addition of NaOH

number of moles of HC2H3O2 = 0.230 - 0.110 = 0.120 moles

number of moles of NaC2H3O2 = 0.170 + 0.110 = 0.280 mole

PH = 4.74 + log(0.280/0.120)

PH = 5.11.