Question

What is the pH of 1L of buffer made from 0.6M HF and 0.54M NaF (pKa of HF = 3.45)? If 0.01M HCl is added (disregard change in volume), what is the new pH? If 0.01M NaOH had been added instead?

Answer 1

HF = 1.0L of 0.6M

number of moles of HF = 0.6M x 1.0L = 0.6 moles

NaF = 1.0L of 0.54M

number of moles of NaF = 0.54M x 1.0L = 0.54 moles

Pka = 3.45

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

PH = 3.45 + log(0.54/0.6)

PH = 3.40

a)

concentration of HCl = 0.01M = 0.01 mol/L

number of moles of HCl = 0.01 mole

number of moles of HF = 0.6M x 1.0L = 0.6 moles

number of moles of NaF = 0.54M x 1.0L = 0.54 moles

after addition of HCl

number of moles of HF = 0.6 + 0.01 = 0.61 moles

number of moles of NaF = 0.54 - 0.01 = 0.53 moles

PH = 3.45 + log(0.53/0.61)

PH = 3.39

b) NaOH = 0.01M = 0.01 mol/L

mnumber of moles of NaOH = 0.01 mole

after addition of NaOH

number of moles of HF = 0.6- 0.01 = 0.59 moles

number of moles of NaF = 0.54 + 0.01 = 0.55moles

PH = 3.45 + log(0.55/0.59)

PH = 3.42.