Question

A)Calculate the pH of 0.100 L of a buffer solution that is 0.27 M in HF (Ka = 3.5 x 10-4 ) and 0.51 M in NaF?B)What is the pH after adding 0.004 mol of HNO3 to the buffer described in Part A?

C)What is the pH after adding 0.004 mol of KOH to the buffer described in Part A?

Answer 1

A)

volume = 0.1 L

molatity of HF = 0.27 M

molarity of NaF = 0.51 M

pKa of HF = 3.46

pH = pKa + log [salt / acid]

     = 3.46 + log (0.51 / 0.27)

     = 3.74

pH = 3.74

B)

HNO3 moles (C) = 0.004

pH = pKa + log [salt - C] / [acid + C]

      = 3.46 + log [0.51 x 0.1 - 0.004 / 0.27 x 0.1 + 0.004]

      = 3.64

pH = 3.64

C)

moles of KOH (C) = 0.004

pH = pKa + log [salt + C / acid - C]

     = 3.46 + log [0.51 x 0.1 + 0.004 / 0.27 x 0.1 - 0.004]

     = 3.84

pH = 3.84