Question

Consider formic acid, HCHO2 (Ka = 1.9x10-4) and its conjugate base, CHO2-.

a) A buffer is prepared by dissolving 25.0 grams of sodium formate in 1.30 L of a solution of 0.660 M formic acid. Assume the volume remains constant after dissolving the sodium formate.

   i) Calculate the pH of this buffer.

   ii) Calculate the pH after 0.100 mole of nitric acid is added to this buffer.

   iii) Calculate the pH after 10.0 grams of potassium hydroxide is added to this buffer.

   iv) What is the buffer capacity of this buffer?

b) 35.0 mL of 0.739 M formic acid is titrated with 0.431 M potassium hydroxide.

   i) What is the pH of the formic acid solution before titration begins?

   ii) How many milliliters of KOH is required for this titration to reach its equivalence point?

   iii) What is the pH of the solution after 10.0 mL of KOH has been added?

   iv) What is the pH of the solution halfway to equivalence?

   v) What is the pH of the solution at the equivalence point?

Answer 1

b)

HCOOH , Ka value = 1.9 x 10^-4

pKa = 3.72

i) What is the pH of the formic acid solution before titration begins?

pH = 1/2 [pKa - logC]

pH = 1/2 [3.72 -log 0.739]

pH = 1.93

ii) How many milliliters of KOH is required for this titration to reach its equivalence point?

here millimoles of acid = millimoles of base

35 x 0.739 = 0.431 x V

V = 60 mL

volume of base = 60 mL

iii) What is the pH of the solution after 10.0 mL of KOH has been added?

millimoles of acid = 35 x 0.739 = 25.86

millimoles of base = 0.431 x 10 = 4.31

HCOOH + KOH ----------------> HCOOK + H2O

25.86            4.31                            0            0

21.55               0                               4.31     -

pH = pKa + log [HCOOK]/ [HCOOH]

pH = 3.72 + log (4.31/21.55)

pH = 3.02

iv) What is the pH of the solution halfway to equivalence?

at halfway equivalence point : pH = pKa

pH = 3.72

   v) What is the pH of the solution at the equivalence point?

here only salt is presnet

[salt] = 25.86 / (35+60)

          = 0.272 M

pH = 7 + 1/2 [pKa + logC] = 7 + 1/2 [3.72 + log 0.272]

pH = 8.58