Question

Determine the pH during the titration of 24.3 mL of 0.310 M formic acid (K_a = 1.8×10^-4) by 0.442 M NaOH at the following points.

(a) Before the addition of any NaOH

(b) After the addition of 4.20 mL of NaOH

(c) At the half-equivalence point (the titration midpoint)

(d) At the equivalence point

(e) After the addition of 25.6 mL of NaOH

Answer 1

a) before addition of NaOH

    pH = 1/2(pka-logC)

pka of formic acid   = -logka = -log(1.8*10^-4) = 3.745

pH = 1/2(3.745-log0.31)
   
     = 2.13

b) . no of mol of formicacid = 24.3*0.31 = 7.533 mmol

   no of mol of NaOH = 4.2*0.442 = 1.86 mmol

PH = PKa +log(base/acid-vase)

     = 3.745+log(1.86/(7.533-1.86))

     = 3.26

c) At the half-equivalence point, pH = pka = 3.745

d) no of mol of formicacid = 24.3*0.31 = 7.533 mmol

   no of mol of NaOH = 7.533 mmol

volume of NAoh = 7.533/0.442 = 17 ml

concentration of salt = 7.533/(24.3+17) = 0.182 M

pH = 7+1/2(pka+logC)

     = 7+1/2(3.745+log0.182)

     = 8.5

e) no of mol of formicacid = 24.3*0.31 = 7.533 mmol

   no of mol of naOH = 25.6*0.442 = 11.31 mmol

   concentration of excess NaOH = (11.31-7.533)/(25.6+24.3) = 0.076 M

   pH = 14- (-log0.076) = 12.9