Question

Calculate the pH during the titration of 40.00 mL of a 0.1000 M propanoic acid (Ka = 1.3 x 10^-5) after each of the following volumes of 0.1000M NaOh has been added. Use ICE tables to show your work.
a) 0.00 mL
b) 25.00 mL
c) 40.00 mL
d) 50.00 mL

The answers are:
a. 2.96
b. 5.12
c. 8.80
d. 12.05

Show work on how you arrive to each solution.

Answer 1

pk Of propanoic acid = -logka

              = -log(1.3*10^-5) = 4.89

a)

pH = 1/2(pka-logC)

      = 1/2(4.89-log0.1)

      = 2.945

b) no of mol of propanoic acid =40*0.1 = 4 mmol

   no of mol of NaOH = 25*0.1 = 2.5 mmol

pH = pka + log(salt/acid)

      = 4.89+log(2.5/(4-2.5))

     = 5.112

c) no of mol of propanoic acid =40*0.1 = 4 mmol

   no of mol of NaOH = 40*0.1 = 4 mmol

concentration of salt = 4/80 = 0.05 M

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

     = 7+1/2(4.89+log0.05)

     = 8.8

d) no of mol of propanoic acid =40*0.1 = 4 mmol

   no of mol of NaOH = 50*0.1 = 5 mmol

concentration of excess NaOH = (5-4)/90 = 0.011 M

pH = 14-(-log0.011)

     = 12.05