Question

Find the pH during the titration of 20.00 mL of 0.1500 M benzoic acid, C6H5COOH (Ka = 6.3 10-5), with 0.1500 M NaOH solution after the following additions of titrant. (a) 0 mL (b) 10.00 mL (c) 15.00 mL (d) 20.00 mL (e) 25.00 mL

Answer 1

millimoles of benzoic acid = 20 x 0.15 = 3

pKa = 4.2

after adding 0 mL :

[H+] = sqrt (Ka x C)

       = sqrt ( 6.3 x 10^-5 x 0.15)

       = 3.07 x 10^-3 M

pH = -log [H+] = -log [3.07 x 10^-3 ]

      = 2.51

pH = 2.51

b) after adding 10.00 ml NaOH :

millimoles of NaOH = 10 x 0.15 = 1.5

C6H5COOH + NaOH ---------------> CH3COONa + H2O    

     3                 1.5                                  0               0

1.5                 0                                   1.5             1.5

it is half- equivalence point. so pH

pH = pKa

pH = 4.20

c) after adding 15 ml NaOH :

millimoles of NaOH = 15 x 0.15 = 2.25

C6H5COOH + NaOH ---------------> CH3COONa + H2O    

     3                 2.25                                  0               0

0.75                0                                   2.25            2.25

pH = pKa + log[salt/acid]

pH = 4.20 + log[2.25/0.75]

pH = 4.68

a) after adding 20.00 mL :

    NaOH millimoles = 20 x 0.15 = 3

it is half equivalence point . so here salt only remains. so

salt concentration = 3 / 20 + 20 = 0.075 M

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (4.2 + log 0.075)

pH = 8.54

e)

25 ml NaOH :

millimoles = 25 x 0.15 = 3.75

here strong base reamined. so

[OH-] = 3.75 - 3 / (20 + 25) = 0.0167 M

pOH = -log [OH-] = -log [0.0167] = 1.78

pH =12.2