Question

4. Suppose you have a 20ml solution of 0.5M acetic acid (Ka = 1.8x10-5). What is the pH of the solution after adding 10ml of .4M sodium hydroxide? 15mL? 35mL? At the equivalence point? Draw the expected titration curve with equivalence point indicated

Answer 1

4)

millimoles of acetic acid = 20 x 0.5 = 10

Ka = 1.8 x 10^-5

pKa = 4.74

a) after adding 10ml of .4M sodium hydroxide :

millimoles of NaOH = 10 x 0.4 = 4

CH3COOH +   NaOH    ----------------> CH3COONa   + H2O

    10                   4                                      0    

     6                    0                                       4

pH = pKa + log [salt / acid]

     = 4.74 + log [4 /6 ]

pH = 4.56

b)

millimoles of NaOH = 15 x 0.4 = 6

CH3COOH +   NaOH    ----------------> CH3COONa   + H2O

    10                   6                                      0    

     4                    0                                       6

pH = pKa + log [salt / acid]

      = 4.74 + log [6/4]

pH = 4.92

c)

millimoles of NaOH = 35 x 0.4 = 14

CH3COOH +   NaOH    ----------------> CH3COONa   + H2O

    10                   14                                      0    

     0                     4                                       10

here strong acid remains.

[OH-] = 4 / (20 + 35) = 0.0727 M

pOH = -log (0.0727) = 1.14

pH = 12.86

d)

At equivalence point :

millimoles of acid = millimoles of base

10 = 0.4 x V

V = 25 mL

here salt only remains.

salt concentration = 10 / 25 + 20 = 0.22 M

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

      = 7 + 1/2 (4.74 + log 0.22)

     = 9.04

pH = 9.04