Question

Consider the titration of 45.0 mL of 0.222-M of KX with 0.127-M HCl. The pKa of HX = 8.88. Give all pH values to 0.01 pH units. a) What is the pH of the original solution before addition of any acid? pH = b) How many mL of acid are required to reach the equivalence point? VA = mL c) What is the pH at the equivalence point? pH = d) What is the pH of the solution after the addition of 27.5 mL of acid? pH = e) What is the pH of the solution after the addition of 94.4 mL of acid? pH =

Answer 1

a)

mmoles of X- = 45 x 0.222 = 9.99

pKa = 8.88

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

     = 7 + 1/2 (8.88 + log 0.222)

pH = 11.11

b)

At equivalence point :

mmoles of KX = mmoles of HCl

9.99 = 0.127 x V

V = 78.66

volume of acid = 78.7 mL

c)

volume at equivalence point = 78.7 mL

here weak acid remains.

concentration of acid = 9.99 / 45 + 78.7 = 0.08078 M

pH = 1/2 (pKa - log C)

     = 1/2 (8.88 - log 0.08078)

pH = 4.986

d)

mmoles of HCl = 27.5 x 0.127 = 3.4925

X-       +    HCl    -----------> HX

9.99        3.49                     0

6.4975          0                     3.4925

pH = pKa + log [salt / acid]

    = 8.88 + log [6.4975 / 3.4925]

pH = 9.15

e)

here strong acid reamins.

pH = 1.84