Question

Consider the titration of 40.0 mL of 0.243-M of KX with 0.097-M HCl. The pK_a of HX = 7.09. 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?

V_A =  mL

c) What is the pH at the equivalence point?

pH =

d) What is the pH of the solution after the addition of 68.1 mL of acid?

pH =

e) What is the pH of the solution after the addition of 120.2 mL of acid?

pH =

Answer 1

pKa of HX = 7.09

Consider the titration of 40.0 mL of 0.243-M of KX with 0.097-M HCl. The pK_a of HX = 7.09. Give all pH values to 0.01 pH units.

a)

It is a salt of strong base and weak acid. so pH > 7

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

      = 7 +1/2 (7.09 + log 0.243)

      = 10.24

pH = 10.24

b) at equivalence point :

millimoles of salt = millimoles of acid

40 x 0.243 = 0.097 x V

Volume of acid = 100 mL

c)

KX    +     HCl <-------------->    KCl +    HX

9.72        9.72                           0           0

0            0                             9.72       9.72

here only HX is remained.

concentration of [HX] = 9.72 / 40 + 100 = 0.0693 M

pH = 1/2 (pKa - log C)

     = 1/2 (7.09 - log 0.0693)

pH = 4.12

d)

millimoles of acid = 68.1 x 0.097 = 6.606

KX    +     HCl <-------------->    KCl +    HX

9.72       6.606                           0           0

3.114       0                             6.606      6.606

pH = pKa + log [salt / acid]

     = 7.09 + log [3.114 / 6.606]

    = 6.76

pH = 6.76

e)

millimoles of acid = 120.2 x 0.097 = 11.66

KX    +     HCl <-------------->    KCl +    HX

9.72       11.66                          0           0

0     1.94                         9.72         9.72

[HCl] = 1.94 / (40 + 120.2) = 0.0121

pH = -log [H+] = -log (0.0121)

pH = 1.92