Consider nicotinic acid, HC6H4NO2 (HNic Ka= 1.4x10^-6) and its conjugate base, C

ID: 1021065 • Letter: C

Question

Consider nicotinic acid, HC6H4NO2 (HNic Ka= 1.4x10^-6) and its conjugate base, C6H4NO2^-1 (Nic^-1 Kb = 7.1x10^-10).

A buffer is prepared by dissolving 47.13 g of NaC6H4NO2 (NaNic) in 1.25L of a soution of 40.0 mL of 0.575 M HNic is titrated with .335 M KOH.

A. Calculate the pH of the solution before titrating.

B. What is the pH of the solution halfway to the equivalence point?

C. Calculate the pH after 50.0 mL of KOH has been added.

D. Calculate the volume of KOH required to reach the equivalence point?

E. Calculate the pH of the solution at the equivalence point?

Thank you for your help!

Titration

A. pH before titration

HNic <==> H+ + Nic-

Ka = 1.4 x 10^-6 = x^2/0.575

x = [H+] = 8.97 x 10^-4 M

pH = -log[H+] = 3.05

B. pH at halfway to equivalence point

pH = pKa = 5.85

C. pH after 50 ml KOH added

[Nic-] formed = 0.335 M x 0.04 L/0.09 L = 0.186 M

[HNic] remained = (0.575 M x 0.04 L - 0.335 M x 0.04 L)/0.09 L = 0.065 M

pH = 5.85 + log(0.186/0.065) = 6.30

D. At equivalence point

Volume of KOH required = 0.575 M x 0.04 L/0.335 M = 0.066 L

E. At equivalence point

[Nic-] formed = 0.575 M x 0.04 M/0.106 L = 0.21 M

Nic- + H2O <==> HNic + OH-

Kb = 1 x 10^-14/1.4 x 10^-6 = x^2/0.21

x = [OH-] = 3.88 x 10^-5 M

pOH = -log[OH-] = 4.41

pH = 14 - pOH = 9.58

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