Consider nicotinic acid, HC6H4NO2 (HNic Ka = 1.4x10^-5) and its conjugate base,

Question

Consider nicotinic acid, HC6H4NO2 (HNic Ka = 1.4x10^-5) and its conjugate base, C6H4NO2^-1 (Nic^-1 Kb = 7.1x10^-10)

1. 40.0 mL of .575M HNic is titrated with .335 M KOH.

A. Caluclate 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 .050L 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 so much for your help!

Explanation / Answer

Given that; Kb = 7.1x10^-10

Ka= Kw/Kb

Ka= 1*10^-14/ 7.1*10^-10

Ka = 1.41 x 10^-5

a)

Ka= [H+][Nic-]/[HNic]

1.41 x 10^-5 = x^2 / 0.575-x
x = [H+]= 0.00327 M

pH = - log [H+]

pH = 2.49

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

At half equivalent point the concentration of acid = salt

pH = 4.85 + log salt/ base

=4.85

C. Calculate the pH after .050L of KOH has been added.

moles acid = 0.040 L x 0.575 M=0.023

moles OH- added = 0.005 L x 0.335 M

=0.001675

HC6H4NO2 + OH- = C6H4NO2- + H2O
moles acid = 0.023 - 0.001675

=0.021325
moles salt = 0.001675
total volume = 0.0450 L
[acid]= 0.021325 / 0.0450=0.474M

[salt]= 0.001675/ 0.0450=0.0372M
pH = 4.85 + log 0.0372/ 0.474

=4.85+ (-1.11)

=3.74

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

Consider nicotinic acid, HC6H4NO2 (HNic Ka = 1.4x10^-5) and its conjugate base, C6H4NO2^-1 (Nic^-1 Kb = 7.1x10^-10)

1. 40.0 mL of 0.575M HNic is titrated with .335 M KOH.

HNic + NaOH = NicNa + H2O

moles acid = 0.040 L x 0.575 M=0.023

reaction occurred 1:1 ratio so we need same number of base; 0.01725 moloe

volume of base in L = Number of moles / molarity

= 0.023 moles / 0.335M

= 0.068 L

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

moles NaOH = 0.051 L x 0.335 = 0.023
moles salt = 0.023
total volume = 0.108 L
[salt]= 0.023 / 0.091

=0.252 M

C6H4NO2- + H2O <-----> HC6H4NO2 + OH-

Given that; Nic^-1 Kb = 7.1x10^-10

Kb = x^2 / 0.252-x

7.1x10^-10 = x^/ 0.252

Due to small value of Kb we can write = 0.252-x= 0.252

X^2= 2.93*10^-11
x = [OH-]= 5.4x 10^-6 M

pOH = - log [OH-]
pOH = 5.26

pH = 14 – 5.26

=8.74

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