A 0.02 M solution of pyridinium hydrochl | Class 11 Chemistry Chapter Equilibrium, Equilibrium NCERT Solutions

Question:

A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine

Answer:

Given,  pH  =  3.44

We know that

pH  =  -log [H+]

∴ [H+]  =  3.63x10-4

Then Kh  = ( 3.63x10-4 )2 / 0.02         (∵ Concentration is 0.02M)

⇒ Kh  =  6.6 x 10-6

Now , Kh  =  Kw / Ka

⇒  K=  Kw / Kh

= 10-14 / 6.6 x 10-6

= 1.51 x 10-9

 


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Equilibrium.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

Comments

Comment(s) on this Question

Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 62: A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of py....