+91-8076753736 support@wiredfaculty.com
logo
logo

NCERT Classes

  • Class 7
  • Class 8
  • Class 9
  • Class 10
  • Class 11
  • Class 12

Previous-Year-Papers

  • CBSE Class 10 and 12
  • NEET Class 12
  • IIT JEE-Main Class 12
  • ICSE Class 10 Class 12
  • CLAT Class 12

Entrance Exams

  • IIT-JEE
  • NEET
  • CLAT
  • SSC-CGL
  • SSC-CHSL

State Boards

  • Assam Board
  • UP Board English Medium
  • Up Board Hindi Medium
  • CBSC Board English Medium
  • CBSC Board Hindi Medium
  • Manipur Board
  • Goa Board
  • Gujrat Board
  • Haryana Board English Medium
  • Haryana Board Hindi Medium
  • Uttarakhand Board English Medium
  • Uttarakhand Board Hindi Medium
  • Himachal Board English Medium
  • Himachal Board Hindi Medium
  • ICSE Board
  • Jharkhand Board English Medium
  • Jharkhand Board Hindi Medium
  • Karnataka Board
  • Meghalya Board
  • Nagaland Board
  • Punjab Board
  • Rajasthan Board English Medium
  • Rajasthan Board Hindi Medium

For Daily Free Study Material Join wiredfaculty

Home > Equilibrium

Sponsor Area

Equilibrium

Question
CBSEENCH11006565
Wired Faculty App

The degree of ionisation of a 0·1M bromoacetic acid solution is 0·132. Calculate the pH of the solution and pKa of bromoacetic acid.

Solution
Short Answer
The concentration of hydrogen ion,
    space equals space open square brackets straight H to the power of plus close square brackets space equals space straight C space cross times space straight alpha
equals space 0.1 space straight M space cross times space space 0.132
equals space 0.0132 space straight M
Now space pH space equals space minus log space open square brackets straight H to the power of plus close square brackets
space space space space space space space space space space space space equals negative log space left parenthesis 0.0132 right parenthesis equals space 1.88
Now space straight K subscript straight a space space equals space fraction numerator Cα squared over denominator 1 minus straight alpha end fraction equals space fraction numerator 0.1 space cross times space left parenthesis 0.132 right parenthesis squared over denominator 1 minus 0.132 end fraction
space space space space space space space space space space space space equals space 2.01 space cross times space 10 to the power of negative 13 end exponent
therefore space space pK subscript straight a space equals space minus log space straight K subscript straight a space equals space minus log space left parenthesis 2.01 space cross times space 10 to the power of negative 3 end exponent right parenthesis space equals space 2.7

  • Bookmark
  • Facebook
  • WhatsApp
  • Twitter
  • Linkedin
  • Copy Link
  • twitter button
  • facebook logo
  • watsapp logo

Some More Questions From Equilibrium Chapter

Which measurable property becomes constant in water vapour equilibrium?

Give an example from daily life in which there is gas solution equilibrium?

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

What is the role of forward and backward reactions at equilibrium of a reversible reaction?

What is a forward reaction?

What condition favours a reversible reaction?

What is a backward reaction?

Mock Test Series

Mock Test Series

  • NEET Mock Test Series
  • IIT-JEE Mock Test Series
  • CLAT Mock Test Series
  • SSC-CGL Mock Test Series
  • SSC-CHSL Mock Test Series

NCERT Sample Papers

  • CBSE English Medium
  • CBSE Hindi Medium
  • ICSE

Entrance Exams Preparation

  • NEET
  • IIT-JEE
  • CLAT
  • SSC-CGL
  • SSC-CHSL

ABOUT US

  • Company
  • Term of Use
  • Privacy Policy

NCERT SOLUTIONS

  • NCERT Solutions for Class-8th
  • NCERT Solutions for Class-9th
  • NCERT Solutions for Class-10th
  • NCERT Solutions for Class-11th
  • NCERT Solutions for Class-12th

CONTACT US

Phone Number

+91-8076753736

Email Address

support@wiredfaculty.com

Location

A2/44 Sector-3 Rohini Delhi 110085

Copyright © 2025 wiredfaculty. Powered by -wiredfaculty.com