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RASHTRASANT TUKADOJI MAHARAJ NAGPUR UNIVERSITY 2018
Master of Science (M.Sc.)
Dhantoli, Nagpur, India - 440010
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE
Subjects taught
English, Mathematics, Science
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE
Subjects taught
Science, English, Mathematics
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE
Subjects taught
Science, Mathematics, English
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE, State
Subjects taught
Science, English, Mathematics
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE, State
Subjects taught
Mathematics, Science, English, Hindi
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE, State
Subjects taught
Physics, Mathematics, Chemistry, English, Biology
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
BSc Physics Subjects
Solid State Physics, Electromagnetic Theory
Type of class
Crash Course, Regular Classes
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
BSc Branch
BSc Physics
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE, State
Subjects taught
Physics, Mathematics, English, Chemistry, Biology
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
State, CBSE
Subjects taught
EVS, Science, Mathematics, Social Science, Hindi, English
Taught in School or College
No
Answered on 28/11/2018 Learn CBSE - Class 11/Chemistry/Equilibrium
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Post a Lesson
disassociation of acetic acid takes place as acetate ions (CH3COO-) and hydrogen ions (H+)
CH3COOH ⇔CH3COO- + H+
given: disassociation or ionization constant , K=1.74 ×10 -5
concentration, C =0.05 M
So,degree of disassociation or concentration of acetate ions (α) is
K=α2C ⇒α =√(K/C)
α =√(1.74 ×10-5/0.05) =√ 0.000348=0.01865
also ,[H+] =αC =0.01865 ×0.05=9.325×10-4
So pH = -log10[H+]= -log10(9.325×10-4)=3.03
Hence ,pH of solution is 3.03
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE
Subjects taught
English, Mathematics, Science
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE
Subjects taught
Science, English, Mathematics
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE
Subjects taught
Science, Mathematics, English
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE, State
Subjects taught
Science, English, Mathematics
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE, State
Subjects taught
Mathematics, Science, English, Hindi
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE, State
Subjects taught
Physics, Mathematics, Chemistry, English, Biology
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
BSc Physics Subjects
Solid State Physics, Electromagnetic Theory
Type of class
Crash Course, Regular Classes
Class strength catered to
Group Classes, One on one/ Private Tutions
Taught in School or College
No
BSc Branch
BSc Physics
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
CBSE, State
Subjects taught
Physics, Mathematics, English, Chemistry, Biology
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Board
State, CBSE
Subjects taught
EVS, Science, Mathematics, Social Science, Hindi, English
Taught in School or College
No
Answered on 28/11/2018 Learn CBSE - Class 11/Chemistry/Equilibrium
Ask a Question
Post a Lesson
disassociation of acetic acid takes place as acetate ions (CH3COO-) and hydrogen ions (H+)
CH3COOH ⇔CH3COO- + H+
given: disassociation or ionization constant , K=1.74 ×10 -5
concentration, C =0.05 M
So,degree of disassociation or concentration of acetate ions (α) is
K=α2C ⇒α =√(K/C)
α =√(1.74 ×10-5/0.05) =√ 0.000348=0.01865
also ,[H+] =αC =0.01865 ×0.05=9.325×10-4
So pH = -log10[H+]= -log10(9.325×10-4)=3.03
Hence ,pH of solution is 3.03
Like 1 Answers 1 Comments Ask a Question
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