Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
82 views
in Chemistry by (66.1k points)
closed by
The dissociation constant of a substitude benzoic acid at `25^(@)C` is `1.0xx10^(-4)`. The pH of a `0.01M` solution of its sodium salt is

1 Answer

0 votes
by (62.9k points)
selected by
 
Best answer
Correct Answer - 8
Given, `K_(a)= 1.0xx10^(-4)`
`pK_(a)= -log K_(a)= -log(1.0xx10^(-4))=4`
and `c = 0.01M`
Now, `pH= 7+(1)/(2)(pK_(a)+logc) at 25^(@)C`
`pH= 7+(1)/(2)[4+log(0.01)]`
`=7+1=8`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...