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
863 views
in Ionic Equilibrium by (48.5k points)
closed by

The Ionization constant of nitrous acid is 4.5 x 10-4 . Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

1 Answer

+1 vote
by (43.7k points)
selected by
 
Best answer

Sodium nitrite is a salt of weak acid, strong base. 

Hence, 

Kh = 2.22 x 10-11 Kw/Kb = 10-14 /(4.5x 10-4

[OH-] = ch = 0.04 x 2.36 x 10-5 = 944 x 10-7 

pOH = – log (9.44 x 10-7) = 7 – 0.9750 = 6.03 

pH = 14 – pOH = 14 – 6.03 = 7.97

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.

...