wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

0.2(M) solution of monobasic acid is dissociated to 0.95% calculate its dissociation constant.
C = 0.2 M

A
1.6×104
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.8×105
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2×104
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.7×105
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 1.8×105
As the concentration of H+ from acids is 0.2×0.95×102=1.9×103 M hence H+ from self-ionization of water can be neglected.
Ka=Cα2
=0.2×(0.0095)2
=1.8×105
C=0.2 mole/lit
or
0.2 mole lit1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Degree of Dissociation
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon