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

500mL of 0.2M acetic acid are added to 500mL of 0.30M sodium acetate solution. If the dissociation constant of acetic acid is 1.5×105 then pH of the resulting solution is:

A
5.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
9.0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3.0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4.0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 5.0
CH3COONa - 500 ml x 0.2 M
CH3COOH - 500 ml x 0.3 M Ka=1.5×105
pH=pKa+log[CH3COONaCH3COOH]
pH=log(1.5×105)+log[0.30.2]=log1.5+5+log1.5=5

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
pH and pOH
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon