wiz-icon
MyQuestionIcon
MyQuestionIcon
10
You visited us 10 times! Enjoying our articles? Unlock Full Access!
Question

The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.1 M NaOH will be: (pKa for CH3COOH = 4.74 )

A
4.74
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
8.87
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
9.1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8.57
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 8.87
0.2 M CH3COOH 0.2 M NaOH
Both react as NaOH is strong base
0.22MCH3COO is left in sals
=01MCH3COO is left in sals
CH3COO+H2OCH3COOH+OHK=KwKa
αnkc
=1074Ka×0.1
107×10Ka
Ka104.74 (PKa=4.74)
α=10710104.74
=107×10+2.67
=104.33<0.1
[OH]=Cα
=101×104.33
=105.33
pOH=5.33
pH=14p01+
=8.67


1057643_1070123_ans_5d2455195fa44a10a1067137269a7a02.png

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Rate of Reaction
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon