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

The Kw for water for the equillibrium, 2H2OH3O+OH changes from 1014 at 25 C to 9.62×1014 at 60 C.
Calculate the pH and nature at 60 C

A
pH=5.51, solution is basic
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
pH=6.51, solution is neutral
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
pH=5.51, solution is acidic
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
pH=6.51, solution is basic
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B pH=6.51, solution is neutral
pH=log[H+], Kw=[H+] [OH],
9.62×1014=x2
x=9.62×1014
x=3.18×107pH=log[3.18×107]pH=6.51

pH=log[H+]
pH=log1[H+]
Kw=[H+] [OH]
log[H+]log[OH]=logKw
pH+pOH=pKw
At 25 C, Kw=1014
At 65 C, solution is neutral.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon