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Tardigrade
Question
Chemistry
The ionic product of water at 25° C is 10-14. Its ionic product at 90° C will be,
Q. The ionic product of water at
2
5
∘
C
is
1
0
−
14
. Its ionic product at
9
0
∘
C
will be,
2263
187
Report Error
A
1
×
1
0
−
14
58%
B
1
×
1
0
−
16
12%
C
1
×
1
0
−
20
14%
D
1
×
1
0
−
12
16%
Solution:
as temperature increases Ionic product increases.