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JEE Main Chemistry Class 11 Equilibrium Questions Part 2 Questions
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© examsnet.com
Question : 14
Total: 100
600
mL
of
0.01
M
HCl
is mixed with
400
mL
of
0.01
M
H
2
SO
4
. The
pH
of the mixture is _______
×
10
−
2
. (Nearest integer)
[Given
log
2
=
0.30
,
log
3
=
0.48
log
5
=
0.69
log
7
=
0.84
log
11
=
1.04
]
[30-Jan-2023 Shift 1]
Your Answer:
Validate
Solution:
Total milimoles of
H
+
=
(
600
×
0.01
)
+
(
400
×
0.01
×
2
)
=
14
[
H
+
]
=
14
1000
=
14
×
10
−
3
pH
=
3
−
log
14
=
1.86
=
186
×
10
−
2
© examsnet.com
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