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JEE Advanced 2022 Paper 1
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© examsnet.com
Question : 39
Total: 54
A solution is prepared by mixing
0.01
mol each of
H
2
CO
3
,
NaHCO
3
,
Na
2
CO
3
, and
NaOH
in
100
mL
of water.
pH
of the resulting solution is_____
[Given :
p
K
a
1
and
p
K
a
2
of
H
2
CO
3
are
6.37
and
10.32
, respectively
;
log
2
=
0.30
]
Your Answer:
Validate
Solution:
H
2
CO
3
+
NaOH
⟶
NaHCO
3
+
H
2
C
Milli moles 10 10 -
At end 0 0 10+10 =20
Final mixture has 20 milli moles
NaHCO
3
and 10 milli moles
Na
2
CO
3
pH
=
pKa
2
+
log
Salt
Acid
pH
=
pKa
2
+
log
(
10
20
)
[
Buffer
:
Na
2
CO
3
+
NaHCO
3
]
=
10.32
−
log
2
=
10.02
© examsnet.com
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