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EAMCET Engineering 2006 Solved Paper
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© examsnet.com
Question : 60
Total: 160
The concentration of oxalic acid is '
x
'
mol
L
−
1
.
40
mL
of this solution reacts with
16
mL
of
0.05
M Mcidified
KMnO
4
What is the pH of ' x ' M oxalic acid solution? (Assume that oxalic acid dissociates completely)
1.3
1.699
1
2
Validate
Solution:
Oxalic acid
=
x
mol
∕
L
Oxalic acid
KMnO
4
M
1
V
1
=
M
2
V
2
40
mL
×
x
=
16
mL
×
0.05
x
=
16
×
0.05
40
=
1
50
x
=
1
50
M
Now convert molarity into normality
N
×
eq. wt.
=
M
×
mol. wt. of oxalic acid
N
×
45
=
1
50
×
90
N
=
1
25
This normality represents the hydrogen ion concentration.
So,
[
H
+
]
=
1
25
pH
=
log
1
[
H
+
]
=
log
25
=
1.3
© examsnet.com
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