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Test Index
Ionic Equilibrium Part 2
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Section:
Chemistry
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© examsnet.com
Question : 1
Total: 29
One litre buffer solution was prepared by adding 0.10 mol each of
NH
3
and
NH
4
‌
Cl
in deionised water.
The change in pH on addition of 0.05 mol of HCl to the above solution is
_
_
_
_
×
10
−
2
. (Nearest integer)
Given:
pK
b
of
NH
3
=
4.745
and
log
10
3
=
0.477
[7 Apr 2025 Shift 2]
Your Answer:
Validate
Solution:
Initially
pOH
=
pK
b
+
log
‌
[
NH
4
‌
Cl
]
[
NH
3
]
=
pK
b
+
log
‌
0.1
0.1
‌
pOH
=
pK
b
‌
pH
=
14
−
pOH
⇒
pH
=
9.255
When 0.05 mol HCl is added
NH
3
‌
+
H
+
⇌
NH
4
‌
+
0.1
‌
‌
‌
‌
‌
‌
‌
‌
0.05
‌
0.05
‌
‌
‌
‌
‌
‌
‌
0
‌
‌
‌
‌
‌
‌
‌
‌
0.15
‌
pOH
=
pK
b
+
log
‌
0.15
0.05
=
5.222
‌
pH
=
8.778
Change in
pH
=
0.477
or
47.7
×
10
−
2
© examsnet.com
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