Calculate the hydrolysis constant of ammonium acetate (NH4CH3COO), degree of hydrolysis and ph.Take M=0.01 and 0.04(pka=4.76 and pkb=4.75)
Answers
Answer:
ANSWER
Both NH
4
+
and CH
3
COO
−
ion of CH
3
COONH
4
CH
3
COONH
4
+H
2
O⇋CH
3
COO
−
+NH
4
+
1 0 0
1−h h h
pK
a
(CH−3COOH)=4.76⟹K
a
=1.74×10
−5
pK
b
(NH
4
OH)=4.75⟹K
b
=1.78×10
−5
K
H
for CH
3
COONH
4
=
K
a
K
b
K
w
=
(1.74×10
−5
)(1.78×10
−5
)
10
−14
=3.23×10
−5
also, h for CH
3
COONH
4
=
K
H
=
3.23×10
−5
=5.68×10
−3
CH
3
COOH⇋CH
C
OO
−
+H
+
K
a
=
[CH
3
COOH]
[CH
3
COO
−
][H
+
]
=
Ch
C(1−h)[H
+
]
[H
+
]=K
a
1−h
1
=K
a
K
H
=K
a
K
a
×K
b
K
w
−
K
b
K
w
K
a
=
1.78×10
5
10
−14
×1.74×10
−5
[H
+
]=9.88×10
−8
pH=7.005
[pH and h are independent of initial concentration of Salt]