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Standard X
Chemistry
pH
25 ml of 0.1 ...
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25 ml of 0.1 M acetic acid is titrated with 0.1NaoH solution. The pH of the solution atequivalence point will be (log 5 0.7CH3COOH = 4.76)
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Q.
When 10 mL of 0.1 M acetic acid
(
p
K
a
=
5.0
)
is titrated against 10 mL of 0.1 M ammonia solution
(
p
K
b
=
5.0
)
the equivalence point occurs at pH of:
Q.
When
10
m
L
of
0.1
M
acetic acid (
p
K
a
=
5.0
) is titrated against
10
m
L
of
0.1
M
ammonia solution
(
p
K
b
=
5.0
)
, the equivalence point occurs at pH:
Q.
Calculate
O
H
−
concentration at the equivalent point when a solution of
0.1
M
acetic acid is titrated with a solution of
0.1
M
N
a
O
H
.
K
a
for the acid
=
1.9
×
10
−
5
Q.
(a) When 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCI solution the variation of pH of solution with volume of HCI added will be :
(b) Variation of degree of dissociation
α
with concentration for a weak electrolyte at a particular temperature is best represnted by :
(c) M acetic acid solution is titrated against M NaOH solution. the difference in pH between 1/4 and 3/4 stages of neutralization of acid will be 2 log 3.
Q.
Calculate the pH at the equivalence point in the titration of 25 mL of 0.10 M formic acid with a 0.1 M NaOH solution (given that
p
K
a
of formic acid = 3.74, log 5 = 0.7)
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