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Sagot :
Answer:
Explanation:
To solve this problem, we will use the Henderson-Hasselbalch equation for buffer solutions and the concept of neutralization reactions.
The Henderson-Hasselbalch equation is:
pH
=
p
+
log
(
[
A
−
]
[
HA
]
)
pH=pK
a
+log(
[HA]
[A
−
]
)
Where:
pH
pH is the desired pH of the buffer.
p
pK
a
is the negative logarithm of the acid dissociation constant
K
a
of hypochlorous acid.
[
A
−
]
[A
−
] is the concentration of the conjugate base (hypochlorite ion,
ClO
−
ClO
−
).
[
HA
]
[HA] is the concentration of the weak acid (hypochlorous acid,
HOCl
HOCl).
Given:
Desired
pH
=
7.600
pH=7.600
Concentration of hypochlorous acid (
HOCl
HOCl) in solution = 0.399 M
Volume of hypochlorous acid solution = 125 mL
Concentration of sodium hydroxide (
NaOH
NaOH) solution = 0.310 M
The
K
a
of hypochlorous acid (
HOCl
HOCl) is
3.0
×
1
0
−
8
3.0×10
−8
, thus
p
=
−
log
(
3.0
×
1
0
−
8
)
=
7.52
pK
a
=−log(3.0×10
−8
)=7.52
Calculate the required ratio of
[
A
−
]
[A
−
] to
[
HA
]
[HA]:
Using the Henderson-Hasselbalch equation:
7.600
=
7.52
+
log
(
[
ClO
−
]
[
HOCl
]
)
7.600=7.52+log(
[HOCl]
[ClO
−
]
)
7.600
−
7.52
=
log
(
[
ClO
−
]
[
HOCl
]
)
7.600−7.52=log(
[HOCl]
[ClO
−
]
)
0.080
=
log
(
[
ClO
−
]
[
HOCl
]
)
0.080=log(
[HOCl]
[ClO
−
]
)
1
0
0.080
=
[
ClO
−
]
[
HOCl
]
10
0.080
=
[HOCl]
[ClO
−
]
1.202
=
[
ClO
−
]
[
HOCl
]
1.202=
[HOCl]
[ClO
−
]
Calculate the moles of hypochlorous acid in 125 mL of solution:
Moles of HOCl
=
0.399
M
×
0.125
L
=
0.049875
moles
Moles of HOCl=0.399M×0.125L=0.049875moles
Let
x be the moles of
NaOH
NaOH added. Since
NaOH
NaOH neutralizes
HOCl
HOCl to form
ClO
−
ClO
−
, the moles of
ClO
−
ClO
−
formed will also be
x and the moles of
HOCl
HOCl remaining will be
0.049875
−
0.049875−x:
Using the ratio from step 1:
0.049875
−
=
1.202
0.049875−x
x
=1.202
Solving for
x:
=
1.202
(
0.049875
−
)
x=1.202(0.049875−x)
=
0.0598935
−
1.202
x=0.0598935−1.202x
+
1.202
=
0.0598935
x+1.202x=0.0598935
2.202
=
0.0598935
2.202x=0.0598935
=
0.0598935
2.202
x=
2.202
0.0598935
≈
0.0272
moles
x≈0.0272moles
Calculate the volume of 0.310 M
NaOH
NaOH solution required to get 0.0272 moles:
Volume of NaOH solution
=
0.0272
moles
0.310
M
≈
0.0877
L
=
87.7
mL
Volume of NaOH solution=
0.310M
0.0272moles
≈0.0877L=87.7mL
So, you would need to add approximately 87.7 mL of 0.310 M sodium hydroxide to 125 mL of 0.399 M hypochlorous acid solution to prepare a buffer with a pH of 7.600.
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