(a)
pH = 4.77
; (b)
[
H
3
O
+
]
=
1.00
×
10
-4
l
mol/dm
3
; (c)
[
A
-
]
=
0.16 mol⋅dm
-3
Explanation:
(a) pH of aspirin solution
Let's write the chemical equation as
m
m
m
m
m
m
m
m
l
HA
m
+
m
H
2
O
⇌
H
3
O
+
m
+
m
l
A
-
I/mol⋅dm
-3
:
m
m
0.05
m
m
m
m
m
m
m
m
l
0
m
m
m
m
m
l
l
0
C/mol⋅dm
-3
:
m
m
l
-
x
m
m
m
m
m
m
m
m
+
x
m
l
m
m
m
l
+
x
E/mol⋅dm
-3
:
m
0.05 -
l
x
m
m
m
m
m
m
m
l
x
m
m
x
m
m
m
x
K
a
=
[
H
3
O
+
]
[
A
-
]
[
HA
]
=
x
2
0.05 -
l
x
=
3.27
×
10
-4
Check for negligibility
0.05
3.27
×
10
-4
=
153
<
400
∴
x
is not less than 5 % of the initial concentration of
[
HA
]
.
We cannot ignore it in comparison with 0.05, so we must solve a quadratic.
Then
x
2
0.05
−
x
=
3.27
×
10
-4
x
2
=
3.27
×
10
-4
(
0.05
−
x
)
=
1.635
×
10
-5
−
3.27
×
10
-4
x
x
2
+
3.27
×
10
-4
x
−
1.635
×
10
-5
=
0
x
=
1.68
×
10
-5
[
H
3
O
+
]
=
x
l
mol/L
=
1.68
×
10
-5
l
mol/L
pH
=
-log
[
H
3
O
+
]
=
-log
(
1.68
×
10
-5
)
=
4.77
(b)
[
H
3
O
+
]
at pH 4
[
H
3
O
+
]
=
10
-pH
l
mol/L
=
1.00
×
10
-4
l
mol/L
(c) Concentration of
A
-
in the buffer
We can now use the Henderson-Hasselbalch equation to calculate the
[
A
-
]
.
pH
=
p
K
a
+
log
(
[
A
-
]
[
HA
]
)
4.00
=
−
log
(
3.27
×
10
-4
)
+
log
(
[
A
-
]
0.05
)
=
3.49
+
log
(
[
A
-
]
0.05
)
log
(
[
A
-
]
0.05
)
=
4.00 - 3.49
=
0.51
[
A
-
]
0.05
=
10
0.51
=
3.24
[
A
-
]
=
0.05
×
3.24
=
0.16
The concentration of
A
-
in the buffer is 0.16 mol/L.
hope this helps :)
The Answer is A CH2O
The molecular formula for glucose is C6H12O6. The subscripts represent a multiple of an empirical formula. To determine the empirical formula, divide the subscripts by the GCF of 6 which gives CH2O.
Answer:
it is equal to 125 kPa
Explanation:
move the decimal 3 to the left
Answer:the sodium carboxylate salt
Explanation:
The reaction between the carboxylic acid and the sodium hydroxide yields a sodium carboxylate. This sodium carboxylate is an ionic in nature; RCCOO-Na+. This can effectively interact with water and remain in the aqueous phase since it is composed of the carboxylate ion and sodium ion in solution. The aqueous phase always contains water soluble ionic substances of which the sodium carboxylate is a typical example of such.
Answer:
Here's what I get
Step-by-step Explanation
(a) Effect of dilution
There will be no effect on the volume of NaOH needed.
The amount of HCl will be halved, so the amount of NaOH will be halved.
However, the concentration of NaOH is also halved, so you will need twice the volume.
You will be back to the same volume as before dilution.
(b) Net ionic equation
Molecular: HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)
Ionic: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + Cl⁻(aq) + H₂O(l)
Net ionic: H⁺(aq) + OH⁻(aq) ⟶ H₂O(l)
(c) Proton acceptor
H⁺ is the proton. OH⁻ accepts the proton and forms water.
(d) Moles of HCl
(e) Equivalence point
The equivalence point is the point at which the titration curve intersects the pH 7 line.
(f) Schematic representation
Assume the box for 0.10 mol·L⁻¹ HCl contains four black dots (H⁺) and four open circles (Cl⁻).
The 0.20 mol·L⁻¹ solution is twice as concentrated.
It will contain eight black dots and eight open circles.
