Answer to Question #265289 in Organic Chemistry for Jess
A hypothetical acid HA has a dissociation constant 1.8 š„ 10ą¬æą¬·.
a. Write a balanced dissociation reaction of the acid
b. Write the equilibrium constant expression for the acid
c. If the initial concentration of the acid is 0.5 M, determine its hydrogen ion concentration. d. Calculate the ionization constant
e. Comment on the strength of the acid
Start by writing a balanced chemical equation for theĀ partial ionizationĀ of the acid
HA
(aq]
+
H
2
O
(l]
ā
A
ā
(aq]
+
H
3
O
+
(aq]
Notice that you haveĀ 1
:
1
Ā mole ratiosĀ across the board.Ā For every moleĀ of acidĀ that ionizesĀ in aqueous solution, you getĀ one moleĀ of itsĀ conjugate baseĀ andĀ one moleĀ of hydronium ions,Ā H
3
O
+
.
In other words, the equation producesĀ equal concentrationsĀ of conjugate base and hydronium ions.
Now, you can use theĀ pHĀ of the solution to calculate theĀ equilibrium concentrationĀ of the hydronium ions.
pH
=
ā
log
(
[
H
3
O
+
]
)
ā
[
H
3
O
+
]
=
10
ā
pH
In your case, the pH of the solution is equal toĀ 4
, which means that you'll have
[
H
3
O
+
]
=
10
ā
4
M
By definition, theĀ acid dissociation constant,Ā K
a
, will be equal to
K
a
=
[
A
ā
]
ā
[
H
3
O
+
]
[
HA
]
The expression for the acid dissociation constant is written usingĀ equilibrium concentrations. So, if the reaction produced a concentration of hydronium ions equal toĀ 10
ā
4
M
, it follows that it also produced a concentration of conjugate base equal toĀ 10
ā
4
M
.
Because the initial concentration of the acid is considerably higher than the concentrations of the conjugate base and hydronium ions, you canĀ approximateĀ it to beĀ constant.
This means that the acid dissociation constant for this acid will be
K
a
=
10
ā
4
ā
10
ā
4
0.100
=
1.0
ā
10
ā
7
This is the underlying concept behind anĀ ICE table
HA
(aq]
+
H
2
O
(l]
ā
A
ā
(aq]
+
H
3
O
+
(aq]
I
0.100
C
(
ā
x
)
(
+
x
)
(
+
x
)
E
0.100
ā
x
x
x
HereĀ x
Ā represents the equilibrium concentration for the conjugate acid and hydronium ions. Since you know thatĀ x
=
10
ā
4
, you will have
K
a
=
10
ā
4
ā
10
ā
4
0.100
ā
10
ā
4
Once again, you can use
0.100
ā
10
ā
4
=
0.0999
ā
0.100
to get
K
a
=
10
ā
8
0.100
=
1.0
ā
10
ā
7
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