The ionization constants of several weak bases are given in
[link] and in
Appendix I .
Ionization Constants of Some Weak Bases
Ionization Reaction
Kb at 25 °C
5.9
10
−4
4.4
10
−4
6.3
10
−5
1.8
10
−5
4.3
10
−10
Determination of
Ka From equilibrium concentrations
Acetic acid is the principal ingredient in vinegar (
[link] ); that's why it tastes sour. At equilibrium, a solution contains [CH
3 CO
2 H] = 0.0787
M and
What is the value of
Ka for acetic acid?
Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)
Solution
We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:
Check your learning
What is the equilibrium constant for the ionization of the
ion, the weak acid used in some household cleansers:
In one mixture of NaHSO
4 and Na
2 SO
4 at equilibrium,
= 0.027
M ;
and
Determination of
Kb From equilibrium concentrations
Caffeine, C
8 H
10 N
4 O
2 is a weak base. What is the value of
Kb for caffeine if a solution at equilibrium has [C
8 H
10 N
4 O
2 ] = 0.050
M ,
= 5.0
10
−3M , and [OH
− ] = 2.5
10
−3M ?
Solution
At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:
Check your learning
What is the equilibrium constant for the ionization of the
ion, a weak base:
In a solution containing a mixture of NaH
2 PO
4 and Na
2 HPO
4 at equilibrium, [OH
− ] = 1.3
10
−6M ;
and
The pH of a 0.0516-
M solution of nitrous acid, HNO
2 , is 2.34. What is its
Ka ?
Solution
We determine an equilibrium constant starting with the initial concentrations of HNO
2 ,
and
as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)
We can solve this problem with the following steps in which
x is a change in concentration of a species in the reaction:
We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration):
To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate
the equilibrium concentration of
from the pH:
The change in concentration of
is the difference between the equilibrium concentration of H
3 O
+ , which we determined from the pH, and the initial concentration,
The initial concentration of
is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).
The change in concentration of
is equal to the change in concentration of
For each 1 mol of
that forms, 1 mol of
forms. The equilibrium concentration of HNO
2 is equal to its initial concentration plus the change in its concentration.
Now we can fill in the ICE table with the concentrations at equilibrium, as shown here:
Finally, we calculate the value of the equilibrium constant using the data in the table:
Check your learning.
The pH of a solution of household ammonia, a 0.950-
M solution of NH
3, is 11.612. What is
Kb for NH
3 .
![A diagram is shown that includes six rust-red rectangles. The first rectangle is labeled, “p H,” and there is an arrow that points to the right to the second rectangle. The second rectangle is labeled, “[ H subscript 3 O superscript plus ].” There is an arrow that points right to the third rectangle. The third rectangle is labeled, “Calculate x subscript [ H subscript 3 O superscript plus ].” There is an arrow that points right to the fourth rectangle. The fourth rectangle is labeled, “Calculate x subscript [ H N O subscript 2 ] and x subscript [ N O subscript 2 superscript negative sign ].” There is an arrow that points down to the fifth rectangle. The fifth rectangle is labeled, “Calculate equilibrium concentrations.” There is an arrow that points down to the sixth rectangle. The sixth rectangle is labeled, “Calculate K subscript a.”](/ocw/mirror/col11760/m51118/CNX_Chem_14_03_steps_img.jpg)
![This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, “H N O subscript 2 plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus sign N O subscript 2 superscript negative sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.0516, negative x, [ H N O subscript 2 ] subscript i plus ( negative x ) equals 0.0516 plus sign ( negative x ). The second column is blank in the first row, positive sign, blank for the third row. The third column has the following: approximately 0, x, [ H subscript 3 O ] superscript positive sign plus sign x [ N O subscript 2 ] superscript negative sign plus sign x plus sign 0 plus sign x. The fourth column has the following: 0, x, 0.0046.](/ocw/mirror/col11760/m51118/CNX_Chem_14_03_ICETable1_img.jpg)
