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The ionization constants of several weak bases are given in [link] and in Appendix I .

Ionization Constants of Some Weak Bases
Ionization Reaction K b at 25 °C
( CH 3 ) 2 NH + H 2 O ( CH 3 ) 2 NH 2 + + OH 5.9 × 10 −4
CH 3 NH 2 + H 2 O CH 3 NH 3 + + OH 4.4 × 10 −4
( CH 3 ) 3 N + H 2 O ( CH 3 ) 3 NH + + OH 6.3 × 10 −5
NH 3 + H 2 O NH 4 + + OH 1.8 × 10 −5
C 6 H 5 NH 2 + H 2 O C 6 N 5 NH 3 + + OH 4.3 × 10 −10

Determination of K a 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 [ H 3 O + ] = [ CH 3 CO 2 ] = 0.00118 M . What is the value of K a for acetic acid?

An image shows the label of a bottle of distilled white vinegar. The label states that the contents have been reduced with water to 5 percent acidity.
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:

CH 3 CO 2 H ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + CH 3 CO 2 ( a q )
K a = [ H 3 O + ] [ CH 3 CO 2 ] [ CH 3 CO 2 H ] = ( 0.00118 ) ( 0.00118 ) 0.0787 = 1.77 × 10 −5

Check your learning

What is the equilibrium constant for the ionization of the HSO 4 ion, the weak acid used in some household cleansers:

HSO 4 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + SO 4 2− ( a q )

In one mixture of NaHSO 4 and Na 2 SO 4 at equilibrium, [ H 3 O + ] = 0.027 M ; [ HSO 4 ] = 0.29 M ; and [ SO 4 2− ] = 0.13 M .

Answer:

K a for HSO 4 = 1.2 × 10 −2

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Determination of K b From equilibrium concentrations

Caffeine, C 8 H 10 N 4 O 2 is a weak base. What is the value of K b for caffeine if a solution at equilibrium has [C 8 H 10 N 4 O 2 ] = 0.050 M , [ C 8 H 10 N 4 O 2 H + ] = 5.0 × 10 −3 M , and [OH ] = 2.5 × 10 −3 M ?

Solution

At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:

C 8 H 10 N 4 O 2 ( a q ) + H 2 O ( l ) C 8 H 10 N 4 O 2 H + ( a q ) + OH ( a q )
K b = [ C 8 H 10 N 4 O 2 H + ] [ OH ] [ C 8 H 10 N 4 O 2 ] = ( 5.0 × 10 −3 ) ( 2.5 × 10 −3 ) 0.050 = 2.5 × 10 −4

Check your learning

What is the equilibrium constant for the ionization of the HPO 4 2− ion, a weak base:

HPO 4 2− ( a q ) + H 2 O ( l ) H 2 PO 4 ( a q ) + OH ( a q )

In a solution containing a mixture of NaH 2 PO 4 and Na 2 HPO 4 at equilibrium, [OH ] = 1.3 × 10 −6 M ; [ H 2 PO 4 ] = 0.042 M ; and [ HPO 4 2− ] = 0.341 M .

Answer:

K b for HPO 4 2− = 1.6 × 10 −7

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Determination of K a Or K b From ph

The pH of a 0.0516- M solution of nitrous acid, HNO 2 , is 2.34. What is its K a ?

HNO 2 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + NO 2 ( a q )

Solution

We determine an equilibrium constant starting with the initial concentrations of HNO 2 , H 3 O + , and NO 2 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:

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.”

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):

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.

To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate [ H 3 O + ] , the equilibrium concentration of H 3 O + , from the pH:

[ H 3 O + ] = 10 −2.34 = 0.0046 M

The change in concentration of H 3 O + , x [ H 3 O + ] , is the difference between the equilibrium concentration of H 3 O + , which we determined from the pH, and the initial concentration, [ H 3 O + ] i . The initial concentration of H 3 O + 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 NO 2 is equal to the change in concentration of [ H 3 O + ] . For each 1 mol of H 3 O + that forms, 1 mol of NO 2 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:

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 sign 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, 0.0470. 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 equals 0.0046, 0.0046. The fourth column has the following: 0, negative x, 0.0046.

Finally, we calculate the value of the equilibrium constant using the data in the table:

K a = [ H 3 O + ] [ NO 2 ] [ HNO 2 ] = ( 0.0046 ) ( 0.0046 ) ( 0.0470 ) = 4.5 × 10 −4

Check your learning.

The pH of a solution of household ammonia, a 0.950- M solution of NH 3, is 11.612. What is K b for NH 3 .

Answer:

K b = 1.8 × 10 −5

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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