Weak Bases & KB
Similar to strong acids, there are only 10 strong bases. And, there are dozens, if not hundreds, of weak bases. Weak bases are weak electrolytes that do not dissociate completely in water. Some of these bases are metal hydroxides, like AgOH. Others are nitrogen-hydrogen based compounds that react with water to produce the hydroxide ion in water. From a chemical reaction standpoint, you can have two kinds of equilibria, one for hydroxide compounds:
And a second for nitrogen-hydrogen compounds, like NH3:
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This chemical reaction is written as an equilbrium because all species exist at the same time. The forward and reverse reaction are happening at the same rate, so it indicates that all the materials are there. Since this is an equilibrium, we can write a K expression for it:
Notice in the above example that since all materials (except water) are dissolved in water, or aqueous, they all need to be included into the K expression. Water is not included because it is a liquid. Since this K expression deals with a base, instead of using Kc, we will use Kb, which stands for the base dissociation constant. Each weak base has a dissociation and a Kb.
Now, because these are constants, that means each base has its own value for Kb. These values can be looked up in any chemistry textbook or found online. One such list is here.
Calculations involving Kb
There are several types of calculations that can be done involving the Kb. We will show a couple of those here:
#1) Determining the Kb of a base from its initial concentration and the pH of the solution:
Example: A 0.90 M weak base solution has a pH of 10.8. What is the Kb of the base?
Solution:
First we write a dissociation equation and Kb expression for this unknown acid. Since we do not know the identity of the base, we will simply call it B:
There are several types of calculations that can be done involving the Kb. We will show a couple of those here:
#1) Determining the Kb of a base from its initial concentration and the pH of the solution:
Example: A 0.90 M weak base solution has a pH of 10.8. What is the Kb of the base?
Solution:
First we write a dissociation equation and Kb expression for this unknown acid. Since we do not know the identity of the base, we will simply call it B:
The pH doesn't help us, because we need the pH. Since the pH is 10.8, then the pOH is 3.2 (14 - 10.8 = 3.2). We can undo the pOH to determine the concentration of OH-1 ion at equilibrium (once the max has ionized):
This number is the equilibrium concentration of OH-1. The initial amount of [OH-1] was zero, and the concentration of [B] at the beginning was 0.90 M. So, we can fill in parts of the ICE table for the reaction:
Then, we can use the differences between the initials and the finals to fill in the rest of the table:
Then, using the equilibrium values, we can calculate the Ka
#2) Determining the pH of a known concentration base using its Kb.
Example: What is the pH of 0.10 M hydroxylamine, NH2OH, solution (Kb = 1.1 x 10-8).
To do this, we also need a chemical reaction for the dissociation of the weak base and a Kb expression:
Now, we know that the initial concentration of NH2OH is 0.10 M. We also know that the initial concentration (before the NH2OH starts dissolving) of [OH-1] [NH2OH2+1] are both 0 M:
However, we do not know how much the reactant dissociates, nor do we know how many ions are going to be in solution at equilibrium. We do know, though that the amount of NH2OH has to go down by some unknown amount (-x) and the amount of OH-1 and NH2OH2+1 both have to go up by that same unknown amount (+x). That fills in the rest of the ICE table to look like this:
The Kb for NH2OH is given to be 1.1 x 10-8. Using Kb, and filling in the numbers at equilibrium, we get:
From here, we can solve for x. However, once we get into this we will have to solve a quadratic equation as written. But, we can make the math easier. For the 0.10 - x for the hydroxylamine concentration, whatever we get for x is going to be significantly smaller than 0.10. How significant? Well, as you will see in a second, the value of x we get will not change 0.10 M (to sig figs) by subtracting it. Therefore, we can omit the -x from the math:
The algebra becomes significantly easier, and we can solve for x:
Since x is equal to the [OH-1] at equilibrium, so we can use this number and convert to pOH, and then to pH.