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When a weak base
is added to water, an equilibrium is quickly reached:
- B(aq) + H2O(l) < = > HB+(aq) +
OH-(aq)
- or
- B-(aq) + H2O(l) < = >
HB(aq) +
OH-(aq)
depending on what type of weak base it is. (Anion or
molecule.) the equilibrium constant for these reactions is thus
- Kb = [HB+][OH-]/[B]
- or Kb = [HB][OH-]/[B-]
A common way to express the strength of an base is the pKb, which
is similar in form to the pH or pKa
- pKb = -log10Kb
The lower
the Kb and pKb values for an base, the stronger the base.
Example 1: The fluoride ion, F-, has a Kb equal
to 1.4*10-11. What is the pKb of the fluoride ion?
Solution 1: The pKb is the negative log of Kb,
so just set up the equation.
- pKb = -log10Kb
- pKb= -log10(1.4*10-11)
- pKb = 10.85
Example 2: What is the pOH of a 0.100 M solution of
NaF? Assume that
all the NaF dissolves.
Solution 2: To compute the pOH, you need to find the hydroxide ion
concentration. This is an equilibrium problem at heart- you need to find the concentrations
of the species in an equilibrium reaction.
First, write down the reaction, then write the equilibrium constant
expression for the reaction:
- F-(aq) + H2O(l) < = >
HF(aq) + OH-
-
- Kb = [HF][OH-]/[F-]
Next, write
out the table of initial concentrations and their change when a reaction occurs.
Here, for every fluoride ion that reacts, one molecule of hydroxide and one
molecule of HF are formed. Thus, if we call the amount of fluoride ion used up
x
|
[F-] (M) |
[OH-] (M) |
[HF] (M) |
| Initial |
0.100 |
0.000 |
0.000 |
| Change |
-x |
+x |
+x |
| Equilibrium |
0.100-x |
x |
x | Now we can write out the equilibrium expression
and solve for x
- Kb = [HF][OH-]/[F-]
- 1.4*10-11 = (x)(x)/(0.100-x)
This looks like we might
have to use the quadratic equation. Luckily, if we consider Kb, we
realize that it is very small. Thus, very little fluoride ion will react, and x
will be very small compared to 0.100M. Thus, we can assume that 0.100-x ~ 0.100
without much loss of accuracy.
- 1.4*10-11 = (x)(x)/(0.100)
- x = 1.18*10-5
x is much smaller than 0.100, so this is
not a bad approximation.
Now, we wanted the pOH. Our equilibrium OH- concentration is just
equal to x, so [OH-] = 1.18*10-5 M. Thus, the pOH is
- pOH = -log10([OH-])
- pOH = 4.93
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