Themes > Science > Chemistry > Inorganic Chemistry > Acids and Bases > Acids and Bases Index > Equilibrium Constants for Weak Bases

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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