The $\mathrm{p}K_\text{a}$ is the $-\log_{10}$ of the $K_\text{a}$ value for the disassociation of an acid. So how do you get a $\mathrm{p}K_\text{a}$ value for a base? The base produces no hydrogen ions so can’t you only have a $\mathrm{p}K_\text{b}$?


3 Answers 3


For bases, you can derive the $\mathrm{p}K_\text{a}$ from the $\mathrm{p}K_\text{b}$.

Knowing that

$$\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$$

Let’s examine ammonia.

Ammonia has a $K_\text{b}$ of $1.8 \times 10^{-5}$.

Taking the negative logarithm, we can derive $\mathrm{p}K_\text{b}$:

$$\mathrm{p}K_\text{b} = -\log( 1.8 \times 10^{-5}) = 4.74$$.

Knowing that $\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$:

We can find $\mathrm{p}K_\text{a}$ by manipulating that expression, thus:

$$\mathrm{p}K_\text{a} = 14 - \mathrm{p}K_\text{b} $$

$$\mathrm{p}K_\text{a} = 14 - 4.74 $$

$$\mathrm{p}K_\text{a} = 9.26 $$

Even in very basic solutions, there will be minute quantities of hydronium ions. By taking the anti-logarithm of the $\mathrm{p}K_\text{a}$, you can derive the $K_\text{a}$.

  • 1
    $\begingroup$ Yes, I know that you can manipulate the pKb to get a pKa. However, what meaning does the pKa have for bases? $\endgroup$
    – 1110101001
    Mar 13, 2014 at 3:44
  • $\begingroup$ $pK_a$ is the acid dissociation constant on a logarithmic scale. And the acid dissociation constant $K_a$ is just a measure of the strength of the acid in solution. No base will ever have a $pK_b$ of 14, thus we can find the $pK_a$ of the base. In the example above, it's a very, very small number. $\endgroup$
    – user3735
    Mar 13, 2014 at 3:47

Many bases do not deprotonate (e.g. $\ce{NaOH}$), making the notion of $\mathrm{p}K_\mathrm{a}$ based on deprotonation (Brønsted-Lowry theory) extremely difficult to measure/apply. However, there are some bases which are also acids: for instance $\ce{HSO4-}$ is a base because it can accept a proton:

$$\ce{HSO4- + H2O <=> H2SO4 + OH-}$$

and an acid because it can donate a proton:

$$\ce{HSO4- + H2O <=> SO4^2- + H+}$$

There are many bases which can act as acids and therefore those “bases” have a $\mathrm{p}K_\mathrm{a}$ for the deprotonation. In that case, the “base” would be acting as an “acid”. Measuring $\mathrm{p}K_\mathrm{a}$ for a specific substance assumes that substance undergoes acid dissociation.

Subtracting the $\mathrm{p}K_\mathrm{b}$ of $\ce{HSO4-}$ from 14 does not give the $\mathrm{p}K_\mathrm{a}$ of $\ce{HSO4-}$: it gives the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid $\ce{H2SO4}$ in water.

If you take this conjugate acid $\ce{H2SO4}$ and ask what its $\mathrm{p}K_\mathrm{a}$ is, you are assuming it is behaving as an acid by losing a proton to form $\ce{HSO4-}$. So set up the following chemical equation

$$\ce{H2SO4 + H2O <=> HSO4- + H3O+}$$

If you are asking what is the $\mathrm{p}K_\mathrm{b}$ of $\ce{H2SO4}$, you are assuming it is behaving as a base by gaining a proton to form $\ce{H3SO4+}$, a likely unstable species. Set up the following chemical equation

$$\ce{H2SO4 + H2O <=> H3SO4+ + OH-}$$

Therefore, when calculating concentrations of acid/base/conjugate based on $K_\mathrm{a}$ or $K_\mathrm{b}$ values, you need to make sure you have the appropriate chemical equilibrium written out which matches either protonation or deprotonation of a specific chemical species in a specific solvent (likely water). Use this correct chemical equation to fill in the equilibrium constant expression and go from there.


The "$\mathrm{p}K_\mathrm{a}$" of a base is just the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of that base.

Consider the following reaction:
$$\ce{B + H2O <=> HB+ + OH- }$$

it has the equilibrium constant: $$K_\mathrm{b} = \frac{(\ce{HB+})(\ce{OH-})}{(\ce{B})}.$$

Now consider the reaction with the conjugated acid $\ce{HB+}$:
$$\ce{HB+ + H2O <=> B + H3O+}$$
the equilibrium constant is: $$K_\mathrm{a} = \frac{(\ce{B})(\ce{H3O+})}{(\ce{HB+})}.$$

The product of the two gives:
$$K_\mathrm{b} \times K_\mathrm{a} = \ce{\frac{(HB+)(OH^{-})}{(B)}} \times \ce{\frac{(B)(H3O+)}{(HB+)}} = \ce{(OH^{-}).(H3O+)} = 10^{-14} = K_\mathrm{w}$$ and $$\mathrm{p}K_\mathrm{b}+\mathrm{p}K_\mathrm{a}=\mathrm{p}K_\mathrm{w}$$ thus $$\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w}-\mathrm{p}K_\mathrm{a}=14-\mathrm{p}K_\mathrm{a}.$$


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