# Relationship between pKa and pKb

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $38$. Does it stand for the following equilibrium:

$$\ce{NH4+ <=> NH3 + H+} \tag{1}$$ With $K_\mathrm{a} = 10^{-38}$ for reaction $(1)$, indicating that the conjugate acid of $\ce{NH3}$ is quite weak?

The $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ is $4.74$. Does it stand for the following equilibrium:

$$\ce{NH3 + H2O <=> NH4+ + OH-} \tag{2}$$ With $K_\mathrm{b} = 10^{-4.74}$ for reaction $(1)$, indicating that $\ce{NH3}$ is a weak base?

Also: $$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 42.74 \neq 14$$

• Regarding the last equation, it is only applicable to the sum of the pKa of a species HX and the pKb of its conjugate base, X-. Since both the pK values above are referring to NH3 itself, the formula is not valid. However, if you were to add pKa(NH4+) and pKb(NH3), you would indeed get 14. – orthocresol May 2 '16 at 15:03

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $38$.

This means, the reaction this value is describing is:

$$\ce{NH3 <=> NH2- + H+}$$

The $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ is $4.74$.

This means, the reaction this value is describing is:

$$\ce{NH3 + H+ <=> NH4+}$$

There is a lot of misuse going on with the term $\mathrm{p}K_\mathrm{a}$ since many people are too lazy to say

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$’s conjugate acid is $9.26$

$$\ce{NH4+ <=> NH3 + H+}$$

rather than

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $9.26$.

$$\ce{NH3 <=> NH2- + H+}$$

but that is really what they should be doing. Thankfully, when looking up values in a table the person compiling the table (hopefully) knew what they were doing and used the correct terminology.

Only for a $\mathrm{p}K_\mathrm{a}$ and a $\mathrm{p}K_\mathrm{b}$ that describe the same reaction (from opposite viewpoints) is it true that $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$.