0
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – orthocresol May 2 '16 at 15:03
9
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.