Q. \(pKa\) of \(\mathrm{NH_3}\).

Goal: Find the pKa of \( \mathrm{NH_3} \).

Step 1: Identify what “pKa of \( \mathrm{NH_3} \)” refers to

The value of \( \mathrm{p}K_a \) for a species refers to the equilibrium of its acid form donating a proton.

For ammonia, \( \mathrm{NH_3} \) acts as a base in water, but it can be treated as the conjugate acid/base system as follows.

In water, \( \mathrm{NH_3} \) accepts a proton to form the ammonium ion \( \mathrm{NH_4^+} \). The relevant acid for \( \mathrm{p}K_a \) is the conjugate acid \( \mathrm{NH_4^+} \).

Step 2: Write the relevant acid dissociation equilibrium

Consider the dissociation of ammonium:

\[
\mathrm{NH_4^+ \rightleftharpoons NH_3 + H^+}
\]

The acid dissociation constant for this reaction is \( K_a \), and the corresponding pKa is the value we report as “the pKa of \( \mathrm{NH_3} \)” (because \( \mathrm{NH_3} \) and \( \mathrm{NH_4^+} \) are a conjugate acid/base pair).

Step 3: Use the standard tabulated value

The pKa of \( \mathrm{NH_4^+} \) (at \(25^\circ \mathrm{C}\)) is commonly tabulated as:

\[
\mathrm{p}K_a(\mathrm{NH_4^+}) \approx 9.25
\]

Therefore, the pKa associated with \( \mathrm{NH_3} \) is:

\[
\mathrm{p}K_a(\mathrm{NH_3}) \approx 9.25
\]

Final Answer

\[
\boxed{\mathrm{p}K_a(\mathrm{NH_3}) \approx 9.25}
\]