When to use 1/Ka vs Kb
$$
\begin{align}
\ce{HNO2 + H2O &amp;&lt;=&gt; NO2- + H3O+} &amp;\quad K &amp;= K_\mathrm{a} \tag{1}\\
\ce{NO2- + H2O &amp;&lt;=&gt; HNO2 + OH-} &amp;\quad K &amp;= K_\mathrm{b} \tag{2}\\
\ce{H2O + H2O &amp;&lt;=&gt; H3O+ + OH-} &amp;\quad K &amp;= K_\mathrm{w} \tag{3}
\end{align}
$$

Above are the reactions associated with equilibrium constants commonly called $K_\mathrm{a}$, $K_\mathrm{b}$, and $K_\mathrm{w}$. If you reverse the first reaction, you get:

$$\ce{NO2- + H3O+ &lt;=&gt; HNO2 + H2O} \qquad K = 1/K_\mathrm{a}\tag{1a}$$
(This looks a bit different from your equation because it shows the hydronium ion instead of the hydrogen ion, but it refers to the same reaction.)

As you might know, the three equilibrium constants are related:

$$K_\mathrm{a} \times K_\mathrm{b} = K_\mathrm{w}$$

This is because when adding up the first and the second reaction, you get the third.


  When to use $1/K_\mathrm{a}$ vs $K_\mathrm{b}$


Use the equilibrium constant that matches the reaction you are working on.