I'm really unsure what the difference is. We're using these when calculating $K$ of combined reactions. In general, I am also confused what $K$ is in comparison to $K_\mathrm{a}$ and $K_\mathrm{b}$. For example here is a calculation where for the reaction involving $\ce{NO2-}$, the equilibrium constant is $1/K_\mathrm{a}$ even though it shows $\ce{NO2-}$ acting as a base.

Use the following four aqueous species to write an acid-base reaction with $K > 1$. (An acid-base reaction is one that involves the exchange of a proton. Your reaction will involve all four species.)

$\ce{NO2-(aq)}$, $\ce{F-(aq)}$, $\ce{HF(aq)}$, $\ce{HNO2(aq)}$

$\ce{HNO2}$ $(K_\mathrm{a} = \pu{4.6e-4})$ is a weaker acid than $\ce{HF}$ $(K_\mathrm{a} = \pu{6.6e-4})$, so $\ce{HNO2}$ holds onto its proton more tightly than $\ce{HF}$. So if we write a reaction when $\ce{NO2-}$ takes a proton from $\ce{F-}$, we'll have a reaction with $K > 1$:

$$\ce{HF + NO2- → HNO2 + F-}$$

We can get the $K$ for this reaction that adding the two acid dissocation reactions together as:

$$ \begin{align} \ce{HF &→ H+ + F-} &\quad &K_\mathrm{a}(\ce{HF})\\ \ce{H+ + NO2- &→ HNO2} &\quad &1/K_\mathrm{a}(\ce{HNO2})\\ \hline \ce{HF + NO2- &→ HNO2 + F-} &\quad &\frac{K_\mathrm{a}(\ce{HF})}{K_\mathrm{a}(\ce{HNO2})} = \frac{\pu{6.6e-4}}{\pu{4.6e-4}} = 1.4 > 1 \end{align} $$


closed as unclear what you're asking by Todd Minehardt, Mithoron, airhuff, Jon Custer, aventurin May 1 at 16:32

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$$ \begin{align} \ce{HNO2 + H2O &<=> NO2- + H3O+} &\quad K &= K_\mathrm{a} \tag{1}\\ \ce{NO2- + H2O &<=> HNO2 + OH-} &\quad K &= K_\mathrm{b} \tag{2}\\ \ce{H2O + H2O &<=> H3O+ + OH-} &\quad K &= K_\mathrm{w} \tag{3} \end{align} $$

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:

$$\ce{NO2- + H3O+ <=> HNO2 + H2O} \qquad K = 1/K_\mathrm{a}\tag{1a}$$ (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.


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