How to estimate the equilibrium constant of the reaction between NaF and HCl

Since $\ce{HCl}$ is more acidic than $\ce{HF}$, the left hand side of a reaction has a stronger acid than the right hand side.


Since $\ce{NaF}$ is more basic than $\ce{NaCl}$, the left hand side has a stronger base relative to the right hand side.


$$\begin{align}\text{Stronger Acid/Base} & \ce{->} \text{Weaker Acid/Base}\\ \text{(Less Stable)} & \ce{->} \text{(More Stable)}\end{align}$$


So shouldn't the equilibrium constant be larger than $1$? 


My textbook says the equilibrium constant is less than $1$.




We can entirely ignore the sodium cation; it is only a spectator ion. This means, the reaction we are observing is:


$$\ce{HCl + F- <=> Cl- + HF}\tag{1}$$


The equilibrium constant for this reaction is:


$$K = \frac{[\ce{Cl-}][\ce{HF}]}{[\ce{HCl}][\ce{F-}]}\tag{2}$$


We can expand this equation:


$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]}\tag{2'}$$


And then we realise that that is nothing else than:


$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]} = \frac{[\ce{Cl-}][\ce{H+}]}{[\ce{HCl}]} \cdot \frac{[\ce{HF}]}{[\ce{H+}][\ce{F-}]} = \frac{K\_\mathrm{a}(\ce{HCl})}{K\_\mathrm{a}(\ce{HF})} \tag{2''}$$


So the equilibrium constant is the fraction of the acidity constants of $\ce{HCl}$ and $\ce{HF}$. We know that $\ce{HCl}$ is a much stronger acid than $\ce{HF}$. This is reflected by the inequation:


$$K\_\mathrm{a}(\ce{HCl}) > K\_\mathrm{a}(\ce{HF}) \tag{3}$$


Since the numerator is larger than the denominator, the value of the fraction must be larger than $1$. Therefore, the product side is preferred.


If your book arrives at any other conclusion, it is disregarding experimental results and should be replaced.


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I guess it's late, but better late than never.


Technically, HF should be the stronger acid, as its parts are already strongly partially charged. So strong in fact, that even after dissociating into water the iones are present rather as ion pairs (H3O+.F-) than as seperate ions. This however inhibits fluoridic acids property as acid.


The proper reasoning for these two acids therefore would be simply per electronegativity difference between their compounds, as both acids have only 2 atoms, of them being hydrogen and the other a member of the halogen class.


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