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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 $1$.

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    $\begingroup$ Have you tried to make the calculus? Which values of $\ce{K_a}$ have you got? And obviously it depends on the values of each concentrations. $\endgroup$
    – ParaH2
    Commented Oct 22, 2015 at 21:53
  • $\begingroup$ I am not allowed to use equilibrium constant tables. I am supposed to come up with an answer based on reasoning but for some reason my reasoning is wrong. $\endgroup$
    – Yashas
    Commented Oct 25, 2015 at 19:36
  • $\begingroup$ My textbook doesn't mention anything about concentrations for that question. $\endgroup$
    – Yashas
    Commented Oct 25, 2015 at 19:37
  • $\begingroup$ Thats strange, let me think. $\endgroup$
    – ParaH2
    Commented Oct 25, 2015 at 20:06
  • $\begingroup$ What solvent is the reaction supposed to occur in? The answer will be different if it is say neat HF vs. if it's water. Or maybe it is liquid HCl at cryogenic temperatures? Or acetonitrile? $\endgroup$
    – Curt F.
    Commented Jun 17, 2016 at 21:40

2 Answers 2

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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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  • $\begingroup$ @CurtF. Fair point. Since the equation balances without $\ce{H2O}$, I removed it. $\endgroup$
    – Jan
    Commented Jun 17, 2016 at 21:43
  • $\begingroup$ Sir although i agree eith your calculation , shouldnt in real life it would not have a eqb between HCl and NaF as such HCl is a strong acid so when it is together in water with NaF it will dissociate completely isnt ? So no equlibrium can be established right ? $\endgroup$
    – ProblemDestroyer
    Commented Apr 27, 2022 at 22:10
  • $\begingroup$ @ProblemDestroyer Have you ever taken a bottle of HCl and taken a sniff? There are acidic fumes caused by the equilibria $\ce{H3O+ + Cl- <<=> H2O + HCl(aq) <=> H2O + HCl (g)$ Or in other words: HCl an only leak into the gas phase if there is an ever so minute concentration of undissociated HCl in the aquaeous phase. Therefore, your assumption is wrong. (For practical intents and purposes, however, hte equilibrium is so strongly product-sided that we can ignore undissociated HCl in almost all calculations.) $\endgroup$
    – Jan
    Commented Apr 29, 2022 at 8:14
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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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    $\begingroup$ This is not true, HCl is a stronger acid than HF. $\endgroup$
    – orthocresol
    Commented May 18, 2016 at 21:00
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    $\begingroup$ Electronegativity doesn't matter here chemistry.stackexchange.com/a/34829/9961 $\endgroup$
    – Mithoron
    Commented May 18, 2016 at 21:13
  • $\begingroup$ No, HF is a weak acid with a dissolution equilibrium of $6.8*10^{-4}$ $\endgroup$
    – Yunfei Ma
    Commented May 19, 2016 at 2:08

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