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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 $1$.

Jan
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Yashas
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    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. – ParaH2 Oct 22 '15 at 21:53
  • 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. – Yashas Oct 25 '15 at 19:36
  • My textbook doesn't mention anything about concentrations for that question. – Yashas Oct 25 '15 at 19:37
  • Thats strange, let me think. – ParaH2 Oct 25 '15 at 20:06
  • 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? – Curt F. Jun 17 '16 at 21:40
  • @Yashas how more stable is concluded from the weak acid/base ? – ProblemDestroyer Apr 26 '22 at 23:06
  • @ProblemDestroyer Weak acid/base don't readily react in acid-base reactions unless a strong complementary base/acid is used. – Yashas Apr 27 '22 at 06:04
  • By the way is this reaction equlibrium possible ? I mean as HCl is strong acid it will dissociate completely so there is nothing called eqb which can be established isnt ? @Yashas – ProblemDestroyer Apr 27 '22 at 22:08
  • @ProblemDestroyer In principle, we can have an equilibrium for nearly all reactions. Some reactions might strongly favor the products or reactants but there is still the notion of equilibrium. In this particular case, even though HCl is a strong acid, we can still talk about equilibrium. In fact, that's how we define pH! – Yashas Apr 28 '22 at 11:49
  • Understood @Yashas . – ProblemDestroyer May 02 '22 at 13:47

2 Answers2

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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.

Jan
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  • @CurtF. Fair point. Since the equation balances without $\ce{H2O}$, I removed it. – Jan Jun 17 '16 at 21:43
  • 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 ? – ProblemDestroyer Apr 27 '22 at 22:10
  • @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.) – Jan Apr 29 '22 at 08:14
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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.

Martin Winters
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