# 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 $1$.

• 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. – Hexacoordinate-C 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. – Hexacoordinate-C 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

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.

• @CurtF. Fair point. Since the equation balances without $\ce{H2O}$, I removed it. – Jan Jun 17 '16 at 21:43

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.

• This is not true, HCl is a stronger acid than HF. – orthocresol May 18 '16 at 21:00
• Electronegativity doesn't matter here chemistry.stackexchange.com/a/34829/9961 – Mithoron May 18 '16 at 21:13
• No, HF is a weak acid with a dissolution equilibrium of $6.8*10^{-4}$ – Yunfei Ma May 19 '16 at 2:08