Numerous inorganic compounds undergo autodissociation in their liquid state. In liquid hydrogen fluo — Organic Chemistry Chemistry Question
Autodissociation of Liquid Hydrogen Fluoride and Superacids
Numerous inorganic compounds undergo autodissociation in their liquid state. In liquid hydrogen fluoride (density, ρ = 1.002 g cm-3) the following equilibrium can describe the autoprotolysis:
3 HF ⇌ H2F+ + HF2–
The corresponding equilibrium constant is 8.0×10–12.
Calculate what fraction of the fluorine is present in the cationic species in liquid HF, supposing that only these three species are present in the system.
Model Answer
From the ionic product: [HF2–] = [H2F+] = √(8.0 × 10–12) = 2.8×10–6
c(HF) = ρ / M(HF) = 50.1 mol dm-3
The autodissociation causes a negligible change in concentration of HF, hence the requested fraction is [H2F+] / [HF] = 5.65·10–8.
Various reactions can take place in liquid HF.
Write the equation of the reactions of liquid HF with the following substances: H2O, SiO2, acetone.
Model Answer
2 HF + H2O = H3O+ + HF2–
SiO2 + 6 HF = SiF6– + 2 H3O+
CH3COCH3 + 2 HF = [CH3–COH–CH3]+ + HF2–
In water HF behaves as a medium-strength acid and dissociates only partially. The most important reactions determining the equilibrium properties of the solution are the following:
HF + H2O ⇌ H3O+ + F– (1)
HF + F– ⇌ HF2– (2)
The equilibrium constants of the two equilibria are
K1 = 1.1×10–3
K2 = 2.6×10–1
Calculate the analytical concentration of HF in a solution having a pH = 2.00.
Model Answer
The equations describing the system are:
[HF]total = [HF] + [F–] + 2 [HF2–] (fluorine balance)
[F–] + [HF2–] = [H3O+] (charge balance)
From the charge balance: [F–] = [H3O+] – [HF2–].
Substituting into the fluorine balance: [HF] = [HF]total – [H3O+] – [HF2–].
The equilibrium constants:
K1 = ([H3O+][F–]) / [HF]
K2 = [HF2–] / ([HF][F–])
[H3O+] = 0.01
Solving the equations we obtain: [HF] = 0.0889, [HF]total = 0.0991 mol dm-3.
In early studies of aqueous HF, equilibrium (2) was not considered. However, pH measurements, assuming only equilibrium (1) led to contradictions.
Show that, assuming only equilibrium (1), pH measurements can indeed lead to a concentration-dependent equilibrium constant for (1).
Model Answer
Expressing the equilibrium constant K1 as a function of [H3O+], [HF]total, and [HF2–] gives:
K1 = [H3O+]([H3O+] – [HF2–]) / ([HF]total – [H3O+] – [HF2–])
Not considering equilibrium (2) the calculation goes as:
K1' = [H3O+]2 / ([HF]total – [H3O+])
Clearly, the two expressions will be equal only in very special cases, i.e. the measurements can indeed indicate non-constant K1'.
Two chemists wanted to determine the acidity constant of HF (K1) from one and the same solution with a known concentration. They measured the pH of the solution and then obtained a K1 value by calculation. However, the better chemist (she) knew about equilibrium (2), while the other one did not. So, she was surprised when they both obtained the same K1 value.
What was the concentration of the HF solution?
Model Answer
The two values for the equilibrium constant were found to be the same:
[H3O+]([H3O+] – [HF2–]) / ([HF]total – [H3O+] – [HF2–]) = [H3O+]2 / ([HF]total – [H3O+])
From this equation [HF]total = 2 [H3O+]
Substituting this into the expressions for K1, we obtain K1 = [H3O+].
Thus, [HF]total = 2 K1 = 0.0022
c(HF)total = 0.0022 mol dm–3.
Calculate the equilibrium constant of the following equilibrium:
2 HF + H2O ⇌ H3O+ + HF2–
Model Answer
K = K1 K2 = 2.86×10–4.
The dissociation equilibrium of a solute in a solvent can be significantly shifted by the addition of a suitable substance into the solution.
Propose three different inorganic compounds for increasing the dissociation of HF in water.
Model Answer
E.g. NaOH, CaCl2, Na2CO3, FeCl3, AlCl3, etc.
Suitable compounds can also shift the autodissociation equilibrium of HF in its liquid state by orders of magnitudes. A well-known such substance is SbF5.
Show how SbF5 shifts the autodissociation equilibrium of liquid HF.
Model Answer
2 HF + SbF5 = SbF6− + H2F+
The shift in the autodissociation also implies an important change in the Brønsted acidity of the solvent. In fact, the degree of solvation of the proton produced from the autodissociation essentially affects the Brønsted acidity of the solvent.
How is the Brønsted acidity of a given solvent determined by the extent of proton solvation?
Model Answer
The weaker is solvation the greater is the acidity.
The mixture of HF-SbF5 belongs to the family of superacids, owing to their very high acidity. These acids are able to protonate very weak bases and thus have enabled the preparation of exotic protonated species. These, in turn, have opened new synthetic routes.
Formulate reaction equations for the reaction of methane and neopentane with the HF-SbF5 superacid. Note that in both cases there is a gaseous product.
Model Answer
CH4 + HSbF6 = [CH5+][SbF6−] → [CH3+][SbF6−] + H2