Sulfuric acid because of its low volatility can be used to manufacture more volatile acids from their corresponding salts.
How does volatility affect the production of acids? Isn't it that sulfuric acid being stronger than the salt of weak acid replaces the anion part of salt so the weak acid is produced or does just the low volatility of sulfuric acid matter? Will it be feasible to produce a highly volatile acid even stronger than $\ce{H2SO4}$ by exploiting the low volatile nature of $\ce{H2SO4}$? What is the real process involved?
Answer
Yes, this is indeed the case. The reasoning behind it is using chemical equilibria to their fullest.
If you have a Brønsted acid and a Brønsted base in the same vessel, you will always have an equilibrium of the following kind:
$$\ce{HA + B- <=> A- + HB}$$
It depends on the nature of the acid and the base — i.e. their $\mathrm{p}K_\mathrm{a}/\mathrm{p}K_\mathrm{b}$ values — which side of the equation is favoured in a closed system (i.e. solution). For example, dissolving $\ce{NaHSO4}$ and $\ce{NaSH}$ in the same sample of water will result in the following set of equations:
$$\ce{H2SO4 + Na2S <=>> NaHSO4 + NaSH <=>> Na2SO4 + H2S}$$
Because the $\mathrm{p}K_\mathrm{a}$ values of both compare as follows:
$$\begin{array}{ccc}\hline \text{acid} & \mathrm{p}K_\mathrm{a1} & \mathrm{p}K_\mathrm{a2}\\ \hline \ce{H2SO4} & \approx -3 & 1.9\\ \ce{H2S} & 7.0 & 12.9\\ \hline\end{array}$$
We already have a way of producing $\ce{H2S}$ by means of protonating sulfide ions in solution here. However, $\mathrm{p}K_\mathrm{a1}\left (\ce{HCl}\right ) \approx -6$, which is more acidic than sulphuric acid, so adding $\ce{NaCl}$ to $\ce{H2SO4}$ should result in a net reaction of nearly nothing.
This is where vapour pressure comes in as a second factor. $\ce{HCl}$ is a gas under standard temperature and pressure while $\ce{H2SO4}$ is a liquid. ($\ce{H2S}$ is also a gas. If we added liquid sulphuric acid to sodium sulfide, $\ce{H2S}$ gas would evolve, but that’s what the $\mathrm{p}K_\mathrm{a}$ value predicts anyway.) If we have an open vessel — i.e. gases can diffuse away — then any $\ce{HCl}$ produced in equilibrium will dissociate away in the long run. This is due to a second equilibrium that we could write as follows:
$$\ce{HCl (g) <=> HCl (aq)}$$
The ‘equilibrium constant’ of this physical process is more or less what we call vapour pressure, and for hydrogen chloride both sides are significantly more balanced than for sulphuric acid. Therefore, we should describe the entire system in the following way:
$$\ce{NaCl + H2SO4 <=> NaHSO4 + HCl (diss) <=>> NaHSO4 + HCl (g) ^}$$
Where $\ce{(diss)}$ is a shorthand for dissolved in sulphuric acid, our reaction medium.
Drawing away the hydrogen chloride gas in the rightmost equation, for example by reducing pressure or inducing an air flow, will shift all equilibria to the right since that product is constantly being removed. And thus we can liberate $\ce{HCl}$ gas with sulphuric acid, even though the $\mathrm{p}K_\mathrm{a}$ of hydrogen chloride is lower than that of sulphuric acid.
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