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Probably we can have an aqueous solution of $\ce{SO2}$ by dissolving it in water, because we would have an equilibrium between $\ce{SO2(g)}$ and $\ce{SO2(aq)}$:

$$\ce{SO2(g) <=> SO2(aq)}$$

  • How can I compare pH of an aqueous solution of $\ce{SO2}$ with that of pure water qualitatively?
  • What happens to the pH of an aqueous solution of $\ce{SO2}$ if I aerate it by bubbling air through it?
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  • $\begingroup$ 1. Is $\ce{SO2}$ an acidic, basic, or neutral oxide? 2. Can $\ce{SO2}$ react with any of the gasses in air, and if so, what is the product? $\endgroup$ – Nicolau Saker Neto Jun 13 '15 at 12:26
  • $\begingroup$ @ Nicolau Saker Neto: I know that So2 is acidic.And I could assume that I would react with gases, since it is a gas.But how can I relate this? $\endgroup$ – On the way to success Jun 13 '15 at 12:39
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    $\begingroup$ If $\ce{SO2}$ is an acidic oxide, then which acid does it create when reacting with water, and what is its name? How strong is this acid? Also, notice that this acid is very similar, in formula and name, to another compound which has one extra atom . What is this compound? Could the extra atom come from air? What would the reaction be? $\endgroup$ – Nicolau Saker Neto Jun 13 '15 at 12:51
  • $\begingroup$ @ Nicolau Saker Neto: When SO2 dissolve in water ,H2SO3 will be the acid that is going to form, it is known as sulphurous acid . And is that extra atom is H or any other? $\endgroup$ – On the way to success Jun 13 '15 at 12:57
  • $\begingroup$ Think of the most common elements in air, and try adding them to the formula for sulphurous acid. Which do you think is the most likely outcome? $\endgroup$ – Nicolau Saker Neto Jun 13 '15 at 13:03
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Upon adding $\ce{SO2}$ gas to water, it shows its acidic oxide character by hydrolyzing to sulfurous acid:

$$\ce{SO2(g) + H2O(l) <=> H2SO3(aq)}$$

$\ce{H2SO3}$ is a weak diprotic acid in water, with the following acid dissociation equilibria:

$$\ce{H2SO3(aq) <=> H+(aq) + HSO3^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1}=1.4\times 10^{-2}}$$ $$\ce{HSO3{}^{-}(aq) <=> H+(aq) + SO3{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=6.3\times 10^{-8}}$$

Thus, addition of $\ce{SO2}$ gas to water will produce an acidic solution.

What would happen if air were bubbled into the solution? Air is a mixture of mostly nitrogen and oxygen. It so happens that $\ce{SO2}$ reacts with oxygen to form $\ce{SO3}$:

$$\ce{2SO2(g) + O2(g) -> 2SO3(g)}$$

Here, the sulfur atom in $\ce{SO2}$ is oxidized (oxidation number goes from +4 to +6) while the oxygen atoms in $\ce{O2}$ are reduced (oxidation number goes from 0 to -2). Now, $\ce{SO3}$ also happens to be an acidic oxide, and it hydrolyzes in water to form sulphuric acid:

$$\ce{SO3(g) + H2O(l) <=> H2SO4(aq)}$$

This time, $\ce{H2SO4}$ is a strong diprotic acid:

$$\ce{H2SO4(aq) <=>> H+(aq) + HSO4^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1} \approx 10^6}$$ $$\ce{HSO4{}^{-}(aq) <=> H+(aq) + SO4{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=1\times 10^{-2}}$$

Assuming that not much $\ce{SO2}$ is lost while bubbling air through the sulphurous acid solution, then you're effectively replacing a weak acid with a strong acid, which would cause the pH to decrease.

As a final remark, if the sulphurous acid solution were bubbled with pure nitrogen, then there would be no oxidation of $\ce{SO2}$ into $\ce{SO3}$, and after a while enough $\ce{SO2}$ will have been dragged away for the solution to become a less concentrated sulphurous acid, which would therefore have a higher pH.

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  • $\begingroup$ If I aerate the sulphurous acid only with N2 ,why doesn't it convert SO2 to SO3, is it because of not having lower oxidation states for nitrogen, other-than zero? $\endgroup$ – On the way to success Jun 14 '15 at 2:04
  • $\begingroup$ Nitrogen can reach negative oxidation numbers, but that's irrelevant; without oxygen gas, $\ce{2SO2 + \color{Red}{O2} -> 2SO3}$ can't happen! $\endgroup$ – Nicolau Saker Neto Jun 14 '15 at 2:10
  • $\begingroup$ Here what happened to the oxygen.You have stated that it is reduced to -2 oxidation state.But where is it? $\endgroup$ – On the way to success Jun 14 '15 at 2:12
  • $\begingroup$ The oxygen gas was added to sulphur dioxide $\ce{SO2}$ to make sulphur trioxide $\ce{SO3}$. All oxygen atoms in both these compounds have an oxidation number of -2. $\endgroup$ – Nicolau Saker Neto Jun 14 '15 at 2:16
  • $\begingroup$ Okay, Could you state some other gases behave like this,and can't liquid and solids behave like the way SO2(g) did? $\endgroup$ – On the way to success Jun 14 '15 at 2:18
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When you dissolve $\ce{SO2}$ in water you get $\ce{H2SO3}$. $\ce{SO2(aq)}$ means $\ce{H2SO3}$ therefor,the pH of the solution will be slightly acidic when compared to that of pure water.

to your other question.when you aerate the solution $\ce{SO2}$ gas right above the solution goes away.The equilibrium is then disturbed.Therefore, $\ce{SO2}$ dissolved in the solution comes out.Because of that pH of the solution increases.

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    $\begingroup$ This is incorrect! While bubbling an aqueous solution of a gas does remove some of it from the water, there is another effect to be taken into account in this case. $\endgroup$ – Nicolau Saker Neto Jun 13 '15 at 13:05
  • $\begingroup$ @ Nicolau Saker Neto:Actually SO2 is dissolved in water.I don't see any other effect to be taken in this case.Please let me know what the other effect is. $\endgroup$ – lucy G Jun 13 '15 at 13:08
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    $\begingroup$ @lucyG The other effect is oxidation of $\ce{H2SO3}$ to $\ce{H2SO4}$ by oxygen in the air. $\endgroup$ – bon Jun 13 '15 at 15:26
  • $\begingroup$ @bon Yes that effect can also be there.But according to the question of On the way to success don't you think we have to give more priority to the equlibrium that has been made rather than other effects.In my point of view this question inquires about the basic knowledge of Le Chatlier's principle $\endgroup$ – lucy G Jun 13 '15 at 15:50
  • $\begingroup$ @lucyG I don't see anything in the question that suggests we should ignore the effect of oxidation. $\endgroup$ – bon Jun 13 '15 at 15:57

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