I'd like know if this is right. The reaction between $\ce{NaCl}$ and $\ce{H_2SO_4}$ should yield $\ce{HCl}$ and sodium sulfate. This is a double substitution, true? No oxidation took place here. The thing is, when I tried the reaction in the lab, the resulting solution (gas bubbled into water) seemed to be sulfuric acid. Simply enough, the solution didn't react with aluminum. I presume that $\ce{SO_2}$ was formed from the oxidation the sulfur (it a little bit like it).

What am I doing wrong? Shouldn't the substitution take place instead?


$ \ce{NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl}\\ \ce{NaCl + NaHSO_4 \rightarrow Na_2SO_4 + HCl} $


$ \ce{2\space NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2\space HCl } $

Is this right?

  • $\begingroup$ Did you measure the pH of the resulting solution? $\endgroup$ May 19, 2013 at 5:20
  • 3
    $\begingroup$ Could you include the exact procedures you followed? $\endgroup$
    – Jerry
    May 19, 2013 at 7:32
  • $\begingroup$ @Jerry I added salt to a RBF and then some excess of sulfuric acid. The reaction began almost immediately. There was bubbling and the gas went through a hose to a cylinder with water and crushed ice. $\endgroup$
    – OFRBG
    May 19, 2013 at 19:53
  • $\begingroup$ @NicolauSakerNeto I measured the pH and it was 1 or less. $\endgroup$
    – OFRBG
    May 19, 2013 at 19:55
  • $\begingroup$ This video suggests another way to identify $\ce{HCl}$. Or you could just wet a pH paper and hold it above the acid solution. A test for $\ce{H_2SO_4}$ would be to add a small amount of a soluble barium salt to the acid. It's a little strange for you solution to be $\ce{H_2SO_4}$ though, since pure sulphuric acid is far from volatile, and chloride probably isn't reducing enough to form volatile $\ce{SO_2}$. How long did you wait for the aluminium to react? It can take several minutes to start breaking through the $\ce{Al_2O_3}$ passivation layer. $\endgroup$ May 19, 2013 at 20:26

1 Answer 1


Ideally, that's the reaction. Realistically, there are a few speed bumps.

First, it occurs stepwise, with the acid salt sodium bisulfate as an intermediate. The first stage happens at room temp between equal proportions of salt and sulfuric acid. The second stage requires temperatures of 200*C (in turn requiring water-free concentrations of acid and dry salt) and an additional equivalent of salt that reacts with the bisulfate.

You did say you were using excess sulfuric acid; if true, you'll react all the salt but what'll be left in the flask is sodium bisulfate, not sodium sulfate. Since you're interested in the HCl gas produced and not the sulfate salt, this isn't a big deal, but I just wanted you to be clear that one mole of sulfuric acid to two moles of salt isn't going to get you the expected 2 moles of HCl unless you're heating a "water-free" combination of the reactants.

Second, if you have too much water in the reaction, the HCl won't bubble out; it'll happily sit in solution. So, dry salt and 10M or higher concentration of sulfuric acid are recommended (as in Nicolau's YouTube video, 18M is plenty strong enough to do the job, and they make and sell it up to 98% concentrations). If you're stuck with 1M, you can heat the flask to boiling, which evaporates some water and increases the acid concentration, liberating the HCl gas (sulfuric acid won't boil below 300*C, so you're fine distilling the HCl like this until you're out of water and the thermometer starts to climb).

Following these steps and bubbling the gas produced through water, what you produce should definitely be hydrochloric acid. Obviously the molar strength of it (and thus its ability to react with test materials as was also indicated by Nicolau) will depend on the relative proportions and amount of reactants, the volume of your trap water, and how much gas is actually absorbed by the water.

  • $\begingroup$ Why does the salt react with the bisulfate and not directly with the sulfate? Let me add the reactions to the question. $\endgroup$
    – OFRBG
    May 21, 2013 at 0:41
  • 1
    $\begingroup$ I'm not sure the bisulphate anion should react with salt, since its acidity is significantly reduced compared to its parent acid (as I mentioned here) $\endgroup$ May 21, 2013 at 0:51
  • $\begingroup$ @NicolauSakerNeto But still, $HSO_4^-$ is a strong acid, isn't it? The weak acid would be the sulfate anion. $\endgroup$
    – OFRBG
    May 21, 2013 at 1:21
  • $\begingroup$ @NicolauSakerNeto - The bisulfate anion does indeed react with salt; however, because it's a weaker acid in solution, the reaction is endothermic and so requires heating to an equilibrium temperature of 200*C as I mentioned. $\endgroup$
    – KeithS
    May 21, 2013 at 1:25
  • $\begingroup$ @Fiire - The $\ce{HSO_4^-}$ anion in combination with a cation other than hydronium is an acid salt, and acid salts are weak acids in water. The primary industrial use for this chemical is to acidify like a strong mineral acid would, but in applications where a strong mineral acid would be hazardous (such as in the hands of the average individual). These applications include pool water conditioning, cooking/baking, metal pickling such as in jewelrymaking, soil acidification (though typically a nitrogen salt like ammonium nitrate is used), etc. $\endgroup$
    – KeithS
    May 21, 2013 at 1:34

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