Here is a problem a made up:

We have some aqueous $\ce{HCl}$. What is the maximum possible molarity of the $\ce{HCl}$?

How do I solve this? I don't have any good ideas.


3 Answers 3


The HCl molecule is, at standard conditions, a gas. The reagent used in laboratories is HCl dissolved in water, which is why you'll find in the label that it is around 37% HCl in weight. The other 63% is water.

Considering 37% as the maximum solubility of HCl, you can calculate the molarity using the solution density (1.2 g/mL) and HCl's molar mass (36.46 g/mol).

For 1000 mL of solution, you will have 1200g of weight, of which 37% is HCl: 444g.

In 444g of HCl you have 12.18 moles (444/36.46), which means the concentration of P.A. HCl is around 12.18 moles per liter.

Note that because P.A. HCl is a solution saturated with a gas, vapors are likely to be released and escape the solution (which is why P.A. HCl shouldn't be handled outside fume hoods), which makes this concentration an approximation, and it should never be used to prepare a solution which you need to know the exact acid concentration.


According to "Hydrochloric acid" IARC Monographs volume 54:

The solubilty of HCl in water is:

82.3 grams HCl per 100 grams water at 0 degree C

67.3 grams HCl per 100 grams of water at 30 degrees C

and the density of 39.1% HCl aqueous solution is 1.20.

So for example at 0 degrees C, 1 liter of solution is 1200 grams and contains (82.3/182.3)(1200 grams) = 542 grams which corresponds to 14.9 moles HCl.

For 30 degrees C, this works out to 13.2 moles HCl.

So the maximum concentration is about 15M at 0 degrees C decreasing to 13M at 30 degrees C.

For data outside this temperature range and not limited to atmospheric pressure see A STUDY OF THE SYSTEM HYDROGEN CHILORIDE AND WATER J. Am. Chem. Soc., 1909, 31 (8), pp 851–866

  • 2
    $\begingroup$ Adding hydrogen chloride to water reduces the melting point of the solution, so it's possible to chill further and increase $\ce{HCl}$ solubility without solidification. This phase diagram suggests that hydrogen chloride monohydrate exists as a liquid above -15°C, containing 67% $\ce{HCl}$ by weight. I don't have the density figure so I can't calculate the molarity, but it would likely be above 20 M. $\endgroup$ Commented Mar 25, 2015 at 15:48
  • $\begingroup$ yes, there is density information for that here pubs.acs.org/doi/abs/10.1021/ja01938a001 would be about 1.27 $\endgroup$
    – DavePhD
    Commented Mar 25, 2015 at 16:11

Molx's answer is very good. However, the way I learned it was a little different and may make more sense to some people. Instead of using 444g of HCl, which is perfectly fine, we can assume we have a sample of 100 g, much like we do in empirical formula calculations as to have every percent be out of 100.

To start, just like Molx mentioned, the maximum solubility at room temperature of HCl is 37% by mass. From this we can assume 100 g of HCl solution. From that assumption we know that 37% (the mass in HCl) of 100g of the HCl solution is 37.0 grams. The rest has to be water, as HCl is a gas at room temperature that is dissolved in water to form Hydrochloric acid.

Ultimately, to find the maximum molarity of the HCl in water, we must use the concentration equation for molarity, which is defined as moles of the solute divided by the liters of the solution (M = moles of solute/ liters of solution).

We now know that we have 37.0 g of HCl, which is necessary for the Molarity calculation. When you multiply 37.0 g HCl by (1 mol HCl/36.46 g HCl) as a conversion factor, we get 1.014 mol of HCl.

To find our liters of solution, we can use our density. Multiply 100 g of our HCl solution by our density conversion factor of (1 mL/1.19g) to get 84 mL, and then divide by 1000 to get 0.084 L.

When you substitute 1.014 mol HCl and 0.084 L into the molarity equation, we find that the molarity is around 12.07 M HCl, which is the highest concentration of HCl that we can obtain at room temperature.

Like Molx said, it "fumes" because it is so concentrated with HCl. As such, it should be treated as an approximation like Molx said as well. I hope this version makes sense too.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.