storing muriatic acid

[Michaela521]

If you live in a very cold place like minnesota should you store muriatic acid in a heated place. It gets very cold where I am too.

[chem geek]

The freezing point of Muriatic Acid (31.45% Hydrochloric Acid) is -46C (-50.8F) (for Hasa; other brands quote even lower freezing points) so unless you live in the Artic, I think you are safe to leave your Muriatic Acid outside.

Richard

[Poolsean]

Richard,

How much difference does it lower the freezing point if you had a 4 parts water to 1 part muriatic solution in an outdoor tank? I would assume lower than water, but not quite into the negative temps...

[chem geek]

Sean,

The freezing point depression of water is -1.86C/m (3.35F/m) where "m" is molality or moles solute per kilogram of solvent (water) for a non-electrolyte (for an electrolyte such as HCl that dissociates into two parts, the depression is double this amount). So the amount of freezing point depression is solely a matter of the quantity (number of molecules; not weight) of substance dissolved in the water. So, if you dilute the Muriatic Acid (31.45% HCl) with 4 parts water, you dilute by 1/5th. The amount of freezing point depression will get reduced, but by more than 1/5th due the definition of molality.

31.45% HCl means 31.45 grams of HCl in 100 grams of solution so that means 100-31.45 = 68.55 grams of water. HCl is 36.46 g/mole so the molality "m" is (31.45 g HCl / 36.46 g/mole HCl) / (68.55 grams water / 1000 grams/kilogram) = 12.58 m. So, 12.58 m * (-1.86C/m) * 2 = -46.8C which is about -46 as claimed by Hasa. Diluting the Muriatic Acid by 1/5th is done by volume, so Hasa says that the density of Muriatic Acid is 1.16 g/ml. The example of 100 grams of solution would be 116 ml in volume. You said you would dilute that by 4 parts of water so you would be adding 4*116 = 464 grams of water since pure water is about 1.00 g/ml. So now you would have the same 31.45 grams of HCl in 464+68.55=532.55 grams of water so that gives 1.62 m. So, 1.62 m * (-1.86C/m) * 2 = -6.0C (21.2F). So you are right that the new freezing point will be above 0F.

[EDIT] This calculation was done backwards. 100 grams of Muriatic Acid solution would be 100/1.16 = 86.2 ml in volume. Diluting with 4 parts water would be 4*86.2 = 344.8 ml which is about 344.8 grams of water since pure water is about 1.00 g/ml. So now you have the same 31.45 grams of HCL in 344.8+68.55=413.4 grams of water so that gives 31.45/36.46/(413.4/1000) = 2.09 m. So 2.09 * (-1.86C/m) * 2 = -7.8C (18.0F). [END-EDIT]

Richard

[Poolsean]

Geek, You ROCK. thanks