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Ted Mooney, P.E. RET
Pine Beach, NJ
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How much HCl is consumed by rust? No room for acid additions
I am working with a company that runs a lot of very Rusty product through their hydrochloric acid. The dilution occurs pretty severely on a daily basis. They start with 32% Hydrochloric and after a week they're usually down to about 12% Hydrochloric. At this time they dump the bath. Because of this I was really curious about how much rust does it take on their iron products to generate one gallon of water into their Hydrochloric bath. They make daily additions to their bath but never have enough room even with the drag out to keep the acid up to strength. The metals concentration is not an issue as we have methods of keeping that reduced. there is no drag in whatsoever and the bath is run at 150 °F. I would greatly appreciate any help you could be!
David French- Charlotte, North Carolina, USA
June 29, 2017
A. Hi David,
If you cannot get round your restrictions, then the only thing is to bail out a portion of the acid at the end each day. Work out how much to bail out and top up to maintain the acid within the desired specification.
Alternatively, is there a way you can look at reducing the rusting of the parts? Or are they received into you factory already rusted?
Good luck
Mark
Aerospace - A sunny rock in the irish sea
July 17, 2017
August 4, 2017
A. Hi David,
Nice question, asks for some stoichiometric calculations in chemistry. Linear calculations, not too complicated once you know the trick, but I have to be careful as calculation errors are easily made and I hope people start checking the numbers.
There are some assumptions:
Rust is a mixture of different oxides and hydroxides. Let's calculate with the general assumption that only Iron(III)oxide and Iron(II)Oxide are present on the surface. I'm sure other people will correct me here.
According to the OxyChem hydrochloric acid handbook p32, 12% HCl contains some 125 g/l HCl and 32% contains some 371 g/l. HCl.
So the difference in [HCl] is 246 g/l.
A kg Iron(III)oxide is reduced to Iron by approx. 1370 grams of HCl and that will give you 2,031 grams of Iron(III) chloride and 338 milliliters of water.
That means per kg of this oxide approx. 5.7 liter of your HCl goes from 32% to 12% and it has 6% volume increase from the formed water.
(In # and fl oz: 1# Iron(III)oxide is reduced to Iron by app 1.33# of HCl and that will give you 1.95# Iron(III) chloride and 5 fl oz water.)
A kg of Iron(II)oxide will react with 1015 g HCl to give 1764 gram of Iron(II) chloride and 251 ml of water.
That means per kg of this oxide approx. 4.1 liter of your HCl goes from 32 to 12% and also has some 6% volume increase.
I don't know the volume of your tank, so the rest of the calculations I leave up to you. I guess replenishment of the HCl is the biggest problem here. You probably add way more HCl to bring it back to the right concentration than what is caused by the water formation from the rust + HCl reaction.
Hope this helps a bit to align your thoughts.
Best Regards,
Harry van der Zanden
consultant - Tilburg, Netherlands
I just wanted to sincerely thank you for the complete and concise answer you provided. I believe this is just the information that I required, Harry. This will be essential in eliminating my customer's acid disposal completely...David
- Charlotte, North Carolina, US
August 8, 2017
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