How much acid is needed to dissolve 1kg of CaCO3?

[matthyaouw]

I wonder if someone can help me here. For part of my geography dissertation I'm hoping to completely dissolve some very pure limestone (~99% CaCO3) to get some microfossils out. I'll be using either acetic, monochloric or formic acid (probably dilute) because as far as I'm aware these are the only ones that won't dissolve the fossils themselves. Can anyone tell me how to come up with a rough estimate of how much I would need to dissolve say 1kg of CaCO3? I'll be researching & writing detailed methods once my deadlines and exams have passed, but right now I need to come up with a rough figure just to let the lab technicians know what sort of quantities I'll be needing.

Sorry if this is more appropriate to homework help mods. Feel free to move it if it is.

 

[mrjeffy321]

For a monoprotic acid (symbolized HA), the reaction would be,

HA + CaCO3 --> H2CO3 + CaA2
Where CaA2 is just some salt byproduct. The Carbonic acid will quickly decompose to form Carbon Dioxide and water,
H2CO3 --> H2O + CO2

So the complete, balanced, reaction would be,
2 HA + CaCO3 --> H2O + CO2 + CaA2

For every 1 mole of Calcium Carbonate, you will need 2 moles of the monoprotic acid.

To find out how much acid you will need to dissolve X grams of CaCO3,
Convert to moles of CaCO3 by dividing by the molar mass of Calcium Carbonate, then multiply by the mole ratio of moles of HA needed per mole of CaCO3 (which is 2 in this case). Then you know the number of moles of acid you will need. From this, you can convert to Liters of acid if you knew what concentration you wanted to use.

If you acid concentration was in terms of Molarity,
Concentration = moles of acid / Liters of solution

The molar mass of Calcium Carbonate is: 100.09 g/mol.


What are the fossils made up of that will not be dissolved by the acid?

 

[matthyaouw]

OK, so if I need to dissolve 1kg then I'll need 19.98 moles of acid. Say I use 10% concentration I'd need 1.998L of it? Sorry, this must be seriously basic stuff to you, but I've not done chemistry since I was about 16.
The fossils are made of calcium phosphate.

 

[mrjeffy321]

If you want to use an acid concentration of 10% (by weight), then you will need to get more specific as to just what type of acid you want to use since the molar mass of the acid will be used in the calculations.

If the concentration you are using is a weight percentage of the total,
then 10% would mean that 10% of the total mass of the solution is whatever acid you are using. You previously calculated the number of moles of acid you would require, to find the mass of this acid simply multiply by the molar mass of the acid.
After you know the mass of the acid, you can find the mass of the total solution since,
concentration = mass of acid / mass of solution
10% = mass of acid / mass of solution
mass of solution = mass of acid / 10%

The once you know the mass of the solution can can either take the easy way out and assume it has a density of about 1 g/mL (the density of water), or you can go and look up the actual density of ___ acid at ___ concentration and calculate the volume in that way.
Density = mass / volume
You know the mass and you can look up / assume the density, so solving for the volume is easy.

If you used a molar concentration (moles / Liter), you would not need to worry about the molar mass of the acid as long as you knew the molar concentration you wanted to use, but this is probably less likely.

5% (by weight) vinegar (acetic acid) is easily available in stores. I have even seen 10% vinegar sold as a type of cleaning solution.

The molar mass of acetic acid is 60.05 g/mol.

 

[matthyaouw]

Ok, so if I use acetic acid, for the mass of solution I have a number of 11997.99g.
If so, since acetic acid has a density of 1.049g/ml, at 10% by weight the overall density is 1.0049 I'll need 11939.5ml for 1kg of CaCO3. Have I done this right, or made a silly mistake somewhere?

 

[mrjeffy321]

acetic acid has a density of 1.049g/ml, at 10% by weight the overall density is 1.0049

No, I don’t think you can do it this way.

I presume the density of acetic acid you found (1.049 g/mL) is for glacial acetic acid / 100% acetic acid.
But you want to find the density at a concentration 10% by weight. There is no way I know of to find this out just by looking at the anhydrous density; you usually need to look it up in a table somewhere.

The table on this page,
http://www.chembuddy.com/?left=CASC&right=density_tables
lists a 10% actetic acid aqueous solution to have a density of 1.0126 g/mL.

Use this value in your calculation and you should be good.

It is going to come out to be several Liters of acid. You will probably want to do the dissolving of the CaCO3 in batches of smaller amounts of acid since as the reaction proceeds, the acid will only dilute itself.
And all these calculations are based on ideal situations where everything reacts perfectly and completely. In reality, you should plan on having some spare acid to make sure and dissolve everything.

 

[matthyaouw]

Great stuff. I'll play around with the numbers for the other two acids and for higher concentrations and see what I come up with.
Thanks very much for your help

 

[chemisttree]

In-the-ground-dissolution of limestone in an oilfield is known and practiced.

SPE 26578; The Optimum Injection Rate for Matrix Acidizing of Carbonate Formations; Y. Wang, et al; Oct. 3-6 1993; (pp. 675-687).
SPE 28547; Optimum Injection Rate From Radial Acidizing Experiments; B. Mostofizadeh, et al; Sep. 25-28 1994; (pp. 327-333).
SPE 37312; Reaction Rate and Fluid Loss; The Keys to Wormhole Initiation and Propagation in Carbonate Acidizing; T. Huang, et al; Feb. 18-21 1997; (pp. 1-10).
SPE 37283; Mechanisms of Wormholing in Carbonate Acidizing; M. Buijse; Feb. 18-21 1997; (pp. 683-686).
SPE 52165; Quantitative Model of Wormholing Process in Carbonate Acidizing; M. Gong; Mar. 28-31 1999; (pp. 1-11).
Chemical Engineering Science, vol. 48. No. 1 (pp. 169-178) 1993; Chemical Dissolution of a Porous Medium by a Reactive Fluid-I. Model for the "Wo,mholing" Phenomenon; G. Daccord, et al.
Chemical Engineering Science, vol. 48. No. 1. (pp. 179-186) 1993; Chemical Dissolution of a Porous Medium by a Reactive Fluid-II. Convection VS Reaction, Behavior Diagram; G. Daccord, et al.
AIChE Journal; vol. 34, No. 1, Jan. 1988; Pore Evolution and Channel formation During Flow and Reaction in Porous Media; M. Hoefner, H. Fogler; (pp. 45-54).
Society of Petroleum Engineers; 1993; Advances in matrix Stimulation Technology; G. Paccaioni, M. Tambini; (pp. 256-263).
Journal of Petroleum technology, Feb. 1987; Role of Acid Diffusion in Matrix Acidizing of Carbonates; M. Hoefner, et al; (pp. 203-208).
Oil Well Stimulation; R. Schehter; Prentice-Hall, Inc. 1992; (pp. 6).
Best Practices--Carbonate matrix Acidizing Treatments; Halliburton Energy Services, Inc. Bibliography No. H01276; ; Oct. 1998; (pp. 1-18.
Society of Petroleum Engineers; 1989; Carbonate Acidizing: Toward A Quantitative Model of the Wormholing Phenomenon; G. Daccord, et al; (pp. 63-68).

The treatment of oil that contains powdered or entrained limestone is trivial. Sometimes ground limestone is added to the oil to scavenge sulfur compounds during coking and distillation of heavy crudes obtained from tar sands.

I'm not sure just what Mr. Wren has discovered but it sounds very suspicious. Is he asking you for some up-front money or services?

 

[chemisttree]

Yes. It is called sulfuric acid. It plucks one oxygen from carbonate (and converts it to water) and produces CO2. The product, gypsum, occupies a much smaller volume and thus a pore is produced. This treatment is known as carbonate acidizing and is nothing new. Read the titles from the list of publications I provided above. You will see that this is well known.

Anyone asking you for $500 up front with a promise to buy you a new truck and a million dollar bonus should be treated with the same level of contempt that he shows you.

Leave this guy far behind you...