Raising ph of waste acids
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goldenchild
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I don't think you could get an answer for your situation. I'm betting there is water in your solutions so the actual volume of acids in solution are unknown. Even if you carefully recorded how much lime you needed to neutralize 1 gallon it still wouldn't help. That is of course unless you kept all of your acids in one container is a nice homogenous blend.
Simply add the lime in small increments until there is no more reaction, periodically testing the ph between additions.
Simply add the lime in small increments until there is no more reaction, periodically testing the ph between additions.
MysticColby
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with the facts you gave, there is not enough info to give a reasonable answer.
the reason for this is that there is such a huge range of acid concentrations that give a low pH. could be highly concentrated, could be highly diluted. 12 M HCl will have the same pH as 0.1 M HCl, but they will require far different amounts of lime to neutralize.
remembering what waste you put in there, you should be able to roughly estimate how much acid is diluted to that volume.
after that, a decent guess for lime required to neutralize can be made.
it might be easier to just add some lime, stir for a few minutes, test pH, add more lime, etc. (If you record how much lime you added, you can actually figure out how much acid was in there)
the reason for this is that there is such a huge range of acid concentrations that give a low pH. could be highly concentrated, could be highly diluted. 12 M HCl will have the same pH as 0.1 M HCl, but they will require far different amounts of lime to neutralize.
remembering what waste you put in there, you should be able to roughly estimate how much acid is diluted to that volume.
after that, a decent guess for lime required to neutralize can be made.
it might be easier to just add some lime, stir for a few minutes, test pH, add more lime, etc. (If you record how much lime you added, you can actually figure out how much acid was in there)
samuel-a
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glondor said:
There is glondor. Study the pH scale and how it is masured.
In regards to your question, it really doesn't matter which acid is used, all we care about is the hydronium ions concentration.
Since, in your specific case, your waste solution is already at pH 2, you know that any base addition will raise the pH.
Lino gave you an exact answer based on the pH formula.
I would suggest to strongly embrace MysticColby's advice if you are to test a solution and it comes out pH 0.
p.s. - 12M HCl and 1M HCl will be pH 0, 0.1M HCl would be pH 1
Geo
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where i live, limestone is plentiful. the local quarry here will load a pickup truck with as much lime dust as it will hold for free. farmers use it almost every year on their fields to balance the use of nitrates (well not all but they really should). lime dust will neutralize acid in a hurry.
MysticColby
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samuel-a said:
I've never actually tested 12 M HCl, sorry for misleading. I think I might try that at work tomorrow - could be interesting. I've just always passively observed that pure diluted acids never change pH very much when adding a base, then all of a sudden will jump from 1 to 13 in 1 drop.
maybe it's possible to estimate glondor's acid concentration based on his pH of 2? it must be low concentration for that pH if 0.1 M HCl has a pH of 1. and it sounds like it's a mixture of acids - nitric and hydrochloric?
samuel-a
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MysticColby
pH is just another way to present the hydronium ions in solution:
pH 0 | 1/1 M | -log(1)= 0
pH 1 | 1/10 M | -log(0.1)= 1
pH 2 | 1/100 M | -log(0.01) = 2
and so forth...
As you see, pH is infacet a simplified way to represent the concentration of free H3O+ ions in the solution.
So, pH 2 will represent 0.01 moles (of H3O+) per liter of solution, where in the case of strong acids such as HCl or Nitric will equal to the exact molar conentration of the acid (in between the zone of 0-14)
There are some exeptions to that such in the case of sulfuric acid which carries two H+...
It get even messier when it comes to weak acid or bases.
I don't know if what i wrote make sanse to you or not. Perhaps these links could help clarify the subject even more:
http://en.wikipedia.org/wiki/PH
http://www.phs.d211.org/science/smithcw/Chemistry%20332/Quarter%204%20Unit%202/4%20logs%20and%20pH.pdf
Edit: There's another element which we sould consider in the case of waste solutions, metals salts.
As base is introduced they will use up some of the OH- ions to their favor to turn to oxides and hydroxide....
When all said and done... just eye ball it :mrgreen:
MysticColby said:
pH is just another way to present the hydronium ions in solution:
pH 0 | 1/1 M | -log(1)= 0
pH 1 | 1/10 M | -log(0.1)= 1
pH 2 | 1/100 M | -log(0.01) = 2
and so forth...
As you see, pH is infacet a simplified way to represent the concentration of free H3O+ ions in the solution.
So, pH 2 will represent 0.01 moles (of H3O+) per liter of solution, where in the case of strong acids such as HCl or Nitric will equal to the exact molar conentration of the acid (in between the zone of 0-14)
There are some exeptions to that such in the case of sulfuric acid which carries two H+...
It get even messier when it comes to weak acid or bases.
I don't know if what i wrote make sanse to you or not. Perhaps these links could help clarify the subject even more:
http://en.wikipedia.org/wiki/PH
http://www.phs.d211.org/science/smithcw/Chemistry%20332/Quarter%204%20Unit%202/4%20logs%20and%20pH.pdf
Edit: There's another element which we sould consider in the case of waste solutions, metals salts.
As base is introduced they will use up some of the OH- ions to their favor to turn to oxides and hydroxide....
When all said and done... just eye ball it :mrgreen:
A simple method I use is to note how many moles of acid are in the waste solution when I'm doing the refining reactions. Note the amounts used in your lab book. This is considered good lab technique also.
Some basic molarities of the acids we use are:
(M= moles per liter)
70% HNO3 approximately 16 M
35% HNO3 approximately 8M
32% HCl aproximately 10 M
98% H2SO4 approximately 18M
From the above data you can determine how many moles are needed of the base you are using to neutralize the volume of acid used.
You'll need to write a balanced equation for the acid base reaction to begin with to determine how many moles as the base are require to neutralize the acid used (eg: for the base sodium hydroxide = lye = NaOH):
HNO3 + NaOH --> NaNO3 + H2O ( 1 mole of NaOH is required per mole of Nitric Acid)
HCl + NaOH --> NaCl + H2O ( 1 mole of NaOH is required per mole of hydrochloric acid)
H2SO4 + 2NaOH --> Na2SO4 + 2H2O ( 2 moles of NaOH are required per mole of Sulfuric Acid)
From the above reactions you'll need:
8 moles of NaOH per liter of 35% HNO3 in the waste
10 moles of NaOH per liter of 32% HCl in the waste
36 moles of NaOH per liter of 98% H2SO4 in the waste
Finally determine the weight of NaOH (mass = 40 g per mole) needed:
8 moles x 40g = 320g per liter of 35% HNO3 in the waste
10 moles x 40g = 400g per liter of 32% HCl in the waste
36 moles x 40g = 1440g per liter of 98% H2SO4 in the waste
For other bases be sure the equations are balanced and to used the correct molar mass of the base in the formulas.
In practice some of the acid will be used up in the lab reaction, so the amounts given by the formulas are considered guidelines and not exact amounts needed. In practice the amount required is usually less.
Add bases to acids slowly with stirring while monitoring heating. Adding bases to acids can cause run away reacitons due to the fast heating which can result in severe burns from exposure to the acids and bases. Check pH after soluiton is allowed to cool and settle. Wear all the proper safety gear when handling acids and bases (apron, goggles, gloves, etc.)
Steve
Some basic molarities of the acids we use are:
(M= moles per liter)
70% HNO3 approximately 16 M
35% HNO3 approximately 8M
32% HCl aproximately 10 M
98% H2SO4 approximately 18M
From the above data you can determine how many moles are needed of the base you are using to neutralize the volume of acid used.
You'll need to write a balanced equation for the acid base reaction to begin with to determine how many moles as the base are require to neutralize the acid used (eg: for the base sodium hydroxide = lye = NaOH):
HNO3 + NaOH --> NaNO3 + H2O ( 1 mole of NaOH is required per mole of Nitric Acid)
HCl + NaOH --> NaCl + H2O ( 1 mole of NaOH is required per mole of hydrochloric acid)
H2SO4 + 2NaOH --> Na2SO4 + 2H2O ( 2 moles of NaOH are required per mole of Sulfuric Acid)
From the above reactions you'll need:
8 moles of NaOH per liter of 35% HNO3 in the waste
10 moles of NaOH per liter of 32% HCl in the waste
36 moles of NaOH per liter of 98% H2SO4 in the waste
Finally determine the weight of NaOH (mass = 40 g per mole) needed:
8 moles x 40g = 320g per liter of 35% HNO3 in the waste
10 moles x 40g = 400g per liter of 32% HCl in the waste
36 moles x 40g = 1440g per liter of 98% H2SO4 in the waste
For other bases be sure the equations are balanced and to used the correct molar mass of the base in the formulas.
In practice some of the acid will be used up in the lab reaction, so the amounts given by the formulas are considered guidelines and not exact amounts needed. In practice the amount required is usually less.
Add bases to acids slowly with stirring while monitoring heating. Adding bases to acids can cause run away reacitons due to the fast heating which can result in severe burns from exposure to the acids and bases. Check pH after soluiton is allowed to cool and settle. Wear all the proper safety gear when handling acids and bases (apron, goggles, gloves, etc.)
Steve
MysticColby
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lye is naoh. But he's using lime, which is usually Ca(OH)2 (doesn''t it usually also contain potassium or magnesium hydroxide?)
something to note: calcium hydroxide, having 2 OH groups, can neutralize twice as much acid per mole as sodium hydroxide. it also weighs more per mole (74.093 g/mol, while sodium hydroxide is 39.9971 g/mol). 1.85 times as heavy... almost cancels out, but not quite. the same weight of Ca(OH2) will neutralize a bit more acid than NaOH.
something to note: calcium hydroxide, having 2 OH groups, can neutralize twice as much acid per mole as sodium hydroxide. it also weighs more per mole (74.093 g/mol, while sodium hydroxide is 39.9971 g/mol). 1.85 times as heavy... almost cancels out, but not quite. the same weight of Ca(OH2) will neutralize a bit more acid than NaOH.
I caught the lime part, I was merely using lye as an example that's why I stated that the reader should substitute the appropriate base in the formulas and equations.
For the record lime is Calcium Oxide which is CaO, slaked lime is calcium hydroxide (Ca(OH2)) or lime plus water.
The equations for lime are the same as lye except the valence and atomic mass:
CaO + 2HCl --> CaCl2 + H2O (divide moles of HCl by two to determine number of moles of lime needed: 5 moles in preceeding example)
CaO + 2HNO3 --> Ca(NO3)2 + H2O (divide moles of HNO3 by two to determine number of moles of lime needed: 8 moles in preceeding example)
CaO + H2SO4 --> CaSO4 + H2O (moles of H2SO4 is equal to the number of moles of lime needed: 18 moles in preceeding example)
I've ignored the water in the reaction as it does not play a role in determining the required amount of base to neutralize the solution.
Lime has a molecular weight of 56 grams per mole.
I also want to add that it helps to know how many moles of the base are in a pound of the base.
Simply divide the number of grams in a pound (454g) by the molecular mass of the base:
NaOH = lye = sodium hydroxide = 40g/mole = 454/40 = 11.35 moles per pound
CaO = lime = calcium oxide = 56g/mole = 454/56 = 8.11 moles per pound
Ca(OH)2 = quicklime = calcium hydroxide = 74g per mole = 454/74 = 6.14 moles per pound
etc.
From this information you can determine the number of pounds of base needed per liter of acid to be neutralized.
Steve
For the record lime is Calcium Oxide which is CaO, slaked lime is calcium hydroxide (Ca(OH2)) or lime plus water.
The equations for lime are the same as lye except the valence and atomic mass:
CaO + 2HCl --> CaCl2 + H2O (divide moles of HCl by two to determine number of moles of lime needed: 5 moles in preceeding example)
CaO + 2HNO3 --> Ca(NO3)2 + H2O (divide moles of HNO3 by two to determine number of moles of lime needed: 8 moles in preceeding example)
CaO + H2SO4 --> CaSO4 + H2O (moles of H2SO4 is equal to the number of moles of lime needed: 18 moles in preceeding example)
I've ignored the water in the reaction as it does not play a role in determining the required amount of base to neutralize the solution.
Lime has a molecular weight of 56 grams per mole.
I also want to add that it helps to know how many moles of the base are in a pound of the base.
Simply divide the number of grams in a pound (454g) by the molecular mass of the base:
NaOH = lye = sodium hydroxide = 40g/mole = 454/40 = 11.35 moles per pound
CaO = lime = calcium oxide = 56g/mole = 454/56 = 8.11 moles per pound
Ca(OH)2 = quicklime = calcium hydroxide = 74g per mole = 454/74 = 6.14 moles per pound
etc.
From this information you can determine the number of pounds of base needed per liter of acid to be neutralized.
Steve
Trouble I see is our waste is a brew of several metals and can also be salts of several acids in solutions, the pH can already have been altered by what we have added to the waste like caustic washes, and so on, I do not think our calculations would mean much (except if we were trying to figure how much we should buy), because I think in use with our waste, the calculations could be very far off as to how much we would need to treat our waste with, too many variables, Ph test papers or a Ph meter would simplify matters.
CuCl2 + CaO --> CaCl2 +CuO
Weight from periodic chart:
Cu--63.5g/mol
Cl--35.4g/mol
(Cl2--2x35.4=70.8g/mol)
Ca--40.0g/mol
O--16g/mol
CuCl2 (63.5g/mol + 70.8g/mol = 134.3g/mol) + CaO (40g/mol + 16g/mol = 56g/mol)
-->
CaCl2 (40g/mol + 70.8g/mol = 110.8g/mol) + CuO (63.5g/mol + 16g/mol = 79.5g/mol)
So not counting water, and considering my math and figuring is right,
It should take 56grams of lime (CaO) to convert 134.3grams of copper II chloride into a copper oxide powder and a solution of calcium chloride salt water.
But my waste would be Iron chlorides or sulfate, and a mixture of base metals and acids in solution, also salts and the water in solution, (or even caustic solutions added) to this waste solution.
In my mind the method of adding reagent and checking Ph seems so much simpler, to me than trying to calculate how much I would need.
Also I will watch for color to leave my salt water solutions, and many times I will raise the Ph somewhat higher then drop it back down to around Ph7, so this uses a little more base.
CuCl2 + CaO --> CaCl2 +CuO
Weight from periodic chart:
Cu--63.5g/mol
Cl--35.4g/mol
(Cl2--2x35.4=70.8g/mol)
Ca--40.0g/mol
O--16g/mol
CuCl2 (63.5g/mol + 70.8g/mol = 134.3g/mol) + CaO (40g/mol + 16g/mol = 56g/mol)
-->
CaCl2 (40g/mol + 70.8g/mol = 110.8g/mol) + CuO (63.5g/mol + 16g/mol = 79.5g/mol)
So not counting water, and considering my math and figuring is right,
It should take 56grams of lime (CaO) to convert 134.3grams of copper II chloride into a copper oxide powder and a solution of calcium chloride salt water.
But my waste would be Iron chlorides or sulfate, and a mixture of base metals and acids in solution, also salts and the water in solution, (or even caustic solutions added) to this waste solution.
In my mind the method of adding reagent and checking Ph seems so much simpler, to me than trying to calculate how much I would need.
Also I will watch for color to leave my salt water solutions, and many times I will raise the Ph somewhat higher then drop it back down to around Ph7, so this uses a little more base.
Thanks for the great info. The science is still a bit beyond me however i will look more into what "moles" are and how they all relate to each other.
I have added lime in various forms to this 25 gallons and i am slowly getting results. I am blowing air into the solution at a pretty good rate to keep it stirred up but I do get the lime settling and clumping on the bottom.
I did have a foam over of the barrel contents one night while it was unattended, however the spill was minor.
I have added 25 pounds of dolomite lime and have only raised the PH to 4. I would add more but there is plenty of free lime in the barrel.
I expected that the process would be quicker with better results. The solution is brown, muddy looking. It smells "sweet" like alkali but is still acid according to ph paper. I will keep at it and try to make sense of it all.
Thanks Mike
I have added lime in various forms to this 25 gallons and i am slowly getting results. I am blowing air into the solution at a pretty good rate to keep it stirred up but I do get the lime settling and clumping on the bottom.
I did have a foam over of the barrel contents one night while it was unattended, however the spill was minor.
I have added 25 pounds of dolomite lime and have only raised the PH to 4. I would add more but there is plenty of free lime in the barrel.
I expected that the process would be quicker with better results. The solution is brown, muddy looking. It smells "sweet" like alkali but is still acid according to ph paper. I will keep at it and try to make sense of it all.
Thanks Mike
Barren Realms 007
In Remembrance
Why not take a small test sample and neutralize it and then calculate the ratios needed to do the job. I don't think you need the air in the the system, just stir it every little while.
I had the same issue with lime not reacting very well, so I switched to lye pellets and those problems went away.
Butcher, you are correct about the side reactions. The calculation will get you in the ballpark, pH paper and patience will get the job done. The point of neutral pH is easy to overshoot so it's important to check the solution frequently and allow the base to fully react before adding more.
Steve
Butcher, you are correct about the side reactions. The calculation will get you in the ballpark, pH paper and patience will get the job done. The point of neutral pH is easy to overshoot so it's important to check the solution frequently and allow the base to fully react before adding more.
Steve
MysticColby
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lazersteve said:
It's because CaO isn't very water soluble (1.19 g/L at RT). Ca(OH)2 and Mg(OH)2 are also not very soluble.
NaOH and KOH, on the other hand, are very water soluble.
not very soluble = takes a while to get into solution and react.
10wt_Percent
Active member
I read somewhere: it's better to add acid to base. Make a slurry of lime or lye (or a mixture of both) in an appropriate plastic bucket. Add waste-acid in modest amounts to this slurry (and stir). Keep adding waste-acid in moderate amounts until resultant solution stops fizzing. Strikes me as a reasonable acid-neutralizing protocol.
heliman4141
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10wt_Percent said:
Im bumping your old post because it has great merit, as I am in the process of neutralizing almost 20 gallons of waste acids that hve been sitting in 5 gal buckets 6 months now. Pre mixing a slurry sounds like a darn good idea, I know lime is a "real bugger" when it contacts water, almost like mixing oil with water. A bit of pure Lye sounds good also, Ill add just a tad.... I have pure lye I picked up so Ill be minimual with it.
I sure do like the search eng. on this website!
Dave
