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Homework Help: Calculating amount of H2SO4 required to reduce pH

  1. Jul 1, 2014 #1
    Hi, I am looking for some guidance for calculating the amount of H2SO4 required to reduce the pH of a Reactor outlet stream from 8.7 to somewhere in the 1.5-2.0 region.

    The stream contains a number of components in varying amounts such as C8H8O2, NaOH, H2O, Na2HPO4, NaH2PO4, C8H12N2O3S and C16H18N2O4S

    I am free to select the concentration of H2SO4 I use.

    I am new to pH calculations and I am wondering how to go about the calculations?

    Is it as straight forward as needing to calculate the amount of H2SO4 required to reduce the pH of each individual component then sum for the total amount used?

    I need to have a material balance for my stream post pH reduction so do I need to consider any reactions that will take place when stating the composition of the steam after acid addition?

    i.e. for the NaOH component:

    H2SO4 + 2NaOH = 2H2O + Na2SO4

    So will the stream post pH reduction will have a new material balance based on the products of any reactions?

    Any pointers in the right direction would be appreciated as I seem to be going around in circles on this one. All the research examples I have read deal with 1:1 mixtures and not pH reduction of a multi component stream.

  2. jcsd
  3. Jul 1, 2014 #2


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    Staff: Mentor

    Yes, that will work.

    Not that it is trivial. Especially with things like C8H8O2 (and two other organics) - do you know their identities, or only formulas?

    What are expected amounts of all substances? How variable they are?
  4. Jul 2, 2014 #3
    Hi Borek. Thanks for getting in touch.

    I know all the identities, mass flowrates, molar flowrates, volumetric flowrates, etc. The volumetric flow data for my inlet stream is as follows:

    C8H8O2, Phenylacetic acid = 72.9 litres/hour
    NaOH, Sodium Hydroxide = 10.3 litres/hour
    H2O, Water = 2836.5 litres/hour
    C16H18N2O4S, Penicillin G = 2.8 litres/hour
    C8H12N2O3S, 6-aminopenicillanic acid = 82.2 litres/hour
    Na2HPO4, Disodium phosphate = 1613.8 litres/hour
    NaH2PO4, Monosodium phosphate = 50.5 litres/hour

    Total flow = 4669.01 litres/hour @ 39degC and pH 8.7

    It is assumed these flows are constant and the aim is to reduce the pH to somewhere in the 1.5 - 2.5 region prior to a removal stage which takes out the C8H8O2 component. The 6-aminopenicillanic acid is the final desired product after a number of other process steps.

    I was playing around with some acid strengths as my main concern is the amount that will be required as I don't want to greatly increase the size of the stream.

    If I break it in to the individual components and say used a 2M concentration of H2SO4 how do I tell what the products of any reaction will be? The NaOH is quite a straight forward one (strong acid/strong base reaction) but some of the others have me confused.

    The Na2HPO4 and NaH2PO4 components have been used as buffers in the reaction section to resist changes in pH so to keep reaction pH at an optimum point. To me it looks like either a lot or a high strength of H2SO4 will be needed to deal with these components but I am unsure of what possible reactions I may get when mixing with these components?

    Any thoughts/suggestions?

  5. Jul 2, 2014 #4


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    Varying amounts? Do you mean varying amounts or various amounts?

    For any calculation we'd need molar flow rates.

    Fr constant flow rates time can be factored out and it is just a calculation about a solution composition.

    To achieve a given pH you don't need four components acid, alkali and two forms of phosphate, you need only two, though one of them should be one of the forms of phosphate.

    I'd imagine that the forms of penicillin have such low concentration as not to be significant in the calculation.

    I wonder what the phenylacetic acid is for.

    We'd really need a more complete explanation of your problem and its constraints.
  6. Jul 2, 2014 #5
    Just to clarify what I am after as I feel I have not been 100% clear in my earlier posts:

    A constant stream is leaving a reactor and consists of the following components with molar, mass and volume flowrates per hour stated:

    Penicillin G (C16H18N2O4S) = 11.82mol/h, 3.96kg/h, 2.8l/h
    6-Aminopenicillanic acid (C8H12N2O3S) = 578mol/h, 125kg/h, 82.2l/h
    Water (H2O) = 157454.9mol/h, 2836.5kg/h, 2836.5l/h
    Sodium Hydroxide (NaOH) = 546.5mol/h, 21.9kg/h, 10.3l/h
    Disodium phosphate (Na2HPO4) = 19086.4mol/h, 2709.5kg/h, 1613.8l/h
    Monosodium phosphate (NaH2PO4) = 589.8mol/h, 70.8kg/h, 50.5l/h
    Phenylacetic acid (C8H8O2) = 578mol/h, 78.7kg/h, 72.9l/h

    Totals = 178845.4 mol/h, 5846.4 kg/h, 4669l/h

    The reactor outlet stream is at 39degC and has a pH of 8.7.

    The pH needs to be reduced to between 1.5 - 2.5.

    The reduction is to be done with H2SO4 at a concentration of my choosing.

    After the pH reduction the reactor stream flows on to the phenylacetic acid removal section so a mass flow and composition of the stream post H2SO4 addition is required.

    The flowrates leaving the reactor are constant so what I require is the flowrate of H2SO4 per unit time that is required for the pH reduction.

    My questions are can each component be dealt with individually then the H2SO4 amount be totalised from this? (Borek has already hinted that this is an acceptable method).

    My second issue is working out what the individual components will be in the stream after pH reduction as will there be a number of reactions with the H2SO4 that lead to new components being formed?

    The bulk of the stream is H2O and Na2HPO4.

    Any pointers of how I set about this would be gratefully received.
  7. Jul 2, 2014 #6


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    You can't have these compounds in the solution at the same time. They will react producing disodium phosphate. You can assume reaction went to completion.

    As your target pH range is quite wide, it should be enough to calculate amount of sulfuric acid required to acidify solution the disodium phosphate, everything else is in negligible quantities.

    Somehow I find it quite unlikely that you know composition and flowrates with such a high accuracy.
  8. Jul 2, 2014 #7
    The feed stream has been determined by another student and it is what I have been given to deal with so I will have to adjust their results for this. Thanks for pointing that out.

    I am guessing the reaction will give:

    NaOH + NaH2PO4 = Na2HPO4 + H2O

    So my material balance will alter based on this. The Na2HPO4 is then the main issue to deal with if I assume the effects of the Penicillin G, 6-penicillanic acid and phenylacetic acid are negligible given they are all present in small amounts (all less than 2wt%) in the stream.

    This would simplify the situation to a H2O/Na2HPO4 mixture being treated with H2SO4. Is Na2HPO4 classed as a weak or strong base?
  9. Jul 2, 2014 #8


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    Staff: Mentor

    A weak one.
  10. Jul 2, 2014 #9
    Ok thanks. The one area I am still concerned about is what the reaction products are from adding H2SO4 to Na2HPO4?

    H2SO4 + Na2HPO4 = Na2SO4 + H3PO4......???
  11. Jul 2, 2014 #10


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  12. Jul 3, 2014 #11


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    OK I've guessed you are phenylacetylating penicillin G in a reactor with immobilised enzyme which explains why you are doing it at pH 8.7.

    The way you are talking about it does not give confidence you understand buffered solutions and I agree with Borek you need to mug up on them, also to understand your colleague's calculations (and whether he does himself).

    But in this case you can get away with a fairly simple robust calculation I will explain, and I will mention some niggles.

    Niggle 1 why is the product, penicillin G = phenylacetyl penicillin there? presumably as part of a process that involves recycling of product of incomplete reaction?

    Niggle 2. It seems to me you have two more feeds than necessary. I do not know for sure what the NaOH is about. It is being supplied in mole amounts almost equal to that of the amino penicillinanic acid. It is not stated in what form this is given you - I presume neutral, zwitterionic. Perhaps the idea of the NaOH is to bring that to pH 8.7? The alkali is 95% of the APA; it ought to be very close to 100% though - because the pK of APA is 4.9, nearly 4 units below the pH. A scarcely important quantitative difference for practical purposes here but I wonder what the thinking was.

    The phosphate is there as a pH-stabilising buffer.. I calculate that the amounts of the two phosphates are indeed such as to give you pH 8.7.

    Niggle 3: you do not need to mix these two phosphates to achieve that. I would expect an engineer to just use the cheapest of the two and add more NaOH into the mix than you were doing anyway if NaH2PO4 is the cheaper. Or use NaHPO4and acid.

    You do not say why you need to get to pH 1.5 to 2.5 at the end. I'm going to leave out of it the small amounts of peniclllinates and NaOH. In your phosphate solution the ions are going to be present in essentially the same molarities as the salts you put, so 19086.4 moles/h of HPO42- are going through. With that same amount 19086.4 equivalents/h, i.e. 9543.2 moles/h H2SO4 you bring all the phosphate all to the form H2PO4-. The pK of phosphoric acid, is 2.15. That's about the pH you want. When the pH = pK the phosphate is half in the form H2PO4- and half H3PO4. So you just need to add another 9543.2/2 = 4771.6 mol/h H2SO4 and you will be at pH 2.15 and have a comfortable margiin to add getting on for as much again without going below your required 1.5. (I wonder whether it is the best to be adding the acid as a flow as other things can be imagined.)

    You might need to mug up buffers etc, to follow some of that; it is not very difficult but a bit somehow elusive so that the most common questions here are about them.
    Last edited: Jul 4, 2014
  13. Jul 4, 2014 #12


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    He wants pH in the 1.5-2.0 range, not 1.5-2.5 range. I agree for the later pH=pKa would be good enough.
  14. Jul 4, 2014 #13


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    Oops figures lost in translation between screens. There is another small error near the end too. After we have added the 19086.4 equivalents of acid protonating the HPO42- we end up with 19086.4 + 589.8 = 19676.2 moles of H2PO4-. We need to further add half that number of equivalents of acid (half again that number of moles of H2SO4) to to bring to pH = pK.

    To bring to some other pH you need to use

    [H3PO4]/[H2PO4-] = [H+]/Ka1 = 10-pH/10-pKa1 = 10(pKa1 - pH)

    where [H3PO4] + [H2PO4] = 19676.2

    (In a manner of speaking - molar ratios = mole ratios = mole/h ratios.)

    [H3PO4] moles equal these extra equivalents of acid you need to add to bring to pH.

    Altogether 19000 X 1.75/2 moles/h H2SO4 should do what's needed I guess.
    Last edited: Jul 5, 2014
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