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  1. #1

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    Sulfite, Bisulfite, the chicken or the egg

    According to Dr. Chapman (who used to write about photochemistry in Photo Techniques), a plain aqueous solution of sodium sulfite is alkaline, a solution of sodium bisulfite is approximately neutral (I thought it was mildly acidic but anyway), and a solution of sodium metabisulfite is mildly acidic. Ok so far.

    Now, mix things together. Chapman says regardless of whether the initial solute was sodium sulfite, bisulfite, or metabisulfite, the final solution alkalinity or acidity determines which ion predominates. So for example, even if the initial compound dissolved in water was sodium metabisulfite, if you then add sodium hydroxide to make the solution alkaline, the sulfite ion predominates, not the metabisulfite ion. Similarly if you started with a solution of sodium sulfite but then added some acetic acid to make the solution neutral, the bisulfite ion predominates. Etc. Is this broadly correct?
    Last edited by Michael R 1974; 03-15-2014 at 08:56 PM. Click to view previous post history. Reason: Correction

  2. #2

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    Egg on my face

    All I know is "1/250 sec., f11 and develop the devil out of the neg."

  3. #3

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    The pH of a solution of sodium bisulfite can vary from a low of 3.8 upwards depending on the concentration. Sodium metabisulfite can be thought of as a container for sulfur dioxide.

    Na2SO3 + SO2 --> Na2S2O5

    The term bisulfite is an older name for what is now called hydrogen sulfite as this better shows that the compound is acidic in nature. The hydrogen sulfite ion dissociates into hydrogen ions and sulfite ions according to the equalibrium equation

    [H+] [SO3-] / [HSO3-] = ka where the terms within brackets indicate concentrations and k is a constant.

    Adding an acid to a solution of a hydrogen sulfite decreases the concentration of sulfite ions and increases the concentration of hydrogen sulfite ions as can be seen from the above equation.
    Last edited by Gerald C Koch; 03-15-2014 at 09:27 PM. Click to view previous post history.
    A rock pile ceases to be a rock pile the moment a single man contemplates it, bearing within him the image of a cathedral.

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  4. #4

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    Suppose one has a sodium sulfite solution of known concentration w/v (pH~9) and then an amount of sodium bisulfite is added such that the pH is brought to 7, does the concentration of sulfite ion SO3 increase or decrease?

  5. #5

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    The following equation explains why sulfite solutions are slightly alkaline.

    SO3-2 + H2O <--> HSO3- + OH-

    When you add a solution of bisulfite you remove some of the hydroxyl ions converting them back to sulfite ions so that the sulfite ion concentration increases and the pH decreases.
    A rock pile ceases to be a rock pile the moment a single man contemplates it, bearing within him the image of a cathedral.

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  6. #6

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    On looking back on my previous post the last sentence is a bit muddled with a clause missing. So much for doing things late at night. It is clearer to say that when you add bisulfite ions you drive the reaction to the left thereby reducing the hydroxyl ion concentration and increasing the sulfite ion concentration. The net effect is to lower the pH of the solution.
    A rock pile ceases to be a rock pile the moment a single man contemplates it, bearing within him the image of a cathedral.

    ~Antoine de Saint-Exupery

  7. #7
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    The bottom line is that you can use any of the compounds above (ending in Sulfite) as long as you use the molar equivalent of Sulfur and by adjusting the pH to where you want, things turn out the same!

    That is why I generally use Sodium Sulfite for everything and just adjust pH.

    PE

  8. #8

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    I'm looking at those reaction equations and still having a chicken/egg problem. Bear with me, I'm still having trouble (yes I probably need to go back to basic chem). I think what's troubling me is the extra level of complexity (I think) regarding the use of bisulfite and metabisulfite (disulfite?) as an acid to adjust pH, as opposed to say acetic acid.

    Let's use a practical example, which might help illustrate what I'm trying to get at, and why I am confused by what Chapman is saying (which is in line with PE's last note).

    D-23 contains 100g sodium sulfite and 7.5g Metol per liter. The pH is probably somewhere between 8 and 8.5. Henn added 15g/l sodium bisulfite to D-23, which reduced the pH to ~7 and this is D-25. The idea, as explained in any photochem text, is that since the lower pH leads to longer development times, there is more solvency.

    The problem is, this leaves the layman with the impression all else is unchanged - ie the only difference between D-23 and D-25 is the pH. Said another way, there is still the same concentration of SO3 (from the 100g of sodium sulfite) and all the sodium bisulfite has done is act as an acid. But is this true? Or does the bisulfite ion create acidity by dissociating to H and SO3, resulting in more sulfite being present? This would mean the greater solvency of D-25 relative to D-23 is due not only to the lower activity/longer development time, but also due to an actual increase in sulfite concentration.

  9. #9
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    Adding Sodium Metabisulfite to D-23 gives you both more acid and more Sulfite ion, but the extra Sulfite does not change solvent strength by much (remember the Altmann/Henn paper you discussed here a while back). It does give you a bit stronger buffering than an addition of e.g. Sulfuric Acid or Acetic Acid.
    Trying to be the best of whatever I am, even if what I am is no good.

  10. #10

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    This has been somewhat of a revelation to me. In my opinion it is not properly addressed in the texts I've read.

    Is it the dissociation of HSO3 into H+ and SO3- which lowers the pH? Is all the bisulfite converted to sulfite? Or is there still some HSO3 at a pH of 7? If some sodium hydroxide was added to raise the pH, what would happen? What if some sulfuric acid was added to lower the pH to an acidic condition? What happens to the concentrations of SO3 and HSO3?

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