Chem202 - for DrBob222

Dr. Bob222:

There is a problem you answered .100M HClO2 titrated with .080M NaOH

I can work the problem many times but I am getting 7.30pH and the answer says 7.38 using quad. I used quad and still get 7.30.

I also do know how you go from the neutralization reaction of HClO2 + NaOH-->???

And then the answer says ClO2 turns into ClO and hydrolized from there?

Can you please assist, thsi problem is driving me nuts.

Thank you and thanks for your assistance.

S

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  1. If possible could you also please show the neutralization reaction and the next reaction that follows where ClO2 is involved.

    I get X = 2.00896X 10-^7? is that correct
    pOH from there 6.70
    pH at 7.30 but again my text is saying 7.38 with quadratic.

    thanks!

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  2. I'm afraid I can't be much help.
    1. I worked this from scratch tonight and I also come out with pOH = 6.70 and pH = 7.30 (with or without the quadratic).

    2. I searched the old files and found a number of posts I had made for this same problem but none of them gave an answer; i.e., I just worked through the hydrolysis etc.

    3. I found one old post that asked for NaOH and chlorous acid(HClO2); however, I worked the problem (again worked meaning just the steps) using HClO and not HClO2. Did that reference to form HClO and then hydrolyzed from there come from my post or from your test?

    4. Where did this problem originate?

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  3. Chemistry 12th Ed. problem 17.48b

    I just wasn't sure either how HClO2 got to ClO2 and which Ka you use?

    There was no ref to form HClO

    minus the issue on 7.38 if you could comment furhter on the neutralization reaction or the following one, Id like to understand that better. Thanks for getting back to me.

    You seem to be the chem guru

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  4. Thanks but I'm no guru; Bob Pursley is the all around guru.
    HClO2 + NaOH ==> NaClO2 + H2O for the neutralization.

    At the equivalence point the pH is determined by the hydrolysis of the
    NaClO2 or Kb for ClO2^-.
    ClO2^- + HOH ==> HClO2 + OH^- is the hydrolysis reaction for which Kb = Kw/Ka and for Ka I used 1.1E-2. The use of different Ka values usually is the reason for numbers not quite agreeing; however, since both of us arrived at the same answer we must have used Ka values that were the same or at least quite similar.

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