The overall reaction for the reaction between NaClO and HCl is: $\ce{NaClO + 2HCl -> NaCl + Cl2 + H2O}$

To try and understand how this reaction occurs, I've thought of a simple mechanism for it. However, I don't even understand my own mechanism. Here's what I proposed:

  • $\ce{NaClO + HCl -> NaCl + HClO}$
  • $\ce{HClO + HCl -> Cl2 + H2O}$

The mechanism looks like it could work, but whether it would work is another question.

One problem I have with understanding it is the second step: how exactly would that work? My first thoughts were that since HCl is the stronger acid out of the two, H2ClO+ and Cl- would be produced in the second step, and a third step would have something happen between both of these products to finally produce H2O and Cl2. However, I scrapped that thought since surely an ionic compound would then be produced and nothing would happen afterwards.

Now, I'm lost on what would really happen in the second step.

Maybe I've got it all wrong, but does anyone have any other thoughts on this? Or are there any actual proposed mechanisms for this out there?

  • 3
    $\begingroup$ For step 2, write HClO as HOCl. HOCl is a weak acid (pKa=7.5) that can be protonated on oxygen by HCl to give H2OCl+ (+ charge on oxygen) and Cl-. Have Cl- displace H2O from H2OCl+ to afford Cl2 (think SN2 rxn.) $\endgroup$ – user55119 Jul 23 '18 at 19:11
  • $\begingroup$ @user55119 That's what I was thinking. After you said SN2 reactions, my thoughts are that a lone pair on the oxygen atom accepts a proton from HCl. The Cl- ion then attacks the chlorine atom in H2ClO+, hence Cl2 and H2O are formed. $\endgroup$ – meadz7789 Jul 23 '18 at 20:04

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