0
$\begingroup$

Our water treatment facility used Chlorine Dioxide to disinfect the produced drinking water. The analysis results indicate concentrations of 0.2 mg/l of chlorine dioxide at the source. Can we analys residual chlorine in the distribution network? We dont have water quality standards for chlorine dioxide, only residual chlorine. Can I calculate the concentration of chlorine in the water from the concentration of chlorine dioxide?

$\endgroup$
0
$\begingroup$

Some chemistry, first add H2O2 and NaOH:

$\ce{2 ClO2 + H2O2 + 2 NaOH -> 2 NaClO2 + O2 (g) + 2 H2O }$

Source: https://chemiday.com/en/reaction/3-1-0-1114

Continue adding H2O2, as:

$\ce{2 NaClO2 + 2 H2O2 -> 2 NaClO + 2 O2 (g) + 2 H2O }$

$\ce{2 NaClO + 2 H2O2 -> 2 NaCl + 2 O2 (g) + 2 H2O }$

which implies a net reaction of:

$\ce{2 ClO2 + 5 H2O2 + 2 NaOH -> 2 NaCl + 5 O2 (g) + 6 H2O }$

So, slowly add H2O2 until the bubbling ceases. Collect the volume of oxygen and convert to moles. Divide by 2.5 to determine moles of ClO2 and your chlorine therefrom.

Here is a related reference, Titrimetric and photometric methods for determination of hypochlorite in commercial bleaches, which cites the use of hydrogen peroxide and acidic permanganate.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Please have a look at Is ChemiDay a reliable enough source (for inorganic reactions) to be cited on our site? $\endgroup$ – andselisk Apr 8 at 21:48
  • $\begingroup$ The action of H2O2 on NaOCl has to be one of the better-known reactions. The reaction mechanics is actually more complex than many are aware, but the final products are not in dispute. $\endgroup$ – AJKOER Apr 10 at 0:20
  • $\begingroup$ Also, do not employ concentrated H2O2, dilute solutions are safer especially in forming normal dioxygen (not singlet oxygen). $\endgroup$ – AJKOER Apr 10 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.