Percent active chlorine
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2007-12-18T03:24:17Z
Alchie1
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'''Percent active chlorine''' is a unit of [[concentration]] used for [[Hypochlorous acid|hypochlorite]]-based [[bleach]]es. One gram of a 100% active chlorine bleach has the same bleaching power as one gram of [[chlorine]]. The term "active chlorine" is used because most commercial bleaches contain substantial amounts of chlorine in the form of [[chloride]] ions, which have no bleaching properties.
Liquid bleaches sold for domestic use are typically 3–10% active chlorine, and should be diluted to 1–2% active chlorine before use. Commercial domestic bleaching powder is typically about 40% active chlorine. Values for some industrial solid bleaches are given in the table below.
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|-
! Formula
! Active chlorine
|-
| [[Chloride|Cl<sub>2</sub>]]
| 100,0%
|-
| [[Sodium hypochlorite|NaClO·2½H<sub>2</sub>O]]
| 59,3%
|-
| [[Calcium hypochlorite|Ca(ClO)<sub>2</sub>]]
| 99,2%
|-
| [[Lithium hypochlorite|LiClO]]
| 121,4%
|-
|}
Values can be higher than 100% because [[hypochlorite]] ion have higher oxidation state (+1) than [[chlorine]](0) and so high oxidation power .</br>
Percent active chlorine values have now virtually replaced the older system of [[chlorometric degree]]s: 1% active chlorine is equivalent to 3.16 °Chl. Taking the (reasonable) assumption that all active chlorine present in a liquid bleach is in the form of hypochlorite ions, 1% active chlorine is equivalent to 0.141 mol/Kg ClO<sup>−</sup>(0.141 mol/L if we assume density=1). For a solid bleach, 100% active chlorine is equivalent to 14.1 mol/kg ClO<sup>−</sup>: [[lithium hypochlorite]] has a [[molar mass]] of 58.39 g/mol, equivalent to 17.1 mol/kg or 121% active chlorine.
Active chlorine values are usually determined by adding an excess of [[potassium iodide]] to a sample of bleach solution and [[Titration|titrating]] the [[iodine]] liberated with standard [[sodium thiosulfate]] solution.
::Cl<sub>2</sub> + 2I<sup>−</sup> → I<sub>2</sub> + 2Cl<sup>−</sup>
or
::ClO<sup>−</sup> + 2I<sup>−</sup> + 2 H<sup>+</sup> → I<sub>2</sub> + H<sub>2</sub>O + Cl<sup>−</sup>
then
::2S<sub>2</sub>O<sub>3</sub><sup>2−</sup> + I<sub>2</sub> → S<sub>4</sub>O<sub>6</sub><sup>2−</sup> + 2I<sup>−</sup>
From the above equations it can be seen that 2 mole of thiosulfate is equivalent to 70.9 grams of active chlorine.
[[Category:Bleaches]]