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Quantity of Sanitizer required for Sanitizing Surfaces.
Greetings to all the forum members.
My query today is related to calculation of quantity required for sanitizing surfaces using 70% (v/v) Ethanol based sanitizer. High pressure sprayers are used for surface sanitization. Please provide the basis of calculation, for eg: how much sanitizer is required for an area of 100 square meters. Any supporting document for the above calculation is highly appreciated.
Your wise response is eagerly awaited.
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4 Replies
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Seems like a bit of a waste of Ethanol. Here’s a list of suggestions for surface disinfection: https://www.canada.ca/en/health-canada/services/drugs-health-products/disinfectants/covid-19/list.html#tbl1
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According to CDC’s website, Benzalkonium Chloride is not an effective agent against SARS COVID-19. Most of the products mentioned on the provided link are based on Benzalkonium Chloride.
Secondly, in case of Sodium Hypochlorite and Hydrogen Peroxide, what concentration is to be used for formulating the final sprayable product ?
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MurtazaHakim said:According to CDC’s website, Benzalkonium Chloride is not an effective agent against SARS COVID-19. Most of the products mentioned on the provided link are based on Benzalkonium Chloride.
Secondly, in case of Sodium Hypochlorite and Hydrogen Peroxide, what concentration is to be used for formulating the final sprayable product ?
A registered biocide product we bought suggested a concentration of 1,25% active chlorine. Most likely the concentration is suggested based on tests.
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The availability of 12% NaOCl is not an issue for us. This 12% grade has available chlorine as 140-190 gpl. The density of this 12% grade of NaOCl is 1.22-1.27 kg/lit. The calculation would then be
Available Chlorine =140 gm/1 litre = 0.140 kg/ 1.22 kg = 0.11475 kg/kg = 11.4754 % (w/w).
We need to dilute this 11.47 % to 1.25 %.
The dilution would then be calculated as 11.47/1.25 = 9.176.
Dividing 1000 kg into 9.176+1 = 10.176 parts, we get as follows
For a 1000 kg batch we would then have to mix 901.73 kg (901.73 lit.) of water and 98.27 kg (80.55 lit.) of this NaOCl 12% grade. The final mixture would then have the density of 1000 kg/ 982.28 lit. = 1.018 kg/lit.
Are the above calculations correct ???
Secondly, how to calculate the surface coverage requirement for disinfection application ???
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