BioArmor Launches Benzalkonium Chloride-based Hand Sanitizer

Bioarmor Hand Sanitizer

BioArmor has developed a hand sanitizer that contains benzalkonium chloride, one of the few ingredients reportedly eligible for evaluation under the U.S. Food and Drug Administration's (FDA's) OTC Drug Review for use in consumer antiseptic rub products.

Related: U.S. FDA Hand Sanitizer Solutions Explained

In an FDA statement, benzalkonium chloride is described as being as effective as alcohol-based hand sanitizers. The ingredient has been used as an antiseptic for germ-killing, to prevent infections and as a first-aid treatment to keep skin healthy.

The ingredient kills germs by disrupting the outer cell membranes, which results in a physical kill (lysing) of the microorganism. This physical kill ensures microorganisms do not adapt or become resistant to the antimicrobial product.

According to the company, benzalkonium chloride is preferred over alcohol-based hand sanitizers since formulations that contain > 60% alcohol are known to dry out the hands and destroy not just harmful germs, but also skin's natural flora.

Previously: What is Skin Care-based Hand Sanitizer All About?

Killing the “good" germs also means surfaces become susceptible to issues such as infections and skin damage, and when alcohol-based hand sanitizer has dried or evaporated from the skin, it leaves hands highly susceptible to re-contamination and/or infection.

Based on benzalkonium chloride, BioArmor Silky Gloves hand sanitizer is promoted as containing no toxic chemicals and is comprised of a natural formula to safely protect the consumer and the environment from the spread of harmful germs, bacteria and micro-organism growth. Furthermore, the product includes three additives for improved skin protection, moisturization and skin-soothing benefits; i.e., aloe, vitamin B5 and blueberry extract.

Benzalkonium chloride is reportedly non-toxic, non-flammable and safe. Formulations regulated by the FDA require benzalkonium chloride, on average, at a level of 0.13%.

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