EDTA Disodium Salt 0.0575M Solution *

$56.00

Select size:
1 Liter
1 Liter
4 Liters
Cases
Sale price $56.00
Sale price $202.00
Sale price $126.00
*Custom Product may require additional time to process.
For questions regarding lead time, please contact a member of our Customer Care Team at customercare@laballey.com

About EDTA Disodium Salt 0.0575M Solution

EDTA, Disodium salt, also known as Ethylenediaminetetraacetic Acid, Disodium Salt Dihydrate has the chemical formula C10H14N2Na2O8. It appears as a white odorless crystalline powder. It is soluble in water and reactive with oxidizing agents. It is a chelating agent that sequesters various polyvalent cations like Calcium and its solution appears as a colorless liquid. Lab Alley’s Disodium EDTA, 0.0575M Solution contains 8-10% by weight of Disodium EDTA powder dissolved in a balanced quantity of distilled Water. Chemically pure or Laboratory reagents are two terms often used to describe Lab Grade chemicals. Lab Grade chemicals do not meet any accepted quality or purity requirements such as the ACS Grade, the USP Grade, and the FCC Grade, despite their acceptable purity. Due to its high purity Disodium EDTA, 0.0575M, Lab Grade, is highly recommended for labs and commercial applications.

Common Uses and Applications

  • Chelating Agent
  • Stabilizing agent
  • Ion exchange agent

Industries

  • Laboratories testing
  • Pharmaceuticals
  • Manufacturing

Safety and Shipping Information

Please contact us to request a Safety Data Sheet (SDS) and Certificate of Analysis (COA) for EDTA, Disodium Salt, 0.0575M.

One-Time Purchase

Liquid

Chemistry Table

6381-92-6
C10H14N2Na2O8
336.21
Edetate Eisodium, Disodium Edetate, EDTA Disodium, Titriplex III, Disodium EDTA, Disodium Versene, Endrate Disodium, Sodium Versenate
8759
Disodium;2-[2-[bis(carboxylatomethyl)azaniumyl]ethyl-(carboxylatomethyl)azaniumyl]acetate
Product Manuals

Have a question?

Here are the most commonly asked.

Have a question?


Questions & Answers

Have a Question?

Be the first to ask a question about this.

Ask a Chemistry Question