Question

Make 500ml of TE containing 10nM Tris and 1nM EDTA

Answer 1

as we know V₁S₁=V₂S₂

V₁ =Initial vol of stock solution needed

S₁= initial concentration of stock solution ( 1 M tris cl, 0.5 M EDTA )

V₂= final volume of desired solution

S₂=  final concentration of desired solution( 10nM tris cl, 1nM EDTA )

in case of tris-cl

V₁S₁=V₂S₂   V₁S₁=V₂S₂

V₁ * 10⁹ nM = 5* 10⁸ nl * 10nM V₁ * 5* 10⁸ nM = 5* 10⁸ nl * 1nM

V₁ = 0.5 nl V₁ = 1 nl

so for 500ml TEcontaining 10nM Tris and 1nM EDTA, one have to add 0.5nl tris (1M), 1nl EDTA (0.5M) in 499.9999985 ml distilled water