Answer to Question #332616 in Biochemistry for Maryam

Question

T= Tris

A= Glacial acetic acid

E= EDTA

B= Boric Acid

10x TAE Recipe

For 1L of 10x solution,

• 48.5 g tris

• 11.4 mL glacial acetic acid

• 20 mL 0.5M EDTA (pH 8.0)

TBE Buffer 10x Stock Recipe

• 108 g tris base

• 55 g boric acid

• 900 ml double-distilled H2O

• 40 ml 0.5 M EDTA solution (pH 8.0)

Adjust volume to 1 L find molarity

Answer 1

$n(T)=48.5/121=0.4 mol; c(T)= 0.4/1=0.4 mol/l$

$m(glacial acetic acid) = v*density=11.4*1.0492=11.96 g, c(glacial acetic acid)= 11.96/(1*60)=0.2 mol/l$

$c(T)=108/(121*1)=0.89 mol/l c(boric acid) = 55/(62*1) = 0.89 mol/l$