Question

Describe how to prepare for this desired Tris buffer:

500 ml of 2 M Tris-HCl buffer with a pH of 9.2 . The molecular weight of Tris base is 121.14 g/mol.

Answer 1

Pkb of Tris base = 5.924

pH = 14 - (pkb + log(Tris-HCl/Tris base))

no of mol of tris buffer = V*M/1000

                       = 500*2/1000

           = 1 mol

   tris buffer = Tris base + Tris-HCl(x)

    1 = 1-x + x

9.2 = 14- (5.924+log(x/(1-x))

   x = 0.07

no of mol of tris base must taken = 1 mol

no of mol of HCl must add to tris base to form Tris-HCl

                = x = 0.07 mol

amount tris base must taken = n*mwt = 1*121.14 = 121.14 g

if 1 M standard HCl is availbale,

volume of HCl required = n/M = 0.07/1 = 0.07 L

                          = 70 ml

take 121.14 g of tris base dissolve in limited water and add 70 ml of 1 M HCl solution, then add water up to 500 ml solution.