Question

you need 250ml of a 100 mM of tris buffer at pH of 8.25. the molecular weight of tris is 121.1 and its pKa is 8.06.

A) how much tris do you need to weight out?

B) you have 10 M HCL soulution, how much will you add to get the right pH?

Answer 1

Pka of Tris base = 8.06

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

pH = pka - log(Tris-HCl/Tris base))

total buffer = V*M = 0.25*0.1 = 0.025 mol

   Tris-HCl+Tris base = buffer

8.25 = 8.06-log(x/(0.025-x))

x = 0.0098

No of mol of Tris-HCl = x = 0.0098 mol

A) No of mol of Tris base required = 0.025-0.0098 = 0.0152 mol

amount of Tris base required = 0.0152*121.1 = 1.84 g

B) No of mol of HCl required = Tris-HCl = 0.0098 mol

volume of HCl required = n/M = 0.0098/10 = 0.00098 L

                             = 0.98 ml