A buffer of pH = 7.79 is prepared using Tris HCl and NaOH. Moles of Tris HCl = 31.52 g x (1 mole / 157.59 g) = 0.20 moles.
Moles of NaOH = Molarity x volume = 10.0 M x 6.7 mL / 10^3 L = 0.067 moles
Tris – HCl + OH^- -> Tris + H_2O + Cl^-
Setup the reaction table as follows :
Tris – HCl + OH^- -> Tris + H_2O + Cl^-
0.200 0.067 Initial
-0.067 -0.067 +0.067 Change
0.1333 0 0.067 Final
The buffer is diluted to ! L and half the solution is taken. So half the moles of tris – HCl and tris are present.
Moles of Tris – HCl = 0.133 / 2 = 0.0665 moles
Moles of tris = 0/067 / 2 0.0335 moles
To this solution, 0.0150 mol of H+ are added. H+ reacts with tris giving Tris – HCl. So moles of tris – HCl = 0.0665 + 0.0150 = 0.0815 moles
Moles of tris = 0.0335 – 0.015 = 0.018 moles
When all the 0.018 moles of Tris are consumed, the remaining capacity of the buffer is exhausted.
So moles of HCl to be added = 0.018 moles.
Volume of Hcl = moles /molarity = 0.018 moles / 10.0 M = 0.0018 L = 1.8 mL
