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^-
<span>0.200 0.067 Initial</span>
<span>-0.067 -0.067 +0.067 Change</span>
<span>0.1333 0 0.067 Final</span>
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.
<span>Volume of Hcl = moles
/molarity = 0.018 moles / 10.0 M = 0.0018 L = 1.8 mL</span>
