Question

You are asked to prepare 1.2 l of a 0.05 M tris buffer at pH= 7.8....

You are asked to prepare 1.2 l of a 0.05 M tris buffer at pH= 7.8. You start with the conjugate base form of tris (121 g/mol). How many grams of tris must you weigh out? How many ml of 6 M HCl (a strong acid) must you add to reach pH=7.8? The pKa for tris is 8.1.

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Answer #1

Tris (T)+ HCl -------------> Tris-Hcl(TH)
PH = PKa + log10 ( [T] / [TH])
7.8 = 8.1 + log (T / 0.05*1.2moles)
-0.3 = log [T] - log (0.05*1.2)
-0.3 - 1.221848 = log [T]
10-1.521818 = [T] = 0.03007128 moles of Tris base
0.03007128 mol * 121 g/mol = 3.638625 gms of Tris base is required

7.8 = 8.1 + log [ (0.03007-X) / X ]
-0.3 = log [ (0.03007-X) / X ]
0.501187 = 10-0.3 = (0.03007-X) / X
0.501187X = 0.03007 - X
1.50118723 X = 0.03007
[X] = 0.02003 mol of HCl is used
20.03 mL for 1 M of HCl
For 6M of HCl = 6 mol/lit * 0.02003 lit = 0.12018 mol
Therefore 120.18 mL for 6M of HCl must be added

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