Question

How would you make 2 liters of a 0.5 M Tris buffer, pH 7.6?

MW of Tris, C4H11NO3: 121.1

Answer 1

The molar relationship between the components of the buffer is calculated:

n Salt / n Acid = 10 ^ (pH - pKa) = 10 ^ (7.6 - 8.1) = 0.3

It has:

i) n Salt - 0.3 * n Acid = 0

ii) n Salt + n Acid = M * V = 0.5 M * 2 L = 1 mol

System of equations is applied and you have:

n Salt = 0.23 mol

n Acid = 0.77 mol

The mass of Tris buffer required, and the volume of 1 M HCl are calculated:

m Tris = M * V * MM = 0.5 M * 2 L * 121.1 g / mol = 121.1 g

V HCl = n Acid * 1000 / M = 0.77 * 1000/1 M = 770 mL

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