How much water given 1M HCl and 1M Tris (pKa = 8.1) would you need to combine to make 100mL of a Tris-HCl buffer at a pH of 7.2?
The molar ratio of the buffer components is calculated:
n Salt / n Acid = 10 ^ (pH - pKa) = 10 ^ (7.2 - 8.1) = 0.125
It has:
i) n Salt - 0.125 * n Acid = 0
ii) n Salt + n Acid = 0.1 M * 0.1 L = 0.01 mol
System of equations is applied and you have:
n Salt = 0.0011 mol
n Acid = 0.0089 mol
The volume of Tris, HCl and water required to prepare the buffer is calculated:
V Tris = n Salt * 1000 / M = 0.0011 * 1000/1 M = 1.1 mL
V HCl = 0.0089 * 1000/1 = 8.9 mL
V H2O = 100 - 1.1 - 8.9 = 90 mL
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