Question

Tris- HCl buffer is often used as a buffer for biochemical reactions in pH range from 7.1 to 9.1. It is made by dissolving in water Tris (hydroxyl methyl) amino methane, a weak base, and then adding HCl. The conjugate acid-base pair for Tris is (HOCH2)3 C-NH3 + (here TRISH+ ) and (HOCH2)3 C-NH2 (here TRIS base): its equilibrium is (HOCH2)3 C-NH3 + ↔ (HOCH2)3 C-NH2 + H+ pKa= 8.08 (at 25 o C) Assume you have to prepare 200 mL of the buffer. To do so you take 100 mL of 0.1 M Tris base, add HCl and then adjust the volume with pure water. How many mL of 0.1 M HCl you have to add to it in order to have the final pH of pH = 8.0?

Answer 1

Answer:

You need to add 46 mL of 0,1 M HCl and 54 mL of water.

Explanation:

First, it is necessary to use Henderson-Hasselbalch equation:

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

8,0 = 8,08 + log₁₀ $\frac{[TRIS base]}{[TRISH^+]}$

0,83 = $\frac{[TRIS base]}{[TRISH^+]}$ (1)

If we will to prepare 200 mL, the total buffer concentration must be 0,05 M, thus:

0,05 M = [TRIS base] + [TRISH⁺] (2)

Replacing (2) in (1):

[TRISH⁺] = 0,027 M

And:

[TRIS base] ) 0,023 M

The equilibirum is:

TRISH⁺ ⇄ TRIS base + HCl   ka = 8,32x10⁻⁹

The equlibrium concentrations of TRIS base and HCl are:

[TRIS base] = 0,05 M - x Where x must be 0,023 M, The concentration of TRISH⁺

[HCl] = Y - x Where Y is initial concentration of HCl.

The equilibrium concentration of HCl is the same pH, thus:

[HCl] = 10⁻⁸ M.

Thus, Y is:

0,023 M +  10⁻⁸ M ≅ 0,023M

Which comes from:

0,023 M × 0,200 L = 4,6x10⁻³ mol ÷ 0,1 M = 0,046 L ≡ 46mL

Thus, you need to add from the 100 mL of 0,1M Tris base, 46 mL of 0,1 M HCl and 54 mL of water to complete 200 mL

I hope it helps!