Question

You work for a commercial firm that is evaluting drugs tht modulate glucose flow through core metabolic pathways. Enzymes are being evaluated in Tris-HCl buffer, and the lab has a stock solution of 1 M Tris-HCl (pH 8.25) that was made just this morning. You ahve been asked to prepare 300mL of 80mM Tris-HCl "Reaction buffer" at pH 7.85 from the stock solution. Note that the pKa of Tris-HCl at 25 celcius is 8.08 and you have ample supplies of 150 mM HCl at your disposal.

A. What volume of stock Tris-HCl buffer is needed to prepare your Tris-HCl buffer?

B. How many millimoles of the protonated (Tris-H⁺) and unprotonated (Tris) will be in your final reaction buffer?

C. What volume of acid is needed to obtain the target pH of your reaction buffer?

Answer 1

Outline to solve the problem:

From Henderson-Hasselbalch equation, pH = pKa + Log([salt]/[acid])

Given that the pH of stock solution of 1 M Tris-HCl = 8.25

a) The no. of millimoles of Tris-HCl buffer you need to prepare = 300 mL*80*10^-3 mmol/mL = 24 mmol

(Note: 1 M = 1 mmol/mL)

The volume of given stock solution to make 24 mmol of Tris-HCl buffer = 24 mmol / (1 mmol/mL) = 24 mL

pKa of Tris-HCl = 8.08

b) According to Henderson-Hasselbalch equation,

7.85 = 8.08 + Log(n_Tris / n_Tris-HCl), where n = no. of mmol

i.e. Log(n_Tris / n_Tris-HCl) = 7.85 - 8.08 = -0.23

i.e. n_Tris / n_Tris-HCl = 10^-0.23 = 0.589 ........ equation 1

Here, n_Tris + n_Tris-HCl = 24 .......... equation 2

From equations 1 and 2, n_Tris-HCl = 15.105 mmol and n_Tris = 24 - 15.105 = 8.895 mmol

c) The concentration of ample supplies of HCl at your disposal = 150 mM

To make it 24 mmol, the volume of acid required to obtain target pH = 24 mmol/150*10^-3 mmol/mL = 160 mL