Question

Suppose you want to make 500 mL of a 0.20M Tris buffer at pH 8.0. On the shelf in lab, you spot a bottle of 1.0M Tris at pH 6.8 and realize you can start with that to make his new buffer. Assuming you have 5.0M HCl and 5.0M NaOH at your disposal, how could you make this 0.20M Tris buffer at pH 8.0 from the 1.0M Tris, pH 6.8? (The pKa of Tris is 8.06.)

Answer 1

Given information:

• pK_a of Tris = 8.06
• Desired pH = 8.00

The Henderson Hasselbalch equation is used to determine the molar ratio of the basic and acidic forms of Tris at pH = 8.00. The equation is given as follows:

The total concentration of Tris in the buffer at pH = 8.00 is given as 0.20 M.

Thus,

Hence, the concentration of Tris base is calculated as follows:

The buffer is prepared by adding 5.0 M HCl with 1.0 M Tris. Hydrochloric acid is added until the pH has to be lowered than the pK_a of Tris. Thus, the total volume of the buffer is 500 mL.

The millimoles of Tris acid and Tris base is calculated as follows:

The reaction of Tris base and HCl is given as follows:

According to reaction, 1 mole of Tris base reacts with 1 mole of HCl to form 1 mole of Tris acid. Hence, the millimoles of HCl is 46.5 millimoles.

Moles of HCl = 46.5/1000 mol

= 0.0465 mol

The volume of 5.00 M HCl =  0.0465 mol / 5.0 M

= 0.0093 L = 9.3 mL

The total millimoles Tris base after the addition of HCl is 46.5 + 53.5 = 100 millimoles = 0.100 mol

The volume of 1.0 M tris = 0.100 mol / 1.0 M

= 0.100 L

= 100 mL