Question

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

Answer 1

pK_a of Tris = 8.06

Desired pH = 8.00

Use the Henderson-Hasslebach equation to determine the molar ratio of the basic and acidic forms of Tris at pH = 8.00.

pH = pK_a + log [Tris Base]/[Tris Acid]

======> 8.00 = 8.06 + log [Tris Base]/[Tris Acid]

======> -0.06 = log [Tris Base]/[Tris Acid]

======> [Tris Base]/[Tris Acid] = antilog (-0.06) = 0.87

======> [Tris Base] = 0.87*[Tris acid] …….(1)

Again, the total concentration of Tris in the buffer at pH = 8.00 is 0.20 M. Therefore,

[Tris Base] + [Tris Acid] = 0.20 M

======> 0.87*[Tris Acid] + [Tris Acid] = 0.20 M.

======> 1.87*[Tris Acid] = (0.20 M)/(1.87)

======> 1.87*[Tris Acid] = 0.17 M.

Therefore, [Tris Base] = 0.20 M – [Tris Acid]

= 0.20 M – 0.17 M

= 0.03 M.

The buffer is prepared by adding 5.0 M HCl TO 1.0 M Tris. HCl is added since the pH has to be lowered than the pK_a of Tris.

Total volume of the buffer = 500 mL.

Millimoles Tris Acid = (500 mL)*(0.17 M) = 85.0 mmol.

Millimoles Tris Base = (500 mL)*(0.03 M) = 15.0 mmol.

Tris acid is formed from Tris Base by adding HCl as per the reaction below.

Tris Base (aq) + HCl (aq) ----------> Tris Acid (aq) + Cl^- (aq)

As per the reaction above,

Millimoles Tris base neutralized = millimoles HCl added = millimoles Tris acid formed = 85.0 mmol.

Therefore,

mL of 5.0 M HCl added = (85.0 mmol)/(5.0 M) = 17.0 mL.

Total millimoles Tris base before addition of HCl = (85.0 + 15.0) mmol = 100.0 mmol.

Volume of 1.0 M Tris taken = (100.0 mmol)/(1.0 M)

= 100.0 mL (ans).