Question

Buffers and pH

1.(12 pts) A stock solution of Tris -HCl buffer sits on the shelf (0.6 M, pH 8.35). You remove 80 mL of the stock solution, add 12 mL of 0.75M HCl and dilute to 1 L. The pKa of Tris-HCl is 8.20.

(a) How many millimoles of both forms of Tris -HCl are present after you added the acid?

(b) What is the final buffer concentration in mM?

(c)What is the final pH of your buffer?

Can you show all work to confirm the answers below:

(a)U = 19.1 mmoles, P = 28.9

(b)48mM

(c)pH = 8.02

Answer 1

The concentration of stock solution of Tris-HCl buffer = 0.6 M = 0.6 mol/L = 0.6 mmol/mL, pH = 8.35

The volume of stock solution removed from the buffer = 80 mL

The no. of mmol of Tris-HCl buffer removed = 0.6 mmol/mL * 80 mL = 48 mmol

No. of mmol of HCl added to the buffer = 0.75 mmol/mL * 12 mL = 9 mmol

The total volume made by distilled water = 1 L

pKa of Tris-HCl = 8.2

(a) The total no. of mmol of Tris-HCl buffer = 48 mmol

(b) The concentration of the final buffer = total no. of mmol of buffer / total volume = (19.1 + 28.9) mmol / 1 L = 48 mM

(Note: 1 mol/ 1L = 1 M, i.e. 1 mmol / 1L = 1 mM)

(c) According to Henderson-Hasselbalch equation,

pH = pKa + Log(U/P), where P = protonated form, U = unreacted form

i.e. pH = 8.2 + Log(19.1/28.9)

i.e. pH = 8.2 - 0.18

i.e. pH = 8.02