Question

You need to make a make 250 mL of a 50 mM acetate buffer pH 4.2, containing 15 mM dithiothreitol. Acetate has a pKa of 4.76. You have at your disposal a 4 liter bottle of acetic acid (17.4 M), 500 g bottle of sodium acetate (FW 82.03) a 1.2 M stock solution of dithiothretiol. Show how to make up this buffer.

Please show steps if at all possible. Thank you.

Answer 1

Here NaAc is Sodium acetate

Step 1.

Calculate the mass of NaAc needed for 250 ml of a 50 mM solution

Now,

Molarity =50mM =50Χ10^-3 M

Mol wt of NaAC = 82.03

Desired Volume = 250 mL=250/1000 L

Mass of NaAc = 1.025 g

Now millmoles of NaAC=Molarity*Volume

                                             = 50Χ10^-3 * 250 = 12.5 millimoles

Step 2. Using the Henderson-Hasselbalch equation given below, calculate the concentration of acetic acid needed to bring the pH to 4.2

Acetic Acid in millimoles = 45.45

Millimoles = Molarity of Acetic Acid * volume

Molarity of Acetic Acid =17.4 M

Volume of Acetic Acid (mL) required to adjust pH = 45.45/17.4

                                                            = 2.61 mL

Step 3. dithiothreitol. Conc.

Dilution Calculation

                              ……………………………. (1)

C₁ = concentration of the stock solution of dithiothretiol = 1.2 M

C₂ = desired concentration of the solution, after dilution. = 15mM = 15 Χ10^-3

V₁ = volume about to be diluted

V₂ = final volume after dilution =250 mL

for all dilution problems C₁> C_2, and V₁< V₂.

Putting the values in (1)

V₁ = 15 Χ10^-3 * 250/1.2

V₁ = 3.125 mL

3.125 mL of 1.2 M stock solution of dithiothretiol required to form buffer.

To Make up this buffer

1. Transfer the 1.025 g of NaAc to a beaker which has been filled to the 100 mL of water. Dissolve the NaAc in water.
2. Then, adjust the pH of the NaAc solution by adding 2.61 mL of 17.4 Acetic acid
3. Add 3.125 mL of 1.2 M stock solution of dithiothretiol to this solution
4. Now, bring the total volume to 250 mL with water.