Question

Phosphoric acid has three pKa values 2.2, 7.2, 12.3 at 30 °C. Explain how you can prepare 100 mL of 20 mM phosphate buffer (pH 3) provided you have purified water and1L of each of the following stock solutions available: 0.1 M Na3PO4, 0.25M Na2HPO4, 0.5 M NaH2PO4, and 1M H3PO4 . (Hint: choose only two!!) How could you double the buffer capacity of this buffer?

Answer 1

Given that pKa1, pKa2 and pKa3 values for phosphoric acid = 2.2, 7.2 and 12.3, respectively.

i.e. H₃PO₄ H₂PO₄^-, pKa = 2.2

H₂PO₄^- HPO₄^2-, pKa = 7.2

HPO₄^2- PO₄^3-, pKa = 12.3

We need to prepare 100 mL of 20 mM phosphate buffer, i.e. 20*10^-3 M * 100 mL = 2 mmol

pH of the required buffer = 3

At pH 3, the two predominant forms of phosphate will be H₃PO₄ and H₂PO₄^-.

According to Henderson-Hasselbalch equation:

pH = pKa + Log(n_salt/n_acid), where n = no. of mmol

i.e. 3 = 2.2 + Log(n_H2PO4^-/n_H3PO4)

i.e. n_H2PO4^-/n_H3PO4 = 6.31

Therefore, n_H2PO4^- + n_H3PO4 = 2 mmol and n_H2PO4^-/n_H3PO4 = 6.31

i.e. n_H2PO4^- = 1.7264 mmol and [H₃PO₄] = 2 - 1.7264, i.e. 0.2736 mmol

Hence, to prepare 100 mL of 20 mM H₃PO₄/NaH₂PO₄ buffer, follow the steps shown below.

The volume of 1 M H₃PO₄ solution in 0.2736 mmol of H₃PO₄ = 0.2736 mmol/1, i.e. 0.2736 mL

The volume of 0.5 M NaH₂PO₄ solution in 1.7264 mmol of NaH₂PO₄ = 1.7264 mmol/0.5, i.e. 3.4528 mL

To make the total volume of the buffer mixture to 100 mL:

0.2736 x + 3.4528 x = 100

i.e. x = 26.84 mL of water is required to dilute the each cocentrated component of buffer mixture.

As a result, the volume of 0.0373 M H₃PO₄ solution is 7.34 mL and that of 0.0186 M NaH₂PO₄ is 92.66 mL.

The buffer capacity of this buffer can be doubled by the addition of more amount of H3PO4 (same amount as in buffer mixture).