Question

Describe the preparation of 40 liters of 0.02 M phosphate buffer, pH 6.9, starting from solid Na3PO4 and 1M HCl. The answer is 131.2 g of Na3PO4 + 1330 mL 1M HCl. Please explain in detail how to get this answer.

Answer 1

Ans.

Note: Total moles of A_- or Na₃PO₄ in the buffer = Total molarity x Vol. in liters

                                                            = 0.02 M x 40.0 L

                                                            = 0.80 mol

# Since 1 mol Na₃PO₄ is neutralized by 1 mol HCl, the maximum vol. of HCl required to neutralize all the Na₃PO₄ = 0.80 mol

            So, Max Vol. of HCl that can be neutralized = Moles / Molarity

                                                = 0.80 mol / 1.0 M = 0.80 L = 800.0 mL

# Once you add more than 800.0 mL of 1 M HCl, the solution becomes highly acidic. So, the required volume of HCl can’t be greater than 800.0 mL (be it 1330 mL or any other value). Please cross-check the Molarity and/or required volume of HCl.