Question

You are asked to prepare a buffer solution of H₂PO₄^-/HPO₄^2- with a pH of 6.68. A 153.00 mL solution already contains 0.126 M HPO₄^2-.

Hint: you will need to look up the proper K_a in your textbook

How many grams of NaH₂PO₄ must be added to achieve the desired pH? You may assume that the volume change upon salt addition is negligible. in g

Answer 1

pH of phosphate buffer = pka2 + log(Na2HPO4/NaH2PO4)

pka2 phosphate buffer = 7.21

pH = 6.68

no of mol of HPO4^2- in the solution = M*V

                                      = 153*0.126

                      = 19.28 mmol

                                      = 19.28*10^-3 mol

no of mol of NaH2PO4 required =   x mol

6.68 = 7.21 + log((19.28*10^-3)/x)

x = 0.0653

amount of NaH2PO4 required = n*Mwt

   n = mol of NaH2PO4 required = 0.0653 mol

Mwt = molarmass of NaH2PO4 = 119.98 g/mol

                           = 0.0653*119.98

                           = 7.83 g