You are asked to prepare a buffer solution of H2PO4-/HPO42- with a pH of 6.68. A 153.00...

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You are asked to prepare a buffer solution ofH2PO4-/HPO42-with a pH of 6.68. A 153.00 mL solution already contains 0.126 MHPO42-.

Hint: you will need to look up the proper Ka in yourtextbook

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

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3.6 Ratings (384 Votes)


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


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