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^-/HPO₄^2-with a pH of 6.68. A 153.00 mL solution already contains 0.126 MHPO₄^2-.

Hint: you will need to look up the proper K_a 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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 153*0.126

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 19.28 mmol

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 19.28*10^-3 mol

no of mol of NaH2PO4 required =Â Â x mol

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

x = 0.0653

amount of NaH2PO4 required = n*Mwt

Â Â n = mol of NaH2PO4 required = 0.0653 mol

Mwt = molarmass of NaH2PO4 = 119.98 g/mol

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.0653*119.98

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 7.83 g

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

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90.2K

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Chemistry

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