What mass of Na3PO4 must be added to 82.9 mL of 0.281 M HCL to obtain...

What mass of Na3PO4 must be added to 82.9 mL of 0.281 M HCL to obtain a buffer with a pH of 7.66?

Ka1(H3PO4) = 7.5x10-3

Ka2(H3PO4) = 6.2x10-8

Ka3(H3PO4) = 3.6x10-13

Mass = ___________ g

****PLEASE ONLY ATTEMPT IF YOU KNOW THE ANSWER AS I'M DOWN TO MY LAST ATTEMPT*** THANKS!

Expert Solution

The reaction is Na3PO4 + HCl --> Na2HPO4 + NaCl

Therefore a buffer will be produced from sodium phosphate and sodium hydrogen phosphate.

80 mL of 0.200 M HCl has 16 mmoles of HCl in the solution. The Na2HPO4 will be 16 mmoles too. Its concentration is 16/80=0.2 M.

The dissociation constant (pKa) for this reaction is 7.20, so the Henderson-Hasselbach equation becomes like this:

7.75=7.20+log(Cb/0.2) and log (Cb/0.2)=0.55 and Cb=0.71 M. When in 80 mL of solution, the moles are 56.8 mmoles, so the mass is 56.8 mmol * (69+31+64=194 mg/mmol)=11,019 mg = 11.02 g of Na3PO4 must be added into HCl solution.

Thank you and good luck.

queen_honey_blossom answered 2 years ago

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