You work in a medical research lab with a biochemist who asks you to make a 1.00

Question

You work in a medical research lab with a biochemist who asks you to make a 1.00 L phosphate buffer stock solution with pH = 7.35. The total phosphorus concentration is to be 0.150 M. In the storeroom, there are the following reagents: (1) a 3.00 M phosphoric acid stock solution, (2) solid sodium dihydrogen phosphate, (3) solid sodium monohydrogen phosphate, (4) solid, sodium phosphate, (5) a 3.00 M HCl(aq) stock solution, (6) a 3.00 M NaOH(aq) stock solution, and (7) deionized water (i) To make an optimum buffer, it is important that the pKa of the acid part is close to the target pH of the buffer. Which phosphate species has a pKa value closest to pH = 7.35? (ii)Choose this phosphate species (in i) as the only phosphorus component. What amounts of this substance and any other reagents (grams of solid(s); mL of any solution(s) and water) would you need to make the buffer with pH = 7.35? (iii)Choose this phosphate species (in i) and either its conjugate acid or base as the only two phosphorus components. What amounts of this substance, its conjugate acid or base, and water (grams of solid(s); mL of any solution(s) and water) would you need to make the buffer with pH = 7.35?

Explanation / Answer

moles of phosphate buffer = 0.150 x 1 = 0.150

H2PO4- + OH- ----------------> HPO42- + H2O

pH = pKa + log [HPO42- / H2PO4]

7.35 = 7.21 + log [HPO42- / H2PO4-]

[HPO42- / H2PO4-] = 1.38

[HPO42- + H2PO4-] = 0.150

1.38 H2PO4- + H2PO4- = 0.150

moles of H2PO4- = 0.063

moles of HPO42- = 0.087

mass of H2PO4- = 0.063 x 96.99 = 6.11 g

mass of HPO42- = 0.087 x 95.98 = 8.35 g

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