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 = 7.4. The total phosphorus concentration is to be 0.250 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) (2 pts) 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 = 7.4?

(ii) (4 pts) 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 = 7.4?

(iii)(4 pts) 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 = 7.4?

Explanation / Answer

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

pKa2 for H3PO4 = 7.2

7.4 = 7.2 + log([base]/[acid])

[base] = 1.585[acid]

[acid] + [base] = 0.25

[base] = 0.25 - [acid]

0.25 - [acid] = 1.585[acid]

[acid] = 0.097 M

[base] = 0.153 M

(i) H2PO4- has a pKa of 7.2. that is pKa2 for H3PO4 is close to desired pH

(ii) [H2PO4-] = 0.097 M

moles of NaH2PO4 = molarity x volume = 0.097 x 1 = 0.097 M

so grams of NaH2PO4 required = 0.097 x 119.98 = 11.64 g

[HPO4^2-] = 0.153 M

moles of HPO4^2- = 0.153 mols

grams of Na2HPO4 required = 0.153 x 141.96 = 21.72 g

So we would require, 11.64 g of NaHPO4 in 1 L of water to give a buffer with pH 7.4

(iii) Conjugate base of H2PO4- is HPO4^2-

[HPO4^2-] = 0.153 M

moles of HPO4^2- = 0.153 mols

grams of Na2HPO4 required = 0.153 x 141.96 = 21.72 g

So we would require, 21.73 g of Na2HPO4 in 1 L of water to give a buffer with pH 7.4

The two salts are put in 1 L of water to give a buffer of pH 7.4. Total volume is 1 L.

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