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: a 2.00 M phosphoric acid stock solution, solid sodium dihydrogen phosphate, (MM 120 g/mol) solid sodium hydrogen phosphate, (MM 142 g/mol) solid, sodium phosphate, (MM 164 g/mol) a 1.00 M HCl(aq) stock solution, a 1.00 M NaOH(aq) stock solution, and deionized water

. a) 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?

b) Choose the phosphate species chosen in (2.a) 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?

Please do question B

Explanation / Answer

ans b) sodium hydrogen phosphate 0.250 M 1 lit buffer solution is to be prepared

As pH = pKa + log ([A-]/[HA])

Now , we put   A- = K2HPO4 and HA = KH2PO4 concentration:

7.4 = 7.21 + log(A/HA)
log(A/HA)= 0.19
A/HA = 1.5488

HA +A = 0.250

A/HA = 1.5488

HA + 1.5488 HA = 0.250 ---->   2.5488 HA =0.25 ----> HA = 0.098 M

A= 0.250 - 0.098 = 0.152 M

we can calculate the required amount:

HA+A=0.250
A/HA=1.5488

HA=0.098 M
A=0.152 M

we have to mix 0.098 Moles of NAH2PO4 and 0.0152 moles of Na2HPO4 salts into 1 liters of solution.

NAH2PO4 = 120 g / mol * 0.098 mol= 11.76 g

NaHPo4 = 142 g /mol* 0.152 mol = 21.58 g

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