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pH Dependence of Drug Absorption lab 1. Calculate the amount of Na2HPO4 * 7H2O ( in...

pH Dependence of Drug Absorption lab

1. Calculate the amount of Na2HPO4 * 7H2O ( in grams) needed to make 50 mL of a 0.40 M solution.

2. Calculate the amounts of Na2HPO4 and NaH2PO4 (in grams) needed to prepare 200 mL of a buffer with pH=8.25 so that the sum of concentrations of HPO42- and H2PO4- ions is 0.5 M. pKa1=2.12, pKa2=7.21, pKa3= 12.38

* on this problem I think you use the Henderson-Hasselbalch equation, but after that I'm lost.

3. Write the equation for the ionization of 3-aminophenol. Would this substance be more soulble in water at pH=2 or pH= 8? Would you expect this substance to be absorbed into the blood stream in the stomach or the instestines?

4. How does the initensity of the spot on the silica sheet relate to the solubility of the drug in the aqueous phase?

Please explain the the problems!!

Solutions

Expert Solution

1. First, we calculate the number of moles required:

moles = M x V = 0.40 mol/L x 0.050 L = 0.02 moles

With this and the molecular weight:

MW = (2x23) + 31 + (5x16) + (8x1) = 165 g/mol

m = 165 g/mol x 0.02 mol = 3.3 g

2. You a re right, you need to use the Henderson - Hasselbach equation:

pH = pKa + log [S]/[A]

If you want a pH of 8.25, you need to use the pKa2 = 7.21 because is closer to the pH. Also, If we actually know that [HPO4] + [H2PO4] = 0.5 M, we can say that [HPO4] = 0.5 - [H2PO4]. So the Henderson - Hasselbach equation:

8.25 = 7.21 - log ([HPO4]/[H2PO4])

1.04 = -log ([HPO4]/[H2PO4])

101.04 = ([HPO4]/[H2PO4])

10.96 = ([HPO4]/[H2PO4])

10.96[H2PO4] = [HPO4]

And if [HPO4] + [H2PO4] = 0.5 then:

10.96[H2PO4] + [H2PO4] = 0.5

11.96[H2PO4] = 0.5

[H2PO4] = 0.5/11.96 = 0.042 M

[HPO4] = 10.96 x 0.042 = 0.458 M

With these values of Concentrations, we can determinate the numbers of moles, and then the mass required:

moles HA = 0.042 x 0.2 = 0.0084 moles

moles A = 0.458 x 0.2 = 0.0916 moles

MWha = 23 + 2x1 + 31 + 16x4 = 120 g/mol

MWa = 23x2 + 1 + 31 + 64 = 142 g/mol

mHA = 120 x 0.0084 = 1.008 g of NaH2PO4

mA = 142 x 0.0916 = 13.0072 g of Na2HPO4

For question 3 and 4, please post them in a different question so you can get a better and faster answer. I would gladly answer it for you, once I clear some doubts regarding those questions.


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