Question

Name: BCHM 370: HOMEWORK 1 DUE 1/14/2020 (5 pts) Q1: Label the functional groups on the molecule below. Thermodynamics Water PH NH HN Не «Н "СООН (2 pts) Q2: Is the molecule above soluble in water? Why? (6 pts) Q3: Can the molecule above participate in hydrogen bonding? If so, circle the groups on the molecule that would hydrogen bond with water. (2 pts) Q4: If the pk, values for phosphoric acid are 2.15, 6.82, and 12.38, which form would predominate at pH5 and which would predominate at pH 10? (4 pts) Q5: You have made a buffer that is at pH 7 with a pk, = 6.8. What is it best suited to buffer against? If it is made to pH 6.2 instead, what is it best suited to buffer against now?

Answer 1

ANSWER - Q1:

The functional groups of the molecule are:

Carbamide HN NH COOH Thioether Carboxylic acid

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ANSWER - Q2:

Yes, this molecule is soluble in water. The solubility in water is due to:

• This is a polar molecule, there are two different functional groups that are polar: the carboxilic acid and the carbamide group (thioether group has a small polarity). Water is a polar molecule that has affinity by polar molecules
• A polar molecule could forms different intermolecular interactions with water as: dipole-dipole interactions and hydrogen bondings. This molecule could form both type of intereaction with water (dipole-dipole and H-bonds). These intermolecular interactions are the main responsible of the solubility in water of the molecule

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ANSWER - Q3:

Yes, this molecule can form hydrogen bondings because has several N and O atoms that are the atoms needed to form thi type of bonding. In this case, the hydrogen bondings could be between 2 molecules of the compound or between the water and the molecule.

The atoms of the carboxylic group and carbamide group can form hydrogen bondings with water:

H-Bonding H-Bonding NH H-Bonding H-Bonding OH H-Bonding

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ANSWER - Q4:

The phosphoric acid has the following equilibrium equation:

H3PO4 = H,POT=HPO = PO-"

where

• the first equilibrium: from H₃PO₄ to H₂PO₄^- has a pKa = 2.15
• the second equilibrium: from H₂PO₄^- to HPO₄^-2 has a pKa = 6.82
• the third equilibrium: from HPO₄^-2 to PO₄^-3 has a pKa = 12.38

Predominant form at pH = 5

• At pH = 2.15 there is the same concentration of H₃PO₄ and H₂PO₄^-. Over pH 2.15 the H₂PO₄^- is the predominant form.
• At pH = 6.82 there is the same concentration of H₂PO₄^- and HPO₄^-2. Whereas, at pH below 6.82 the predominant form is H₂PO₄^-
• Then, at pH 5, the predominant form is H₂PO₄^-​​​​​​​​​​​​​​

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Predominant form at pH = 10

• At pH = 6.82 there is the same concentration of H₂PO₄^- and HPO₄^-2. At pH above 6.82 the predominant form is HPO₄^-2
• At pH = 12.38 there is the same concentration of HPO₄^-2 and PO₄^-3. At pH below 12.38 the predominant form is HPO₄^-2​​​​​​​
• Then, at pH 10, the predominant form is HPO₄^-2​​​​​​​​​​​​​​