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Sedaia [141]
3 years ago
13

how to determine the net charge of the tripeptide Asp-Gly-Leu at pH 7. Can someone show in details and tricks on how to solve it

? Thank you!!
Chemistry
1 answer:
Ugo [173]3 years ago
5 0

Answer:

0!

Explanation:

  • You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:            

           

                   -COOH         -NH3              

pH 7------------------------------------------------------              

Asn               -                      +

Gly                -                      +

Leu               -                      +

  • Now that we have our table we'll sketch our peptide's structure:

<em>HN-Asn-Gly-Leu-COOH</em>

This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:

+1 (HN-Asn)

-1 (Leu-COOH)

=0

The net charge is 0!

I hope you find this information useful and interesting! Good luck!

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Benzophenone has a normal freezing point of +48.1 oC, with freezing point depression constant Kfpt = − 9.78 oC/m. A 0.1500 molal
Neporo4naja [7]

Answer:

i = 2.79

Explanation:

The excersise talks about the colligative property, freezing point depression.

Formula to calculate the freezing point of a solution is:

Freezing point of pure solvent - Freezing point of solution = m . Kf . i

Let's replace data given. (i = Van't Hoff factor, numbers of ions dissolved in solution)

48.1°C - 44°C = 0.15 m . 9.78°C/m . i

4.1°C / (0.15 m . 9.78°C/m) = i

i = 2.79

In this case, numbers of ions dissolved can decrease the freezing point of a solution, which is always lower than pure solvent.

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