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A purified protein is in a HEPES buffer at pH 7.0 with 600 mM NaCl. A 1 mL sample of the protein solution is dialyzed against 1.0 L of the same HEPES buffer with 0 mM NaCl. Once the dialysis has come to equilibrium, what is the concentration (in mM) of NaCl in the protein sample

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ajeigbeibraheem

Answer:

Explanation:

Using the dilution formula to determine the final concentration of NaCl in the protein sample.

[tex]M_1V_1 = M_2V_2[/tex]

[tex]600 \ mM \times 1 \ mL= M_2 \times ( 1000\ mL + 1 \ mL)[/tex]

[tex]M_2= \dfrac{600 \ mM \times 1 \ mL}{ ( 1001\ mL)}[/tex]

[tex]M_2= 0.599 \ mM[/tex]

However, when dialysis attains equilibrium, it implies that the sample was dialyzed twice, SO:

[tex]M_2 = \dfrac{600 \ mM\times 1 \ mL}{201 \ mL} = 2.97 \ mM[/tex]

Now, take the concentration as 2.97 mM

Then:

[tex]M_f = \dfrac{2.985 mM \times 1 mL}{201 \ mL}[/tex]

[tex]M_f = 0.01485 \ mM[/tex]

[NaCl] = 0.01485 mM

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