Q. 2. Dialysis. A purified protein is in a sodium phosphate buffer at pH 7.4 wit
ID: 1087005 • Letter: Q
Question
Q. 2. Dialysis. A purified protein is in a sodium phosphate buffer at pH 7.4 with 500mM NaCl. A sample (1mL) of the protein is placed in a dialysis tube and dialyzed against 1L of the same phosphate buffer with 0 mM NaCl. Small molecules and ions such as Nat, Cl and phosphate can diffuse across the dialysis membrane, but the protein cannot. a. Once the dialysis has come to equilibrium, what is the concentration of NaCl in the protein sample? Assume no volume changes occur in the sample during the dialysis If the original 1mL sample were dialyzed twice, successively, against 100mL of the same Na phosphate buffer with 0mM NaCI, what would b. be the final NaCI concentration in the sample?Explanation / Answer
a. Concentration of NaCl in dialysis bag before dialysis = C1 = 500 mM
Concentration of NaCl in dialysis bag after dialysis = C2 = ?
volume of protein sample = V1 =1 ml = 0.001 L
volume of phosphate buffer outside = 1 L
At equilibrium:
C1V1 = C2V2
500*0.001 = C2*1
C2 = 0.500 mM
concentration of NaCl in dialysis bag after dialysis = 0.500 mM
b. In first dialysis:
Concentration of NaCl in dialysis bag before dialysis = C1 = 500 mM
Concentration of NaCl in dialysis bag after dialysis = C2 = ?
volume of protein sample = V1 =1 ml
volume of phosphate buffer outside = 100 ml
At equilibrium:
C1V1 = C2V2
500*1 = C2*1100
C2 = 5..00 mM
concentration of NaCl in dialysis bag after first dialysis = 0.500 mM
After 2nd dialysis:
Concentration of NaCl in dialysis bag before dialysis = C1 = 5.00 mM
Concentration of NaCl in dialysis bag after dialysis = C2 = ?
volume of protein sample = V1 =1 ml
volume of phosphate buffer outside = 100 mlL
At equilibrium:
C1V1 = C2V2
5.00*1 = C2*100
C2 = 0.050 mM
concentration of NaCl in dialysis bag after second dialysis = 0.050 mM
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