2. Two students were given a 1 mL protein sample containing 150 mM ammonium sulf
ID: 209282 • Letter: 2
Question
2. Two students were given a 1 mL protein sample containing 150 mM ammonium sulfate (NH4)504. The students take different approaches to dialyzing their sample against buffer (the dialysis membrane is permeable only to the salt). Student 1 dialyzes the sample against 125 mL buffer to equilibrium. She then repeats the dialysis against a new 125 ml buffer solution to equilibrium (assume the dialysis tube volume remains 1 mL). Student 2 decides to save a trip to the lab and dialyzes the sample in 500 mL of buffer to equilibrium. A) What is the final concentration of salt in the dialysis tubing at the conclusion of each student's dialysis treatment (for Student 1, report the concentration after dialysis 1 and dialysis 2). (Remember, you must show all calculations).(3 pts) B) What yolume of buffer solution should student 2 have used to reach the same final concentration of salt in the dialysis tubing at equilibrium after one dialysis step that student 1 achieved in 2 steps? (2 pts)
Explanation / Answer
First, let us convert the concentration of ammonium sulfate from mM to g/L.
1M = Molecular weight of ammonium sulfate / 1L of solvent = 132.14 g/L
150 mM = 0.15M = 132.14*0.15 = 19.821 g/L = 19.821 mg/ml
A) At equilibrium the concentration of ammonium sulfate in sample and buffer will be equal.
Student 1:
Dialysis 1:
The sample contain 19.821 mg of ammonium sulfate in 1 ml.
Volume of sample (VS) = 1 ml; Volume of buffer (VB)= 125 ml
Quantity of ammonium sulfate present in 1 ml = 19.821 mg
Final quantity of ammonium sulfate after dialysis = Quantity of ammonium sulfate present in the sample/(VS+VB)
= 19.821 / (1+125) = 0.1573
Final concentration of ammonium sulfate after dialysis 1 = 0.1573 mg/ml
Dialysis 2:
Now the quantity of ammonium sulfate in the sample = 0.1573 mg
The sample contain 19.821 mg of ammonium sulfate in 1 ml.
Volume of sample (VS) = 1 ml; Volume of buffer (VB)= 125 ml
Quantity of ammonium sulfate present in 1 ml = 0.1573 mg
Final quantity of ammonium sulfate after dialysis 2 = 0.1573 / (1+125) = 0.0012
Final concentration of ammonium sulfate after dialysis by student 1 = 0.0012 mg/ml
To convert the concentration from mg/ml to M = 0.0012/132.14 = 0.00000908 M = 9.08 µM
Final concentration of ammonium sulfate after dialysis by student 1 = 9.08 µM
Student 1:
The sample contain 19.821 mg of ammonium sulfate in 1 ml.
Volume of sample (VS) = 1 ml; Volume of buffer (VB)= 500 ml
Quantity of ammonium sulfate present in 1 ml = 19.821 mg
Final quantity of ammonium sulfate after dialysis = Quantity of ammonium sulfate present in the sample/(VS+VB)
= 19.821 / (1+500) = 0.0396
Final concentration of ammonium sulfate after dialysis by student 2 = 0.0396 mg/ml
To convert the concentration from mg/ml to M = 0.0396/132.14 = 0.00029968 M = 299.68 µM
Final concentration of ammonium sulfate after dialysis by student 1 = 299.68 µM
B) To find the buffer volume to be used to obtain the final concentration as student 1, let us rearragen the formula used to calculate the final concentration.
Final quantity of ammonium sulfate after dialysis = Quantity of ammonium sulfate present in the sample/(VS+VB)
0.0012 = 19.821 / (1+VB)
VB = (19.821/0.0012)-1 = 16517.5 - 1 =16516.5
Volume of buffer to be used = 16516.5 ml
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