Question

You are removing imidazole from your pooled nickel column samples by dialysis. The
volume of the pooled sample is 32 mL and the imidazole concentration is 0.5 M.

a. What is the concentration of imidazole after exposure to three consecutive dialyses
against 1 L of buffer?
b. If the imidazole at the start is 3.4% (w/v), what is the concentration in
weight percent at the end of the dialysis (three consecutive exposures to one liter
of buffer)?

Answer 1

(a). Volume of sample V1=32ml

Strength of imadazole initially S1=0.5M

Volume after 1st dialysis=1000ml + 32 ml (volume of buffer + sample)=V2

Strength of imadazole after 1st dialysis=S2.

Therefore, S1V1=S2V2=> 32(ml) * 0.5(M)/ 1032(ml)=S2=0.0155 M ( concentration after 1st dialysis)

For 2nd dialysis S1 becomes 0.0155 and V1= 1032 (present volume of sample + buffer)

V2= (1032+ 1000)ml= previous volume + volume of freshly added buffer=2032 ml.

Therefore, S1V1=S2V2=> 0.0155(M) * 1032(ml)/ 2032(ml) => S2= 0.0078 M (concentration after second dialysis)

Similarly, concentration after final and third dialysis= 0.0078 (M) * 2032(ml)/3032 (ml)= 0.005 (M)

(b) (w/v) percentage is defined as the mass of the solute(gms)/ volume of the total solution(ml)

Therefore, initially volume of total solution=32 ml= volume of sample + volume of imidazole in the solution.

Therefore, 3.4%= [mass of solute(gm)/32(ml)] * 100%=> mass of solute=1.088gm.

Therefore mass of solute always remains the same throughout the dialysis. Volume changes only and total volume of the sample after three dialysis= 32+1000+1000+1000= volume of sample+ 3 times the volume of buffer required for three dialysis.

Therefore (w/v) % after three dialysis= (1.088/3032) *100%= 0.036%