Question

Show the calculations and specific steps involved that are necessary to prepare a 145 mL solution that contains 18mM Tris-HCl, 22mM EDTA, 13mM NaCl, and 0.5% BSA. (Note: You have already prepared, shelf-ready stock solutions of 2.3M NaCl, 2.8M Tris-HCl, 0.7M EDTA, and you have a 100g bottle of BSA.)

Answer 1

• For 18mM Tris- HCl, We have stock solution of 2.8M Tris- HCl, For 145 ml, we are using MiVi = MfVf, Mi = 2.8 M, Vf = 145 ml, Mf = 18×10-3 M. Substitute the value Vi = 18 × 10-3 × 145 /2.8 = 0.932 ml will required
• For 22mM EDTA, we have Vf = 145 ml, Mf = 22mM, Mi = 0.7M. Substitute value in above equation. Vf = 22× 10-3 × 145/0.7= 4.55 ml will required.
• For 13mM NaCl, we have Vf = 145 ml, Mf = 13mM, Mi = 2.3 M. Substitute the value in equation. Vi = 13× 10-3 ×145 / 2.3 = 0.891 ml.
• 0.5% BSA in 145 ml, mean 100ml contain 0.5g of BSA, so 145 ml solution contain = 0.5× 145 / 100 = 0.725g

For final solution add 0.725 g of BSA followed by 0.932

ml of 2.8M Tris- HCl then 4.55 ml of 0.7 M EDTA then 0.891 ml of 2.3 M of NaCl. This will gives = 0.932 + 4.55+ 0.891 = 6.308ml. Now makeup volume to 145 ml.