Question

To make 10X gel electrophoresis running buffer, 30.0 g Tris base, 144.0 g glycine, and 10.0 g SDS must be dissolved into 1000 mL water. The working stock is a 1X concentration. a. What is the concentration in % w/v of each of the 3 reagents in the 10X stock? b. Describe how you would make 1 L of a 1X working stock solution. c. If 50 mL of the 1X working stock are used to run a gel, how many grams of each of the 3 reagents are in the 1X solution?

Answer 1

a. Volume of water = 1000 ml

Mass of water = volume*density = 1000*1.0 = 1000 g

Total mass of solution = mass of water+mass of tris+mass of glycine+mass of SDS = 1000+30.0+144.0+10.0 = 1184.0 g

Considering density as 1.0 g/ml of solution, volume of solution = 1184.0 ml

% w/v of tris base = (mass of tris base/volume of solution)*100 = (30.0/1184.0)*100 = 2.5 %

% w/v of glycine = (144.0/1184)*100 = 12.2 %

% w/v of SDS = (10.0/1184)*100 = 0.8 %

b. Concentration of prepared solution = M1 = 10X

Volume of 10X solution needed = V1

Concentration of stock solution = M2 = X

Volume of 1X of stock solution = V2 = 1000 ml

M1V1 = M2V2

10X*V1 = X*1000

V1 = 10 ml

10 ml of 10 X solution is to be dissolved in 990 ml of water to prepare 1X stock solution.

c. 1184 ml of 10X solution contains 30.0 g Tris base, 144.0 g glycine, and 10.0 g SDS.

1000 ml of 1X solution contains 10 ml of 10X solution.

So, that 10 ml would contain:

Tris base = (10/1184)*30 = 0.25 g

Glycine = (10/1184)*144 = 1.22 g

SDS = (10/1184)*10 = 0.08 g

That 10 ml is dissolved in 100 ml solution. So, 100 ml solution would also contain this much grams.

50 ml of each of that solution would contain:

Tris base = (50/1000)*0.25 = 0.013 g

Glycine =(50/1000)*1.22 = 0.061 g

SDS = (50/1000)*0.08 = 0.004 g