Question

Because everyone in the world is using lung cell culture medium, there is a shortage. But, no problem, you can make a homemade version using basic components. The final concentration of the components in culture medium are 130 mM sodium chloride, 2.5 mM potassium chloride, 25 mM sodium bicarbonate, and 100 μg/mL BSA. You need to prepare a 10 x concentrate of culture medium. You have the following stock solutions: 5 M sodium chloride, 0.5 M potassium chloride, 2.5 M sodium bicarbonate, and 30% (w/v) BSA. How would you make up 200 mL of this 10 x concentrate?

Answer 1

As per the given information,

Composition of the Stock solution( per liter)	Original concentration of the media( 1x)	The resultant concentration of the components in 10X
NaCl- 5 M	NaCl- 130mM	NaCl- 1300mM = 1.3M
KCL- 0.5M	KCl- 2.5 mM	KCl- 25Mm=0.025M
NaHCO3-2.5 M	NaHCO3-25mM	NaHCO3-250Mm=0.25M
BSA- 30% i.e. 300g of BSA dissolved in 1 L of water, which can be written as- 300g/Litre of BSA	BSA- 100µg/ml	BSA- 1000µg/ml        =1mg/ml        =1000mg/L        =1g/L

Now for NaCl- let the amount of Nacl solution(in ml) that should be taken from 5M solution to prepare 1.3M NaCl solution in 200 ml be ‘x’ , then

                            5M * x= 1.3M * 0.2l

                         Or, x=

                           So, X= 0.052 L or, 52 ml

Now, for KCl- - let the amount of KCl solution(in ml) that should be taken from 0.5M solution to prepare 0.025M KCl solution in 200 ml be ‘x’ , then,

                          0.5M * X = 0.025M * 0.2L

                       Or, X =

                         So, X= 0.01 L or10 ml

Now for NaHCO3, - let the amount of NaHCO3 solution(in ml) that should be taken from 2.5M solution to prepare 0.25M NaHCO3solution in 200 ml be ‘x’ , then,

                                  2.5M * X = 0.25M * 0.2 L

                               Or, X =

                                   So, X = 0.02 L or, 20ml

Now for BSA, let the amount of BSA solution (in ml) that should be taken from 300g/L solution to prepare 1g/L BSA solution in 200 ml be ‘x’ , then

                              300g/L * X = 1g/L * 0.2L

                           Or, X =

                             So, X = 0.0006 L or 0.6 ml

So, finally according to the results, the final volume of the components that is required from the provided stock solutions to prepare 10x of the media in 200 ml are-

52ml of NaCl , 10ml of KCl , 20 ml of NaHCO3 and 0.6 ml of BSA solution

The rest of 200 ml will be Distilled water , i.e 117.4 ml