Question

Calculate that volumes of each stock solution you will need to assemble the following cysteine-detection reaction: 30 mM TEA, 15 mM DTNB, 75 µM cysteine in a 0.8 mL total volume. Stock concentrations are 0.5M TEA, 100 mM DTNB, and 1 mM cysteine. Don’t forget to include the volume of water you will need to reach the final total volume.

Answer 1

According to law of dilution   MV = M'V'

Where M = Molarity of stock

V = Volume of the stock

M' = Molarity of dilute solution

V' = Volume of the dilute solution

Plug the values we get , V = M'V' /M

(1) Preparation of 0.8 mL of 30 mM TEA from  0.5M TEA

M = 0.5 M , V = ? , M' = 30 mM = 30x10^-3 M , V' = 0.8 mL

So V = M'V'/M = 0.048 mL

So take 0.048 mL of stock solution and diluted with water( volume of water = 0.8 - 0.048 = 0.752 mL) to get 0.8 mL.

(2) Preparation of 0.8 mL of 15 mM DTNB from  100 mM DTNB

M = 100 mM = 100x10^-3 M , V = ? , M' = 15 mM = 15x10^-3 M , V' = 0.8 mL

So V = M'V'/M = 0.12 mL

So take 0.12 mL of stock solution and diluted with water( volume of water = 0.8 - 0.12 = 0.68 mL) to get 0.8 mL.

(3) Preparation of 0.8 mL of  75 µM cysteine from  1mM cysteine

M = 1 mM = 1x10^-3 M , V = ? , M' = 75 µM = 75x10^-6 M , V' = 0.8 mL

So V = M'V'/M = 0.06 mL

So take 0.06mL of stock solution and diluted with water( volume of water = 0.8 - 0.06 = 0.74 mL) to get 0.8 mL.