Question

Describe how you would prepare 2 mL of solution containing NaCl, MES buffer, and water. NaCl stock solution is 1M and the working concentration required is 5mM. MES buffer stock solution is 100 M and the working concentrations required is 20 mM. To make 2 mL of solution calcualte the amount of water you would add.

Please Show Work

Answer 1

Ans. # Required volume of NaCl:

Given, Stock [NaCl] = 1 M                           ; Working [NaCl] = 5 mM = 0.005 M

            Final volume of working solution = 2.0 mL

Now, using –

            C1V1 (Stock NaCl solution) = C2V1 (Working solution)

            Or, 1.0 M x V1 = 0.005 M x 2.0 mL

            Or, V1 = (0.005 M x 2.0 mL) / 1.0 M = 0.010 mL

Hence, required volume of stock NaCl solution = 0.010 mL = 10.0 uL

# Required volume of MES buffer:

Given, Stock [MES] = 100 mM                    ; Working [MES] = 20 mM

            Final volume of working solution = 2.0 mL

Now, using –

            C1V1 (Stock MES buffer) = C2V1 (Working solution)

            Or, 100 mM x V1 = 20 mM x 2.0 mL

            Or, V1 = (20 mM x 2.0 mL) / 100 mM = 0.400 mL

Hence, required volume of stock MES buffer = 0.400 mL = 400.0 uL

# Required volume of water:

Given, volume of working solution = 2.0 mL = 2000.0 uL

Required volume of water = Volume of working solution – Volume of (stock NaCl + MES)

                                                = 2000.0 uL – (10.0 uL + 400.0 uL)

                                                = 1590.0 uL

# Therefore, the constitution of working solution is-

            1.0 uL stock of NaCl solution

            400.0 uL of stock MES buffer

            1590.0 uL of water.