Answer to Question #9278 in Biochemistry for Carter

Question

How would you make a buffer of 50 mM (pKa =7.15) at pH 7, with .708 mole MOPS Base and 1.00 mole MOPS acid and 34.2 L solution, using 500 mM solution MOPS acid and 1.00 M solution NaOH?
mL of 1.00 M NaOH
mL of 500 mM MOPS acid
mL Water

Answer 1

For this purpose zou need to prepare to independent solutions (.708 mole MOPS Base and 1.00 mole MOPS acid), which further you should mix in one graduated
flask and adjust a total volume to 34,2L with water.
1)For preparing 1.00 mole MOPS acid: we can calculate a necessary volume of stock 500mM MOPS acid.
500mM=0,5M, it means 0,5 mole in 1L, therefore you need 2L 500mM MOPS acid for
1,00 mole.
2) For preparing .708 mole MOPS Base. First of all let's calculate a necessary volume of stock solution MOPS acid: 0,5 mole for 1L, that's why for
0.708 mole you need 1,4L. As MOPS acid has one basic capasity, for MOPS base we
need to mix the equivalent volumes of MOPS acid and NaOH. However, a
concentration of NaOH (1M) is twice higher than a concentration of MOPS acid
(0,5M) that is why we should take twice less volume of NaOH: 1,4/2=0,7L. You mix
1,4L 500mM MOPS acid with 0,7L 1M NaOH and in this way have 2,1L 0,708 mole MOPS
base.
3)MIx 2L 500mM Mops acid+2,1L MOPS Base= 4,1L and adjust it with 30,1LWater to Total 34,2L

In General for total volume 34,2L:
700 mL of 1.00 M NaOH
3700 mL of 500 mM MOPS acid
34200 mL Water