Question

I am trying to calculate the how amount of Tris-bicarbonate should be added in the reaction. I would like to have final volume of bicarbonate 10mM in 2ml reaction.

I made Tris-base (1M, 100ml) and instead of HCl, CO2 was bubbling to set pH7.5.
0.1L*0.1M= 0.01 moles of Tris

……………..Tris + HCO3- ==> Tris*HCO3-
initial mols...0.010....0........0
change............-x....x........x
final.........0.010-x...x....... x

7.4 = 8.06 + log [(0.01-x)/(x)]
x = 0.0082 moles HCO3-

I am confusing from here. If bicarbonate concentration is 0.082moles in 0.1L, it’s concentration 0.82M. right?
So, if I want to make 10mM bicarbonate concentration in 2ml, is this correct that I have to add 1/82*2ml=0.024ml, right?

please help me. Thanks!

Answer 1

I'm confused, too. You show 1M, 100 mL but you then have L x M = (0.1L x 0.1M). So do you mean 100 mL of 1 M or 100 mL of 0.1M?
Near the end you have 0.0082 mols/0.1L and that is 0.082 and not 0.82.

Answer 2

sorry Dr.Bob for confusion. Please consider 0.1M, 0.1L Tris!
so, it should be 0.0082mols/0.1L and 0.082M right?? so, if I want to make 10mM on 2ml, 1/82*2ml= 0.024ml. Is this correct?

Answer 3

OK. So we go with the 0.01 in the initial ICE chart as correct. You have 100 mL of 0.082 M. You want 2 mL of 10 mM (0.01 M)
The dilution formula is
mL1 x M1 = mL2 x M2
mL x 0.082 = 2 mL x 0.01M
mL of 0.082M needed= 2*0.01/0.082 = 0.244 mL.