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!
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.
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?
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.
To determine how much Tris-bicarbonate should be added to achieve a final bicarbonate concentration of 10mM in a 2ml reaction, we need to calculate the volume of Tris-bicarbonate solution required.
First, let's calculate the number of moles of bicarbonate (HCO3-) required for the desired final concentration. The final concentration is given as 10mM (millimolar), which means 10 millimoles per liter.
Since the final volume is 2ml, we need to convert the final concentration from moles per liter to moles per milliliter:
10mM / 1000 = 0.01 moles/ml
Now, let's find the number of moles of Tris that reacts with bicarbonate to form Tris-bicarbonate.
From the given information, we have 0.01 moles of Tris. Since the mole ratio between Tris and bicarbonate is 1:1, 0.01 moles of bicarbonate are also present.
Next, let's determine the concentration of bicarbonate in the Tris-bicarbonate solution.
Given that the Tris solution was prepared as a 1M solution, which means 1 mole of Tris per liter of solution, we have 1 mole of Tris in 0.1L (100ml) of solution.
Since the mole ratio between Tris and bicarbonate is 1:1, the concentration of bicarbonate in the Tris-bicarbonate solution is also 1M.
Now, we can calculate the volume of Tris-bicarbonate solution needed.
The number of moles of bicarbonate required is 0.01 moles, and the concentration of bicarbonate in the Tris-bicarbonate solution is 1M.
Using the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume, we can rearrange the formula to solve for V2:
V2 = (C1V1) / C2
V2 = (1M * 0.01 moles) / (1M)
V2 = 0.01L = 10ml
Therefore, you need to add 10ml of the Tris-bicarbonate solution to the reaction to achieve a final bicarbonate concentration of 10mM in a 2ml reaction.
The calculation you have made, where you converted 0.082 moles of bicarbonate in 0.1L to concentration, is incorrect. It should have been 0.82M, not 0.82M. But since you have already calculated the number of moles required, you need to use that value to determine the volume of solution needed, which is 10ml.