- #1

DavitosanX

- 9

- 1

I had a problem recently while preparing an ATP solution. I needed to prepare a 200 mM solution, and afterwards adjust its pH to 7 using KOH. Anticipating that the concentration would change both by adding KOH and by taking aliquots to measure the pH, I prepared a 250 mM solution, so that later I could adjust the final concentration to 200 mM. I kept track of the volumes added and extracted, but unfortunately I ended up with a 180 mM concentration after the pH had been adjusted.

I knew that I could add more ATP to the final solution, but that would alter the volume as well, so calculating the final concentration would be tricky. At this point I considered that this could be represented as a differential equation, since final concentration changes with volume added, and the rate at which this happens changes with concentration. I plotted the behavior (Final concentration vs Added volume) and obtained a clear sigmoid curve. Boundary conditions can also be easily established, since at added volume = 0, concentration is the same as the original solution, and at added volume = infinity, the concentration tends to be the same as the added solution.

My math skills are quite poor, so I took the chemistry route. To solve this, I made an example: Let's say that I have 20 ml of a 1M NaOH solution (solution a) and I add 15 ml of 8M NaOH (solution b) to it. The resulting concentration would be 4M. The calculation goes like this:

Concentration of Solution a C

_{a}- 1M

Volume of solution a V

_{a}- 20 ml

Concentration of Solution b C

_{b}- 8M

Volume of solution b V

_{b}- 15 ml

Final concentration C

_{f}- ?

Final volume V

_{f}- 35 ml

Since concentration is moles/volume of solution, the only problem here is to calculate the number of moles in the final volume.

Moles in solution a - 1M = 1 mole/L, which is 0.02 moles/20 ml

Moles in solution b - 8 M = 8 moles/L which is 0.12 moles/15 ml

Moles in final solution = 0.02 + 0.012 = 0.14 moles

Final concentration = 0.14 moles/35 ml = 4M

This can be expressed as: [itex]C_f = \frac{V_a C_a + V_b C_b}{V_f}[/itex]

[itex]C_f = \frac{(20ml)\left ( \frac{1mol}{1000ml} \right ) + (15ml)\left ( \frac{8mol}{1000ml} \right )}{35ml}=\frac{0.14mol}{35 ml}=4M[/itex]

Using this result I can simulate my original problem using NaOH. Let's say that I have 20 ml of NaOH and I would like to increase its concentration to 4M using an 8M NaOH stock solution. What would be the required volume necessary to achieve this?

Concentration of Solution a C

_{a}- 1M

Volume of solution a V

_{a}- 20 ml

Concentration of Solution b C

_{b}- 8M

Volume of solution b V

_{b}- ?

Final concentration C

_{f}- 4M

Final volume V

_{f}- ?

Now we have two unknowns, and this was the part that stumped me for a while. Fortunately, we have a second equation in there:

[itex]V_f = V_a + V_b[/itex]

so between that and

[itex]V_b = \frac{C_f V_f - V_a C_a}{C_b}[/itex]

we get

[itex]V_b = \frac{C_f (V_a + V_b) - V_a C_a}{C_b}[/itex]

and ultimately

[itex]V_b = \frac{C_f V_a - V_a C_a}{C_b - C_f}[/itex]

which solves the problem for any two solutions of known concentration.

(By the way, my chemistry training was almost 14 years ago, but I'm pretty sure that the 'formula' I arrived at is already known and commonly used, but I can't seem to find it. Does anyone know if it has a name or something? If not, dibs!)