# Increasing the concentration of a solution - Diff Eq?

1. Mar 3, 2017

### DavitosanX

I have a question regarding a problem in chemistry. I ultimately solved it, but I wanted to know if the problem itself could be represented as a differential equation, and if the solution I found is the solution to that equation. The explanation is a little long, but I don't think I can make it shorter without losing clarity.

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 Ca - 1M
Volume of solution a Va - 20 ml
Concentration of Solution b Cb - 8M
Volume of solution b Vb - 15 ml
Final concentration Cf - ?
Final volume Vf - 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: $C_f = \frac{V_a C_a + V_b C_b}{V_f}$
$C_f = \frac{(20ml)\left ( \frac{1mol}{1000ml} \right ) + (15ml)\left ( \frac{8mol}{1000ml} \right )}{35ml}=\frac{0.14mol}{35 ml}=4M$

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 Ca - 1M
Volume of solution a Va - 20 ml
Concentration of Solution b Cb - 8M
Volume of solution b Vb - ?
Final concentration Cf - 4M
Final volume Vf - ?

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

$V_f = V_a + V_b$

so between that and

$V_b = \frac{C_f V_f - V_a C_a}{C_b}$

we get

$V_b = \frac{C_f (V_a + V_b) - V_a C_a}{C_b}$

and ultimately

$V_b = \frac{C_f V_a - V_a C_a}{C_b - C_f}$

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!!)

2. Mar 3, 2017

### Staff: Mentor

You basically get a system of linear equations to be solved simultaneously. As always, you need as many equations as unknowns to have a unique solution (if it exists).

Since your system was of size 2, you could easily solve it by hand. For bigger systems, there exists some systematic methods for that, which you can lookup in a textbook on linear algebra.