Increasing the concentration of a solution - Diff Eq?

• DavitosanX
In summary, the conversation was about a chemistry problem where the final concentration needed to be adjusted to 200 mM. The person had prepared a 250 mM solution and had to add more ATP to reach the desired concentration, but this would alter the volume. They considered representing this problem as a differential equation and plotted a sigmoid curve to find the final concentration. They then used a known formula to solve for the final volume needed to achieve the desired concentration. The formula is commonly used and involves solving a system of linear equations.
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!)

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.

1. What is the purpose of increasing the concentration of a solution?

Increasing the concentration of a solution involves adding more solute to a given amount of solvent. This can be done for various reasons, such as to make a solution more potent or to achieve a desired reaction rate.

2. How is the concentration of a solution calculated?

The concentration of a solution is typically calculated by dividing the amount of solute by the volume of the solution. This can be expressed in different units such as molarity, molality, or mass percent.

3. What is the effect of increasing the concentration of a solution on the rate of reaction?

Increasing the concentration of a solution can lead to a higher rate of reaction. This is because the higher concentration of solute particles increases the chances of collisions between reactants, leading to a faster reaction rate. However, there is a limit to how much the concentration can increase before it becomes saturated and does not further affect the reaction rate.

4. How does temperature affect the concentration of a solution?

Temperature can affect the concentration of a solution in several ways. For some solutions, an increase in temperature can lead to an increase in solubility, thus increasing the concentration. However, in other cases, temperature can cause a decrease in concentration if the solute becomes less soluble at higher temperatures.

5. What are some methods for increasing the concentration of a solution?

Some common methods for increasing the concentration of a solution include evaporation, distillation, and reverse osmosis. These techniques involve removing some of the solvent from the solution, leaving behind a more concentrated solution. Another method is to simply add more solute to the solution until the desired concentration is reached.

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