If you had the same stock of the blue solution as in the serial dilution simulation you did (1M solution), and you needed 10 mL of a solution that was 1.0x10-6 M, what volume of the blue stock solution would you need, and what volume of diluent would you need to reach 10 mL of the desired concentration

We start by doing a 1/10 serial dilution, using 100 µL of 1M solution into 900 µL of water, until we get a 1.0x10^-5M (0.00001M) solution. Then use 1 mL of this 1.0x10^-5M solution into 9 mL of water.

Explanation:

To answer this question, we must use the law of conservation of mass, which states: In every chemical reaction mass is conserved, this means the total mass of the reactants is equal to the total mass of the products. The law implies that mass can neither be created nor destroyed, but it can be transformed. For example, in chemical reactions, the mass of the chemical components before the reaction is equal to the mass of the components after the reaction. Therefore, during any chemical reaction and low-energy thermodynamic processes in an isolated system, the total mass of the reactants or starting materials must be equal to the mass of the products. This law is quite accurate for low-energy processes, such as chemical reactions.

So, if the solution to be used is 1M , and a 10 mL of a 1.0x10^-6M solution is needed, we use the following equation:

Initial concentration x initial volume = final concentration x final volume.

The initial concentration is 1M, the final concentration is 1.0x10^-6M and the final volume required is 10 mL.

1M x initial volume = 1.0x10^-6M x 10 mL

initial volume=  1.0x10^-5 mL= 0.01 µL of 1M solution.

Since the final volume is 10 mL, we have to add the difference in volume with water, which is 10 mL - 1.0x10^-5 mL= 9.99999 mL.

However, since 0.01 mL is a very small volume that is difficult to take, the best option in this case is to make serial dilutions.

Usually, we start from a concentrated solution and prepare a series of dilutions to the tenth (1:10) or half (1:2). In this way a series of solutions is obtained, related for example by a dilution factor of 10, i.e. 1/10; 1/100; 1/1000 and so on.

Here we can prepare a series of dilutions to the tenth, from 1M to 1.0x10^-6.

We start by doing a 1/10 dilution, using 100 µL of 1M solution into 900 µL of water. This is a 0.1M solution. Then we take 100 µL of it into 900 µL of water to get a 0.01M solution. We continue doing that until we get a 1.0x10^-6M (0.000001M) solution. This final solution is the desired concentration, however we need 10 mL of it, and actually we have 1 mL. So we can just take 1 mL of the 1.0x10^-5M solution into 9 mL of water:

1.0x10^-5M x 1 mL = 1.0x10^-6M x 10 mL.

the answer is learned