An experiment requires 12.0 mL of a 0.100 M solution of HCl. If you have a 12.0 M HCl stock solution in the laboratory, how would you prepare enough 0.1 M solution to run at least 3 experiments?

Answer 1

Hello, Noah,

Each experiment requires 12.0 ml, so three experiments will require a total of 36.0 ml of 0.100M HCl. Assuming a plentiful supply of 12.0M HCl stock solution, we can calculate the amount required to prepare a minimum of 36.0 ml of 0.100 HCl solution by using the relationship:

M₁V₁ = M₂V₂

We want V₁, the amount of 6.0 M HCl required for 36.0 ml of 0.100M HCl.

V₁ = M₂V₂/M₁

V₁ = (0.100M)(36.0ml)/(6.0M)

V₁ = 0.6 ml of the 6.0M solution

30.0 ml of water should be added to 0.6 ml of the 6.0M solution to result in 36.0 ml of 0.100M HCl.

Bob

Answer 2

So for this question you need to use the following formula: M1V1 = M2V2. We know the volume (12.0 mL) and molarity (0.100 M) that we need for one experiment. However, because we need to do at least 3 experiments, we need to triple our volume to 36.0 mL, but our molarity does not change (0.100 M). I will use these two values for V1 and M1, respectively. Then, for M2, I will use the molarity of the standard solution, which is 12.0 M. Now I will plug these values into M1V1 = M2V2 to get the following: (0.100 M)(36.0 mL) = (12.0 M) (V2). Now I will solve for V2 to get 0.3 mL. What does this mean? Well, if I want to make 36.0 mL of a 0.100 M HCl solution, I have to do the following: add 0.3 mL of 12.0 M HCl to a beaker or graduated cylinder, and then fill the rest of the container with pure water UNTIL I get to a total volume of 36.0 mL (so you're adding 35.7 mL of pure water to get to 36.0 mL total). That's it! Hope I was able to help!