Standard solution, serial dilution

Question

Can someone please explain this for me?

I have made a standard solution using 325mg of solute and dissolving it in water to make it up to 25ml. the conc = 13mg/ml

Then I need to transfer 1ml of my standard solution to measure the absorbance of my known concentration.

Then it asks me to dilute my standard solution and make at least 4 known concentrations using 1 ml samples of each. So I guess that means 0.1ml of my stock and 0.9ml of distilled water to tube 2 and then from tube 2 0.1 ml + 0.9ml distilled water...so forth

However, I have already removed 1ml from the standard solution to a cuvette since I used it to measure my absorbance of known conc. Does that mean that when I dilute my standard now my V1=24ml? since I poured 1 ml out? How would I work out the new concentrations..?

I am a bit confused with this...

sorry and thank you.

Answer 1

The concentration of the original stock solution is 13 mg/ml. That doesn't change regardless of how much volume you remove. It is still 13 mg/ml, but you just have fewer mls of it. So, there are 2 ways to make dilutions of the original stock. One way is as you said, take 0.1 ml and add 0.9 ml to make solution #2. Then take 0.1 ml of solution #2 and add it to 0.9 ml H2O to make solution #3, etc., etc. This is called serial dilutions.

Another way is to take different volumes of the stock solution and dilute it that way. For example, you could take 0.1 ml original stock and add it to 0.9 ml H2O. Then take 0.2 ml original stock and add it to 0.8 ml H2O, etc. You could also take 0.05 ml of original stock and dilute it to 1.0 ml, etc.,etc. There are an infinite ways of making such dilutions. The choice is yours.

I hope the explains what you need to know. The bottom line and the answer to you question is, the concentration of the original stock DOES NOT CHANGE unless you add water to it, But as long as you just keep taking out of it, the concentration remains at 13 mg/ml.