# How to Make a Standard Additions Curve?

1. Mar 20, 2013

### phpbyte

Hello,

I am wondering how to make a standard additions curve for this experiment. In this experiment, we used cyclic voltammetry to determine the original concentration of an elixir. We diluted the elixir to 0.75mM based on a claimed concentration of acetaminophen (instructed) and an elixir density of 1.1g/mL. We used 10mL of this dilution and measured the peak anodic current after each 20uL addition of 10mM standard solution of acetaminophen.

I know that I have to plot the points and create a linear regression that I will then use to solve for the x-intercept (concentration of acetaminophen present). Current is on the y-axis, but how do I calculate the x values of my points?

I know of two options that I could use for this particular experiment:
1. Use concentration
2. Use volume

I do not quite know how to do #1...I tried to multiply the original elixir molarity by 10mL to obtain moles of elixir and divide by 10mL + incremented volume (e.g. 1x = 75uL, 2x = 150uL, etc.), but since the additions were small, my molarity barely changed. I don't think this is right..?

#2 was fairly easy because I just used the incremented volumes. However, the x-intercept of this equation tells me simply the original volume of acetaminophen present. How do I find the starting concentration from this?
I have considered:
1. Multiplying the original elixir molarity by volume (0.75mM*10mL - with correct units, of course) and then dividing by the x-intercept value to obtain molarity. Does this work since I started with moles of elixir?
2. Dividing the x-intercept by 10mL (original volume) to get the ratio, which may give me the relative moles of acetaminophen? Then divide these moles by 10mL to obtain molarity (with correct units, of course). This was hinted to be the correct method by the professor, but I am not sure if it is valid.

Could someone please go through and explain both methods to me? I would like to know for this homework as well as for future reference in case I need to do this later on. Thank you in advance!

2. Mar 20, 2013

### Staff: Mentor

First of all, sometimes you speak of a concentration of "elixir" (whatever it is) and sometimes of a concentration of acetaminophen. Even if it is not confusing to you, it is confusing to me - I am not sure I know what you are doing and what you are trying to determine.

3. Mar 21, 2013

### phpbyte

I am very sorry for the confusion. We are trying to determine the concentration of acetaminophen present in children's pain reliever (the elixir in this case is the viscous, liquid form of medication that they can drink). The other concentrations of acetaminophen concern the 10mM standard solution of acetaminophen that we made so that we could add known quantities of known concentration of acetaminophen to the elixir. From this, extrapolating backwards should obtain the original concentration of acetaminophen present.

I have looked at https://www.physicsforums.com/showthread.php?t=533017 beforehand but it didn't describe the molarity method and I attempted something similar for volume, but alas, I ended up with pretty much the same values for each calculation (if I recall correctly).

4. Mar 21, 2013

### Staff: Mentor

So if I understand you correctly you have a V0 of liquid containing C0 of active substance, then you add V1 of solution containing C1 of active substance, and you want to know what is the final CF?

Check dilution and mixing explanation to see if it helps.

While you don't know C0 you should be able to express new concentrations using C0 as unknown, that will be enough to find it.

5. Mar 21, 2013

### phpbyte

Yes, that is correct.

Hmm...so I can simply use the M1V2=M2V2 (or c0V0 = cfVf).. I had applied it before but had gotten a very small number, so I thought it was the incorrect approach. I played around more with it and did it again, resulting in a much closer value to the theoretical.

I suppose the first M1V1=M2V2 accounted for the new molarity due to the additions and then the second application of the formula accounted for the dilution, since I diluted it to 0.5mM.

Thank you so much for your help, Borek!