[Manie28]

Hi, I want to prepare a colloid solution with the following parameters:

- Volume: 10 ml
- Colloid concentration: 0,05 g/l
- ionic strength: 0,01 mol/l (adjusted with NaCl)
- Europium concentration: 10^-6 mol/l
- pH = 8 (adjusted with 0,01 M NaOH)

I work with a ICP Europium Standard. The Eu3+ concentration in the standard is 1000 mg/l which I calculated as 6,581 mmol/l. There is 2-3 % HNO3 in the standard.

If I neednt add Europium to the solution I would prepare the solution like that:

1) Fill 5 ml of a 0,1 g/l colloid suspension in a flask
2) Fill 5 ml of a 0,02 M NaCl Solution in the same flask
Thus, the colloids and the NaCl are corretly diluted.
3) Now I would add 0,01 M NaOH/HCl depending on the wanted pH.

My question is, how can I add the Eu3+ correctly? How is its influence on the ionic strength? Do I have to consider the Nitric acid? 

Please help me!!!!

[Borek]

Nothing very complicated - just use ionic strength definition. Note, that you have to use about 1.5 ?L of europium solution per 10 mL sample - you may safely ignore it and assume that it won't change a thing about ionic strength.

[Manie28]

I dont understand how you come to the 1.5 µl Europium per 10 ml sample. And what about the influence of the nitric acid on the ionic strength?
In my first post I did a list on how I would prepare the solution? Can you tell me how the Europium solution fits in there?

[Borek]

I dont understand how you come to the 1.5 µl Europium per 10 ml sample.

Have you tried to do volume calculations by yourself? They are on the HS level, you just use C=n/V (or classic dilution approach - C₁V₁=C₂V₂).

10 mL of 10^-6 M solution contains 10^-8 mole of Eu³⁺ - that's amount of the Europium you have to use. Your standard concentratoin is 0.006851 M. C=n/V so V=n/C, thus you need 10^-8/0.006851 = 1.5*10^-6 L or 1.5?L.

As for the ionic strength - I have gave you link to the definition. All you have to do now is to plug Europium concentration (10^-6) into it and check how does it compare with the 0.01 (required ionic strength).

[Manie28]

Have you tried to do volume calculations by yourself? They are on the HS level, you just use C=n/V (or classic dilution approach - C₁V₁=C₂V₂).

10 mL of 10^-6 M solution contains 10^-8 mole of Eu³⁺ - that's amount of the Europium you have to use. Your standard concentratoin is 0.006851 M. C=n/V so V=n/C, thus you need 10^-8/0.006851 = 1.5*10^-6 L or 1.5?L.

Thanks a lot for your answers until now.
Sorry for my (stupid) questions, but I don´t understand how you can say that 10 ml of 10^-6 M solution contains 10^-8 mole of Eu³⁺. The 10 ml solution should contain 10^-6 M Eu3+! How do you get to the 10^-8 mole Eu3+?

I would calculate that like this:

10 ml * 10^-6 M = x ml * 0,006851 M

x = 1,5 * 10^-3 ml = 1,5 µl

So we have the same result. With that small amount, can I just "titrate" it on the 10 ml solution? It would be only a very small change? And the ionic strength would have a very small change, too.

[Borek]

Sorry for my (stupid) questions, but I don´t understand how you can say that 10 ml of 10^-6 M solution contains 10^-8 mole of Eu³⁺. The 10 ml solution should contain 10^-6 M Eu3+! How do you get to the 10^-8 mole Eu3+?

You are missing conversion factors between mL and L.

I would calculate that like this:

10 ml * 10^-6 M = x ml * 0,006851 M

x = 1,5 * 10^-3 ml = 1,5 µl

So we have the same result. With that small amount, can I just "titrate" it on the 10 ml solution? It would be only a very small change? And the ionic strength would have a very small change, too.

Luckily for you with this approach conversion factors cancel out, that's why you get right answer.

No idea what do you mean by 'titrate' - so small volumes should be added with micropipette (Eppendorf pipette or how they are called in English) if not after dilution.

[Manie28]

Luckily for you with this approach conversion factors cancel out, that's why you get right answer.

No idea what do you mean by 'titrate' - so small volumes should be added with micropipette (Eppendorf pipette or how they are called in English) if not after dilution.

I meant that I would pipet the 1,5 µl with a micropipet "on" the 10 ml suspension. The error would be still quite small?!?!

[deinos]

Hi, Everyone can see good information about colloids in:
https://sites.google.com/site/joseluislopezfontan/