Preparing solution of low solubility organics in water

1. Nov 19, 2012

FaNgS

I'm making a stock solution of xylene in water.

The solubility of xylene in water is 200 mg/L (Andrews and Keefer 1949). The density of xylene is 864 mg/mL.

Making 1 L stock solution is too much and a waste.

0.23mL (or 230µL) of xylene is equivalent to 198.72mg. To make the stock solution, if I add the 0.23mL of xylene to 50mL of water the concentration is 3974.40 mg/L.

This is where I'm confused and I feel silly.

I'm getting confused between the 198.72mg of xylene and the stock concentration 3974.40mg/L. Which one should I be looking at?

2. Nov 19, 2012

Staff: Mentor

Something is wrong. You can't have a stock solution with concentration higher than the solubility.

3. Nov 19, 2012

FaNgS

I think I figured out that I should be looking at the concentration of the solution rather than the mass of the analyte.

After a bit of number crunching this is what I got:

If I add 57µL which is equivalent to 49.248mg of xylene and make it up to 250mL of water I get a solution concentration of 196.99mg/L.

Is this right?

4. Nov 19, 2012

Staff: Mentor

Looks OK, but I wonder if it makes sense.

Are you aiming at the saturated solution? If so, take some excess of xylene (don't waste your time measuring exactly 57 µL) - whatever dissolves, dissolves, whatever is left, will float on the surface. Just use separating funnel to make sure you have just solution without xylene on the surface, otherwise taking samples will be difficult.

5. Nov 19, 2012

FaNgS

You have an excellent point.

Now here's something else that is weird.

Lets say I make a 250mL stock solution of xylene with a concentration of 190mg/L (slightly lower than the max solubility!).

If I want to make a 500mL solution with a concentration of 100mg/L using the stock solution. Using C1 * V1 = C2 * V2 where C1 and V1 are the concentration and volume of the stock solution, respectively, and C2 and V2 are the concentration and volume of the standard solution which I want to make.
According to my calculations I would need 263mL of stock. Is the only way around this to make a stock solution of higher volume (> 250mL)?

6. Nov 19, 2012

Staff: Mentor

Yes. Nothing weird here.

7. Nov 19, 2012

FaNgS

Thank you. I'm just having one of those nights.

One last thing. Is there any other way besides using a burette to accurately transfer large liquid volumes between 100-400mL?

8. Nov 19, 2012

Staff: Mentor

How accurately?

A typical large burette is 100 mL. It is possible to buy a larger one, but it can be either difficult or expensive. These volumes are more often measured with a cylinder.

9. Nov 20, 2012

FaNgS

I've calculated how much of stock solution I would need to make lower concentration solutions and looking at the volumes and glassware capacities at my disposal I would need to transfer liquids several times to get the required amount before diluting which would add errors.

Would it be better to make solutions with the desired concentrations I require rather than diluting a higher concentrated stock solution?

10. Nov 20, 2012

Staff: Mentor

No simple answer to this question, as it depends on the concentrations you need, glass at hand and accuracy required. But as long as your stock concentration is not much higher than the concentration of the final solutions (say you need 100 mg/L and your stock is 200 mg/L) making stock can be a waste of time.

11. Nov 21, 2012

FaNgS

I haven't done this before but what do you think about measuring the density of my stock solution and calculating the mass required to be diluted?

12. Nov 21, 2012

Staff: Mentor

Density of your stock solution will be almost exactly that of water.

13. Nov 21, 2012

FaNgS

Is it frowned upon to use mass rather than volumes to transfer liquids?

14. Nov 21, 2012

AGNuke

Volume of liquid is something you can measure easily compared to mass, so I guess...

15. Nov 21, 2012

Staff: Mentor

No, but - as AGNuke mentioned - volume measurements are usually easier.