The protocol specifies to add sodium citrate to TiCl3 (0.5g, 3.24mmol). But this has confused me. How much TiCl3 do I use? I have a stock solution that is 15% TiCl3 (Sigma, product number 1401). The instructions are quite confusing to me.
I could calculate the volume of my stock solution that would contain 0.5g of TiCl3, but I dont understand where the 3.24mmol comes in.
Any help would be greatly appreciated.
The description that I am referencing has been attached as an image and is from the following paper:
Titanium(IV) Citrate Speciation and Structure under Environmentally and Biologically Relevant Conditions | Inorganic Chemistry (acs.org)
TiCl3 molar mass is 154.225 g/mol. In order to convert grams to moles you have to divide the mass (grams) by molar mass. (0.5 /154.225=0.00324 moles). Your quantity is expressed in milimole not mole (1 mole=1000 milimole), this means you have 0.00324 x 1000=3.24 mmol of TiCl3
You have got a solution with a concentration expressed as w/w. 15% w/w means your solution contains 15g of TiCl3 (0.082 mol) for 100 g of solution. You need to know the density to convert it to mol/L, and then you can calculate the volume you need to get 3.24 mmol. If density is - as it should - 1.21g/mL (at 20 °C), then you have that 100 g of your solution correspond to a volume of 82.6 mL (82.6·10–3 L). Hence, your molar concentration is C = 8.2·10–2 mol / 82.6·10–3 L = 0.99 M. Since C = n/V, the volume corresponding to 3.24 mmol is V = 3.24·10–3 mol/0.99M = 3.27 mL.
This having been said, of course your recipe requires a weighed mass of 0.5 g of solid TiCl3. By the way, sigma-Aldrich also sells solid TiCl3 (cat. No. 514381).
I hope this ah help!
You just need to make the concentration ratio of Cit:Ti equal to 2 to 2.1 . Donot worry about the concentration of TiCl3 you used.
- Badges
- Science topic
- Similar topics
- Physical Sciences
- Materials Science
- Materials Chemistry
- Polymer Chemistry