Ti(III)citrate preparation
Titanium(III) citrate was prepared by adding 5 ml of a 15 percent titanium(III) chloride solution (Merck) to 50 ml of a 0.2M sodium citrate solution and neutralized with a saturated sodium carbonate solution. This agent was added in the required quantities to the oxygen-free medium.
https://science.sciencemag.org/content/sci/194/4270/1165.full.pdf
Dear Iram Aziz ,
I would say the calculation is pretty basic. The MW TiCl3 is 154,225 g/mol using the numbers as given in the method the calculation comes down to:
-5 ml of a 15 % titanium(III)chloride means 5 x 0.15 = 0,75 gram thus
0,75 : 154,225 = 0,0049 mol
-In a total of 55 mL (5 plus 50 mL) this means 0.0049 : 0.055 = 0,089 mol/L so roughly 0,1 M (so, there is a two-fold excess of citrate).
Best regards.
Hi Iram,
You may use the web tool to calculate
https://www.alfa-chemistry.com/solution-dilution-calculator.htm
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