Dear fellow researchers:
I need to prepare 1.6% chromic acid for the purpose of AAO etching. The amount of information available online to make chromic acid is overwhelming. I also could not find any balanced equations which can help me to calculate how much of potassium dichromate has to be reacted with how much sulfuric acid to make a 100mL solution. Any suggestions or input will be helpful.
Please note that 1 mol of potassium dichromate in presence of at least 1:2 mol/mol of sulfuric acid can be taken as equivalent to 2 mol of chromic acid referred to as given by the formula CrO3: K2Cr2O7 + 2 H2SO4 → 2 CrO3 + 2 KHSO4 + H2O. Supposing that chromic acid concentration is given as referred to CrO3 formula ― indeed, the considered formula should be specified ― 1.6 wt% should refer to 1.6 g of CrO3 hypothetically added to (100 - 1.6) g of aq. solution, or (1.6 / MWCrO3) mol of CrO3 added to that solution. Indeed, no CrO3 is actually added, but rather half its molar amount as K2Cr2O7: i.e. x = ½(1.6 / MWCrO3)·MWK2Cr2O7 g of K2Cr2O7. This quantity should thus be added to 100 - x g of solution. This solution should consist on (1.6 / MWCrO3) mol of H2SO4 dissolved in water, or y = (1.6 / MWCrO3)·MWH2SO4 g of H2SO4, supposing that no excess of H2SO4 should subsist in the final solution. Hence, the concentration of the sulfuric acid solution should be given by [y/(100 - x)]·100 wt%. The density of H2SO4 aq. sol. can be easily found (cf. e.g. Perry's Chemical Engineers' Handbook), so that we can also obtain the equivalent volume of this initial solution, in which x g of K2Cr2O7 should be dissolved. Note also that the final volume of the solution obtained as mentioned would not exactly match 100 cm3, since the suggested calculation considered a 100 g basis for that solution.
Thanks Dr. Carlos. I realized that if I start with chromium oxide (CrO3), the process will be much simple. Thus I decided to just make an aqueous solution of CrO3 which will be chromic acid. Nevertheless, thank you very much for the informative comment.
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