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Dear Robert,

We typically use HF at a 1:6 to 1:8 molar ratio, depending upon the element. Therefore, convert the ppm of elements present to millimoles and multiply by the preferred coordination number to get a good estimate of the total HF added. For example, lets look at 1000 ug/mL of Ti and 1000 ug/mL of Mo.

Step 1:  Go to Inorganic Ventures' Interactive Periodic Table and click on each of these elements. You'll see that the preferred coordination number found for both Ti and Mo is 6.

Step 2:  A 1000 ug/mL solution of an element is equal to 1000 mg/L of that element, or it's equal to 1000/47.88 = 20.9 millimolar Ti and 1000/95.94 = 10.42 millimolar Mo. This yields a total of 31.3 millimoles of Ti + Mo, requiring 6 times the number of millimoles of HF = 187.8 millimoles HF or a total of 6.5 mL of conc (28.8M HF) HF present with each liter of 1000 ug/mL Ti and Mo multis.

Step 3:  The free amount of HF present can be estimated from the conditional formation constants of fluoride with the Ti and Mo. The amount of free HF should typically be at least 0.01 times the amount added assuming a rather weak 5^th and 6^th formation constant between the HF and the Mo and Ti. Please note that equilibrium calculations must take into account side reactions such as hydrolysis and the presence of excess hydronium ion (typically from nitric acid).

We have designed our HF Multi-element standards to contain a minimum of free HF. Please feel free to contact me with further questions.

Serving you in chemistry,
Paul R. Gaines, Ph.D.
CEO of Inorganic Ventures & Fellow Chemist

DISCLAIMER: Advice offered by the chemists at Inorganic Ventures is intended for the individual posing the question. Feel free to contact us to verify whether these suggestions apply to your unique circumstances.