Chromatography Forum

I recently found a recipe for borate buffer as follows:
Borate buffer, pH 8.5
A) Boric acid H3BO3, 12.4g/l (MW 61.
B) Sodium tetraborate, 19.1g/l (MW 381.4)

Add 50ml of A) to 14.5 ml of B) and bring up to 200ml with deionised water. Adjust to pH 8.5 with NaOH (3M).

All of this is fine but I can't work out the concentration of the buffer due to the tetraborate - I have beeb receiving different information about what the concentration is.

Can anyone shed any light on the subject?
Thanks

"Tetraborate" is, as you might expect, a tetramer of boric acid. It forms spontaneously when you crystallize a sodium borate solution. In fact, "borate" in solution exists as an equilibrium mixture of oligomers. So, if the molarity of borate is important, moles = weight(Na2B4O7 10H2O)*4/381.4.

Personally, I think that the buffer recipe is far too much work. Just compute the moles of boron and sodium in the final solution and weigh out the required amount of H3BO3 and add the required amount of 50% NaOH. With a bit of care, you can hit the pH every time without any titration.

Thanks for that - I'll give it a try

But I have to warn you. Since we started using of borate buffers for HPLC we observe more frequent failures of HPLC pumps (inlet and outlet valves).
The mobile phase is filtered before using and system is carefully flushed with water/organic 1:1 after analytical sequence.
I suppose that microcrystals are formed in bottle during analysis.

Regards

Unfortunately, the quality of sodium tetraborate is often poor. Making it from boric acid and sodium hydroxide is faster and has fewer impurities.

Borate salts are quite abrasive. If you have piston seal wash on your pump, use it. If not, upgrade if possible. It is good practice to flush any nonvolatile salts out of the pump before turning it off.