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Is there a simple reaction to yield boric acid ($\ce{H3BO3}$) from boron trifluoride ($\ce{BF3}$)?

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    $\begingroup$ "Quick questions" without elaboration effort are not very welcome, and may be closed. $\endgroup$ – Poutnik Oct 28 '20 at 11:03
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Yes, hydrolysis of boron trifluoride.

$$\ce{4 BF3 + 3 H2O → 3 HBF4 + B(OH)3}$$

The reaction commences with the formation of the aquo adduct, $\ce{H2O−BF3}$, which then loses $\ce{HF}$ leading to formation of the products.

The heavier trihalides do not undergo analogous reactions, possibly due to the low stability of the tetrahedral ions $\ce{BCl4-}$ and $\ce{BBr4-}$.

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    $\begingroup$ I can imagine a less weird way to prepare boric acid. :-) $\endgroup$ – Poutnik Oct 28 '20 at 14:21
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Wikipedia gives several methods to get boric acid. The first one uses a relatively common natural source of boron. Curiously compared with the other answer, boron trifluoride is excluded among the halides.

Boric acid may be prepared by reacting borax (sodium tetraborate decahydrate) with a mineral acid, such as hydrochloric acid:

$\ce{Na2B4O7·10H2O + 2 HCl → 4 B(OH)3 [or H3BO3] + 2 NaCl + 5 H2O}$

It is also formed as a by product of hydrolysis of boron trihalides and diborane:[1]

$\ce{B2H6 + 6 H2O → 2 B(OH)3 + 6 H2}$

$\ce{BX3 + 3 H2O → B(OH)3 + 3 HX (X = Cl, Br, I)}$

Upon further review, boron trifluoride can be used to give boric acid as well, as shown in the other answer, but at reduced yield as most of the boron forms $\ce{HBF4}$ instead. Using a heavier halogen favors all the boron going to boric acid.

Cited reference:

1. Housecroft, C. E.; Sharpe, A. G. (2008). "Chapter 13: The Group 13 Elements". Inorganic Chemistry (3rd ed.). Pearson. p. 340. ISBN 978-0-13-175553-6.

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