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I have been searching around and have found varying answers to the question. What exactly would reaction of $\ce{H3BO3}$ and $\ce{NaOH}$ give? Shouldn't it be a neutralisation reaction as $\ce{H3BO3}$ is an acid and $\ce{NaOH}$ is a base?

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This isn't something you could know except by experiment, but boric acid doesn't exist as solely $\ce{H3BO3}$ in solution. In addition to being an Arrhenius acid, $\ce{H3BO3}$ is also a Lewis acid. This means it is capable of accepting a lone pair, and will undergo the following reactions in aqueous solution:

$$\ce{B(OH)3 + H2O <=> B(OH)4- + H+}\tag1\label{Eq1}\\$$

$$\ce{4B(OH)4- + 2H+ <=> B4O7^{2−} + 9H2O}\tag2\label{Eq2}\\$$

By combining \eqref{Eq1} and \eqref{Eq2}, we get the full reaction:

$$\ce{4B(OH)3 <=> B4O7^{2−} + 2H+ + 5H2O }\tag3\label{Eq3}\\$$

Boric acid does not produce tetraborate exclusively, however, and many other species species are also present in solution. The system is actually so complex that it remains difficult to characterize, one paper mentioning that there are some ten different equilibrium reactions in solution$^{[1]}$. In performing an acid-base calculation, however, it is okay to simply treat it as a solution of $\ce{H2B4O7}$, neutralizing according to the equation:

$$\ce{H2B4O7 + 2NaOH <=> Na2B4O7 + 2H2O}\tag4\label{Eq4}\\$$

Titration confirms that $\ce{H3BO3}$ reacts with $\ce{NaOH}$ in a $\text{2:1}$ ratio, and crystallization of the product confirms $\ce{Na2B4O7}$ is produced.


$[1]$ Trejo G., Frausto-Reyes C., Gama S.C., Meas Y., Orozco G. Raman Study of Benzylideneacetone on Silver. Int. J. Electrochem. Sci. 2012, 7, 8436–8443.

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  • $\begingroup$ Please note that the sodium tetraborate solution actually contains the $\ce{[B4O5(OH)4]^2-}$ anion. $\endgroup$ – vapid Oct 18 '16 at 7:37
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All reaction are given here:http://chemiday.com/search/?q=H3bo3+%2B+naoh&lang=en&m=stuff

  1. $$\ce{H3BO3 + NaOH → Na[B(OH)4]}$$

Orthoboric acid react with sodium hydroxide to produce sodium tetrahydroxoborate(III). Sodium hydroxide - saturated solution.

  1. $$\ce{4H3BO3 + 2NaOH → Na2B4O7 + 7H2O}$$

Orthoboric acid react with sodium hydroxide to produce sodium tetraborate and water. Sodium hydroxide - diluted solution.

  1. $$\ce{4H3BO3 + 2NaOH + 3H2O ⇄ Na2B4O7•10H2O}$$

Orthoboric acid react with sodium hydroxide and water to produce decahydrate sodium tetraborate.

  1. $$\ce{H3BO3 + NaOH ->[350-400 C] NaBO2 + 2H2O}$$

Orthoboric acid react with sodium hydroxide to produce sodium metaborate and water. This reaction takes place at a temperature of $\pu{350-400\!^\circ C}$.

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There are conflicting interpretations for the origin of the acidity of aqueous boric acid solutions. Raman spectroscopy of strongly alkaline solutions has shown the presence of B(OH)− 4 ion,[3] leading some to conclude that the acidity is exclusively due to the abstraction of OH− from water:[3][4][5][6]

or more properly expressed in the aqueous solution:

This may be characterized as Lewis acidity of boron toward OH−, rather than as Brønsted acidity.

However other sources say that boric acid is also a tribasic Brønsted acid, with successive ionization steps:

Since the value of Ka1 is comparable to that of the reaction with OH−, the concentrations of BO(OH)− 2 and B(OH)− 4 are similar.

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  • $\begingroup$ could someone help format my answer $\endgroup$ – Vedant Oct 18 '16 at 8:06
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    $\begingroup$ I think you'll find this meta post quite useful. $\endgroup$ – ringo Oct 18 '16 at 8:20

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