Question

Boric acid, $\mathbf{B}(\mathbf{OH}{)}_3$, is an acid that acts differently from the usual Brønsted-Lowry acids. It reacts with water according to

localid="1663752682120" $$\begin{gathered} \mathbf{B}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\left(aq\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\ell \right)\mathbf{\rightleftarrows }\mathbf{B}\mathbf{(}\mathbf{OH}{\mathbf{)}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)} \\ {\mathbf{K}}_{\mathbf{a}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}} \end{gathered}$$

(a) Draw Lewis structures for$\mathbf{B}(\mathbf{OH}{)}_3$and$\mathbf{B}(\mathbf{OH}{)}_4^{-}$. Can these be described as Lewis acids or Lewis bases?
(b) Calculate the pH of a 0.20 M solution oflocalid="1663752366373" $\mathbf{B}\mathbf{(}\mathbf{OH}{\mathbf{)}}_{\mathbf{3}}\left(aq\right)$

Short answer

1. The Lewis structure for $\mathrm{B}(\mathrm{OH}{)}_3$is,

b. The Lewis structure for role="math" localid="1663752747631" $\mathrm{B}(\mathrm{OH}{)}_4^{-}$is,

The pH of a 0.20 M solution of $\mathrm{B}(\mathrm{OH}{)}_3$is 4.97.

Step-by-step solution

Step 1:Subpart (a)

The Lewis structure for $\mathrm{B}(\mathrm{OH}{)}_3$is,

It is a Lewis base because it can donate electron pairs to acid.

The Lewis structure for $\mathrm{B}(\mathrm{OH}{)}_4^{-}$is,

It is a Lewis base because it can donate electron pairs to acid.

Step 2:Subpart (b)

The reaction is:

$\mathrm{B}(\mathrm{OH}{)}_3+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{B}(\mathrm{OH}{)}_4^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$

The$K_a$ is:

$K_a=\frac{\left[\mathrm{B}(\mathrm{OH}{)}_4^{-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[\mathrm{B}{\left(\mathrm{OH}\right)}_3\right]}$

When we put values, we get:

$$\begin{gathered} 5.8\times {10}^{-10}=\frac{{\mathrm{x}}^2}{0.2-\mathrm{x}} \\ {\mathrm{x}}^2+5.8\times {10}^{-10}\mathrm{x}-1.16\times {10}^{-10}=0 \\ \mathrm{x}=1.08·{10}^{-5}\mathrm{M}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \end{gathered}$$

The pH is:

$$\begin{gathered} \mathrm{pH}=-\log \left(\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\right) \\ \mathrm{pH}=-\log \left(1.08·{10}^{-5}\right) \\ \mathrm{pH}=4.97 \end{gathered}$$