The autoprotonation constant for the solvent ethanol, CH3 CH2 OH(l), is 8 ×10^-20 M^2 at 25^∘ C.

step 1 First, we'll start with the equilibrium expressing the auto -protonation of ethanol. This has it

The autoprotonation constant for the solvent ethanol, CH3...

The autoprotonation constant for the solvent ethanol, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)$, is $8 \times 10^{-20} \mathrm{M}^{2}$ at $25^{\circ} \mathrm{C}$. The autoprotonation equilibrium equation is $2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \leftrightharpoons$ $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+}(a l c)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}(a l c) $$ where (alc) denotes a solute in alcohol solution. (a) We define a $\mathrm{pH}$ scale in alcohol by the equation $$ \mathrm{pH}=-\log \left(\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{*}\right] / \mathrm{M}\right) $$ Calculate the $\mathrm{pH}$ of a neutral alcohol solution at $25^{\circ} \mathrm{C} .$ (b) Calculate the $\mathrm{pH}$ of a $0.010$ -M solution of sodium ethoxide, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} \mathrm{Na}^{+}(a l c)$, in alcohol at $25^{\circ} \mathrm{C}$. Assume that $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} \mathrm{Na}^{+}(a l c)$ is completely dis- sociated in alcohol.

The autoprotonation constant for the solvent ethanol, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)$, is $8 \times 10^{-20} \mathrm{M}^{2}$ at $25^{\circ} \mathrm{C}$. The
autoprotonation equilibrium equation is
$2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \leftrightharpoons$
$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+}(a l c)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}(a l c)
$$
where (alc) denotes a solute in alcohol solution.
(a) We define a $\mathrm{pH}$ scale in alcohol by the equation
$$
\mathrm{pH}=-\log \left(\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{*}\right] / \mathrm{M}\right)
$$
Calculate the $\mathrm{pH}$ of a neutral alcohol solution at $25^{\circ} \mathrm{C} .$
(b) Calculate the $\mathrm{pH}$ of a $0.010$ -M solution of sodium ethoxide, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} \mathrm{Na}^{+}(a l c)$, in alcohol at $25^{\circ} \mathrm{C}$.
Assume that $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} \mathrm{Na}^{+}(a l c)$ is completely dis-
sociated in alcohol.

Key Concepts

Autoprotonation Constant

The autoprotonation constant (or autoprotolysis constant) defines the equilibrium concentration of ions produced when a solvent self-ionizes. In this context, it is analogous to the ion product of water (Kw) in aqueous solutions but is specific to ethanol, indicating how much the neutral ethanol molecules contribute to forming positively and negatively charged species at equilibrium.

pH Scale in Non?Aqueous Solvents

The pH scale in a non?aqueous solvent like ethanol is defined in a way similar to aqueous solutions, by taking the negative logarithm of the concentration of the protonated solvent species. However, because the autoprotonation constant in ethanol is much smaller than in water, the neutral point on this pH scale is shifted, reflecting the unique acid–base properties and ionization behavior of the solvent.

Equilibrium and Proton Transfer in Alcohol

In alcohol solvents, the equilibrium between the protonated and deprotonated species arises from the self-ionization reaction of the solvent. This proton transfer equilibrium is governed by the autoprotonation constant, and any changes in the concentrations of the species involved will shift the equilibrium according to Le Chatelier's principle. Understanding this equilibrium is crucial for predicting the response of the solution when external acids or bases are introduced.

Influence of Strong Base Addition

The addition of a strong base, such as sodium ethoxide in ethanol, disrupts the established proton transfer equilibrium by providing additional deprotonated species. This causes the equilibrium to shift in order to counteract the change in concentration, significantly altering the pH of the solution. Analyzing the equilibrium shift in response to strong base addition highlights the interplay between added solutes and the inherent self-ionization properties of the solvent.

Transcript

Please give Ace some feedback