Question

\(\mathrm{CH}_{3} \mathrm{OH}\left(p K_{a}=15.5\right)\) would never exist as a methoxide ion \(\left(\mathrm{CH}_{3}-\mathrm{O}^{-}\right)\)dominantly at \(\mathrm{pH}\) (A) 2 (B) 7 (C) 10 (D) 14

Short answer

The pH at which methanol would dominantly exist as the methoxide ion is pH 14. Using the Henderson-Hasselbalch equation, only at pH 14 does the methoxide ion concentration surpass the methanol concentration. So, the correct answer is (D) 14.

Step-by-step solution

Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: pH = pKa + log (\(\frac{[A^-]}{[HA]}\)) Where pH is the pH of the solution, pKa is the acidity constant of the weak acid (in this case, CH3OH), [A-] is the concentration of the conjugate base (CH3O-), and [HA] is the concentration of the weak acid (CH3OH).

Analyze each pH option

Our goal is to determine the pH at which methanol would dominantly exist as methoxide ion, meaning the concentration of CH3O- is greater than CH3OH. Using the Henderson-Hasselbalch equation, we can determine the ratio of [A-] to [HA] at each pH value. Option A: pH = 2 Henderson-Hasselbalch equation: 2 = 15.5 + log (\(\frac{[CH_3O^-]}{[CH_3OH]}\)) At pH = 2, log (\(\frac{[CH_3O^-]}{[CH_3OH]}\)) is very negative, so the [CH_3O-] is much less than [CH_3OH]. Option B: pH = 7 Henderson-Hasselbalch equation: 7 = 15.5 + log (\(\frac{[CH_3O^-]}{[CH_3OH]}\)) At pH = 7, log (\(\frac{[CH_3O^-]}{[CH_3OH]}\)) is still negative, so the [CH_3O-] is less than [CH_3OH]. Option C: pH = 10 Henderson-Hasselbalch equation: 10 = 15.5 + log (\(\frac{[CH_3O^-]}{[CH_3OH]}\)) At pH = 10, log (\(\frac{[CH_3O^-]}{[CH_3OH]}\)) is still slightly negative, so [CH_3O-] is less than [CH_3OH], but the difference is not as significant as in previous options. Option D: pH = 14 Henderson-Hasselbalch equation: 14 = 15.5 + log (\(\frac{[CH_3O^-]}{[CH_3OH]}\)) At pH = 14, log (\(\frac{[CH_3O^-]}{[CH_3OH]}\)) is positive, so [CH_3O-] is greater than [CH_3OH].

Determine the correct pH

From our analysis, only at pH = 14 does the methoxide ion concentration surpass the methanol concentration, meaning the CH3OH would exist dominantly as CH3O- at this pH. So the correct answer is (D) 14.

Key concepts

pKa and pH Relationship

The relationship between pKa and pH is a cornerstone of understanding acid-base balance in chemistry. pKa is the negative logarithm of the acid dissociation constant (Ka), which indicates the strength of an acid in solution. A lower pKa value signifies a stronger acid that dissociates more completely in water.

In contrast, pH measures the acidity or basicity of a solution on a scale from 0 to 14, with lower values being more acidic. The Henderson-Hasselbalch equation bridges these two concepts by helping to determine the degree of ionization of an acid in a particular pH environment:

\[\text{pH} = \text{pKa} + \log \left(\frac{[A^-]}{[HA]}\right)\]

This equation is instrumental when comparing the concentrations of the acid \([HA]\) and its conjugate base \([A^-]\). For instance, a higher pH than the pKa results in a greater concentration of the base \([A^-]\), whereas a lower pH leads to a more predominant acid concentration.

Conjugate Acid-Base Pair

In any acid-base reaction, the concept of a conjugate acid-base pair is essential. A conjugate base is what remains after an acid donates a proton (H+), whereas a conjugate acid is formed when a base accepts a proton.

Conjugate acid-base pairs are linked through the removal or addition of a single proton. For example, in the exercise, methanol (\(\mathrm{CH}_{3} \mathrm{OH}\)) acts as the acid, and when it loses a proton, it forms the methoxide ion (\(\mathrm{CH}_{3}-\mathrm{O}^-\)), which is its conjugate base.

This concept is pivotal because it helps us understand how reagents transform in chemical reactions, influencing equilibrium and reaction dynamics. As pH changes, these pairs shift back and forth between their acid and base forms, which is seen in buffer solutions where a weak acid and its conjugate base resist drastic changes in pH.

Organic Chemistry Acid-Base Reactions

In organic chemistry, acid-base reactions often appear in the form of proton transfer reactions. These reactions are fundamental, affecting the properties and behaviors of organic molecules.

One key aspect is the ability of different functional groups in organic compounds to donate or accept protons, influencing the overall reactivity. For example, alcohols like methanol can act as weak acids, as demonstrated in the exercise. Methanol's conversion to the methoxide ion involves losing a proton and forming a more reactive species.

Understanding these reactions in the context of the Henderson-Hasselbalch equation allows chemists to predict the form that such compounds will take under varying pH conditions. As evident in the exercise, the ambient pH determines whether methanol remains largely un-ionized or whether it converts to its methoxide form. This dynamic is central to many practical applications, such as designing drugs, optimizing chemical syntheses, and developing materials that function under specific pH conditions.