Problem 46 - Chemistry The Science in Context

Question

The \(K_{\mathrm{a}}\) of proline is \(2.5 \times 10^{-11}\) in water, \(2.8 \times 10^{-11}\) in an aqueous solution that is \(28 \%\) ethanol, and \(1.66 \times 10^{-8}\) in aqueous formaldehyde at \(25^{\circ} \mathrm{C}\) a. In which solvent is proline the strongest acid? b. Rank these compounds on the basis of their strengths as Bronsted-Lowry bases: water, ethanol, and formaldehyde.

Step-by-Step Solution

Answer

Answer: Proline is the strongest acid in aqueous formaldehyde. The ranking of Bronsted-Lowry bases, from strongest to weakest, is Formaldehyde, Ethanol, and Water.

Step 1: Compare the Kₐ values of proline in the three different solvents: water, 28% ethanol, and aqueous formaldehyde. Ultimately, the solvent with the highest Kₐ value will produce proline with the strongest acidic properties. In water: Kₐ = 2.5 x 10⁻¹¹ In 28% ethanol: Kₐ = 2.8 x 10⁻¹¹ In aqueous formaldehyde: Kₐ = 1.66 x 10⁻⁸ #Step 2: Identify the solvent with the highest Kₐ value#

The highest Kₐ value for proline is 1.66 x 10⁻⁸, which occurs in aqueous formaldehyde. Therefore, proline is the strongest acid in aqueous formaldehyde. Now we will answer part a: In which solvent is proline the strongest acid? - Proline is the strongest acid in aqueous formaldehyde. #Step 3: Rank compounds as Bronsted-Lowry bases#

Step 2: To rank water, ethanol, and formaldehyde as Bronsted-Lowry bases, we need to examine how their acidity (Kₐ values) changes when proline is dissolved in each solvent. The greater the increase in Kₐ, the stronger the Bronsted-Lowry base. Water: (2.5 x 10⁻¹¹) / (2.5 x 10⁻¹¹) = 1 28% ethanol: (2.8 x 10⁻¹¹) / (2.5 x 10⁻¹¹) = 1.12 Aqueous formaldehyde: (1.66 x 10⁻⁸) / (2.5 x 10⁻¹¹) = 6640 #Step 4: Compare the ratios and rank the Bronsted-Lowry bases#

The higher the ratio, the stronger the Bronsted-Lowry base: 1) Formaldehyde (6640) 2) Ethanol (1.12) 3) Water (1) Now we will answer part b: Rank these compounds on the basis of their strengths as Bronsted-Lowry bases. - The ranking, from strongest to weakest, is Formaldehyde, Ethanol, and Water.

Key Concepts

Bronsted-Lowry Base Strength

In the context of acid-base chemistry, the strength of a Bronsted-Lowry base is determined by its ability to accept protons. Strong bases are efficient proton acceptors and typically possess a high affinity for protons, while weak bases have a lower propensity for proton acceptance.

When a base accepts a proton, it is converted into its conjugate acid. The strength of a base can inversely relate to the strength of its conjugate acid; a very strong base has a weak conjugate acid. This is because, in equilibrium, the stronger the base, the less likely it will give up the proton it has gained.

Using the given rankings:

• Formaldehyde
• Ethanol
• Water

we see that formaldehyde is the strongest Bronsted-Lowry base among the three. This is attributed to its higher ratio when analyzing the changes in Ka values upon the addition of proline. The larger increase in Ka signifies a stronger base, as more of the proline is dissociated into its conjugate acid form.

Ka (Acid Dissociation Constant)

The acid dissociation constant (Ka) plays a pivotal role in understanding the strength of an acid. It quantifies the extent to which an acid donates protons to water, forming its conjugate base and hydronium ions (H3O+). Mathematically, for a generic acid HA, the equation is represented as:\[\mathrm{HA} \rightleftharpoons \mathrm{A}^- + \mathrm{H}_3\mathrm{O}^+\]and the Ka is given by:\[K_a = \frac{\left[\mathrm{A}^-\right]\left[\mathrm{H}_3\mathrm{O}^+\right]}{\left[\mathrm{HA}\right]}\]where [A-], [H3O+], and [HA] are the equilibrium concentrations of the conjugate base, hydronium ions, and the undissociated acid, respectively.

The larger the Ka value, the stronger the acid—that is, a greater proportion of the acid molecules donate protons. Conversely, a smaller Ka value denotes a weaker acid, with fewer proton donations. Thus, by examining the Ka values of proline in different solvents, we deduce its relative acid strength in those environments.

Proline Acidity

Proline, an amino acid, exhibits varied acidity depending on the solvent environment. This property is critical in biochemistry, as amino acids may act differently in different parts of cells or tissues, impacting their biological activity.

The acidity of proline is expressed through its Ka value, a measure of how readily proline donates a proton to the solvent, forming its conjugate base. As shown:

• Water: Ka = 2.5 x 10⁻¹¹
• 28% Ethanol: Ka = 2.8 x 10⁻¹¹
• Aqueous Formaldehyde: Ka = 1.66 x 10⁻⁸

These values indicate that proline has the highest acidity in aqueous formaldehyde, because the Ka value is the highest. The interaction between proline and the solvent molecules significantly affects its ability to act as an acid, highlighting the role of molecular environment on acidity.

Solvent Effects on Acidity

Solvent effects on acidity are crucial for understanding the behavior of acids and bases in different chemical environments. The polarity and ability of the solvent to stabilize ions can significantly impact an acid's dissociation tendency. For instance, in a highly polar solvent, an acid's protons are more likely to dissociate due to the solvent's ability to stabilize the resulting ions.

Comparing the Ka values of proline in different solvents, we observe the impact of solvent composition on acidity. The trend we note:

• Aqueous Formaldehyde: Ka = 1.66 x 10⁻⁸
• 28% Ethanol: Ka = 2.8 x 10⁻¹¹
• Water: Ka = 2.5 x 10⁻¹¹

suggests that the ability of the solvent to stabilize the conjugate base of proline dictates its acidity. Aqueous formaldehyde, with the highest Ka value, indicates a stronger interaction that stabilizes the conjugate base more effectively than water or ethanol, thus enhancing the acidity of proline.