Question

Proline has \(\mathrm{pK}_{\mathrm{a} 1}=1.99\) and \(\mathrm{p} K_{\mathrm{a} 2}=10.60 .\) Use the HendersonHasselbalch equation to calculate the ratio of protonated and neutral forms at \(\mathrm{pH}=2.50 .\) Calculate the ratio of neutral and deprotonated forms at \(\mathrm{pH}=9.70\)

Short answer

The ratio at pH 2.50 is 3.24, and at pH 9.70 it's 7.69.

Step-by-step solution

Identify Given Values

We have the following values: \(\mathrm{pK}_{\mathrm{a}1} = 1.99\), \(\mathrm{pK}_{\mathrm{a}2} = 10.60\). We will use each \(\mathrm{pK}_a\) value with the corresponding pH (\(2.50\) and \(9.70\) respectively) in the Henderson-Hasselbalch equation.

Calculate Ratio for Protonated and Neutral Forms

Using the equation: \[ \text{pH} = \mathrm{pK}_{\mathrm{a}1} + \log \left( \frac{[\text{neutral form}]}{[\text{protonated form}]} \right)\]Substitute \(\mathrm{pH} = 2.50\) and \(\mathrm{pK}_{\mathrm{a}1} = 1.99\):\[ 2.50 = 1.99 + \log \left( \frac{[\text{neutral form}]}{[\text{protonated form}]} \right) \]Solve for the ratio:\[ \log \left( \frac{[\text{neutral form}]}{[\text{protonated form}]} \right) = 2.50 - 1.99 = 0.51 \]\[ \frac{[\text{neutral form}]}{[\text{protonated form}]} = 10^{0.51} \approx 3.24 \]

Calculate Ratio for Neutral and Deprotonated Forms

Using the equation: \[ \text{pH} = \mathrm{pK}_{\mathrm{a}2} + \log \left( \frac{[\text{deprotonated form}]}{[\text{neutral form}]} \right)\]Substitute \(\mathrm{pH} = 9.70\) and \(\mathrm{pK}_{\mathrm{a}2} = 10.60\):\[ 9.70 = 10.60 + \log \left( \frac{[\text{deprotonated form}]}{[\text{neutral form}]} \right) \]Solve for the ratio:\[ \log \left( \frac{[\text{deprotonated form}]}{[\text{neutral form}]} \right) = 9.70 - 10.60 = -0.90 \]\[ \frac{[\text{deprotonated form}]}{[\text{neutral form}]} = 10^{-0.90} \approx 0.13 \]So, \[ \frac{[\text{neutral form}]}{[\text{deprotonated form}]} \approx \frac{1}{0.13} \approx 7.69 \]

Key concepts

pKa Values

In the realm of acid-base chemistry, understanding \( pKa \) values is essential. The \( pKa \) represents the acid dissociation constant and is a vital indicator of how easily a molecule donates hydrogen ions (H\(^+\)). Lower \( pKa \) values mean the acid more readily loses its proton. For example, in our exercise, Proline has \( mathrm{pK}_{ ext{a}1} = 1.99 \) and \( mathrm{pK}_{ ext{a}2} = 10.60 \). A \( pKa \) of 1.99 suggests stronger acidic properties than a \( pKa \) of 10.60.
Knowing the \( pKa \) values, students can predict the degree of ionization of acids and bases at different pH levels, which is crucial for calculating the ratio of protonated to deprotonated forms using the Henderson-Hasselbalch equation.
Using these values effectively allows chemists to determine how a substance will behave in varying environments, which is critical in fields like pharmacology and biochemistry.

Protonated and Deprotonated Forms

Protonated and deprotonated forms describe the state of a molecule in relation to hydrogen ions (H\(^+\)). A protonated form means that a molecule has accepted H\(^+\), while a deprotonated form means that it has lost H\(^+\). This concept is fundamental in acid-base chemistry because it directly impacts the chemical behavior of molecules.
In the context of our exercise, understanding whether a molecule like Proline is in a protonated or deprotonated state can help predict its interaction with its environment. For instance, at different pH levels, Proline shifts between these forms, affecting not only the molecule's charge and reactivity but also the resulting solution's properties.

• In acidic environments (lower pH), Proline is more likely to be protonated.
• In basic environments (higher pH), it tends to be deprotonated.

Demonstrating these transitions through calculations offers a practical insight into real-world chemical reactions.

pH Calculations

pH calculations are a cornerstone of acid-base chemistry, allowing us to describe the acidity or basicity of a solution. When we calculate pH, we are essentially determining the concentration of hydrogen ions in a solution.
To accurately predict the behavior of molecules, like those transitioning between protonated and deprotonated forms, we often utilize the Henderson-Hasselbalch equation. This equation is \( ext{pH} = ext{pK}_a + ext{log} rac{[ ext{conjugate base}]}{[ ext{acid}]} \)
For our exercise, this calculation helped derive the ratio between forms of Proline at different pH levels, using its known \( pKa \). Consistently using this formula allows students to translate theoretical knowledge into practical problem-solving skills, providing clarity on how biological systems maintain their acid-base balance.

Acid-Base Chemistry

Acid-base chemistry examines the nature and interactions of acids and bases. It is a broad field that touches on numerous practical applications, such as metabolic processes and industrial chemical reactions.
At its core, acid-base chemistry focuses on how substances donate or accept protons, which is essential for understanding biological pathways. This is where the Henderson-Hasselbalch equation becomes an indispensable tool, allowing chemists to quantify and visualize these proton exchanges.
In our scenario with Proline, recognizing the significance of its \( pKa \) values and its ability to exist in protonated and deprotonated forms elucidates how this compound may behave inside living organisms or other systems. By mastering these concepts, students not only deepen their understanding of chemistry but also enhance their ability to anticipate and manipulate chemical reactions in various contexts.