Problem 16 - Biochemistry: A Short Course

Question

For an acid HA, the concentrations of HA and \(\mathrm{A}^{-}\) are 0.075 and \(0.025,\) respectively, at \(\mathrm{pH} 6.0 .\) What is the \(\mathrm{p} K_{\mathrm{a}}\) value for \(\mathrm{HA} ? 6\)

Step-by-Step Solution

Answer

The pK_a value is approximately 6.48.

Step 1: Identifying the formula

To find the pK_a value, we can use the Henderson-Hasselbalch equation: \[ \mathrm{pH} = \mathrm{pK_a} + \log \left(\frac{[A^-]}{[HA]}\right) \] where \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the acid.

Step 2: Substitute known values

We are given \( \mathrm{pH} = 6.0 \), \([HA] = 0.075\), and \([A^-] = 0.025\). Substitute these values into the equation to get: \[ 6.0 = \mathrm{pK_a} + \log \left(\frac{0.025}{0.075}\right) \]

Step 3: Calculate the log term

Calculate the log of the concentration ratio: \[ \log \left(\frac{0.025}{0.075}\right) = \log \left(\frac{1}{3}\right) \approx -0.4771 \]

Step 4: Solve for pK_a

Substitute the log value back into the equation and solve for \( \mathrm{pK_a} \): \[ 6.0 = \mathrm{pK_a} - 0.4771 \] \[ \mathrm{pK_a} = 6.0 + 0.4771 \] \[ \mathrm{pK_a} = 6.4771 \]

Step 5: Result

The calculated \( \mathrm{pK_a} \) value for the acid HA is approximately 6.48.

Key Concepts

Acid-Base Chemistry

Acid-base chemistry is a fascinating subject that revolves around the interaction between acids and bases. In simple terms, an acid is a substance that can donate a proton (H+ ion), while a base is one that can accept a proton. This exchange of protons is what characterizes acid-base reactions.

In any aqueous solution, you'll commonly encounter acids and bases in two forms: as molecules (like HA) and their corresponding ions (like \( ext{A}^- \)). These represent the acid and its conjugate base. In our example, HA is the acid that donates a proton to become \( ext{A}^- \).

Understanding the equilibrium state between acids and bases is key. It helps us comprehend the balance and shifting nature of these reactions. This is what ultimately leads us to predict the pH of solutions, among other crucial parameters. Acid-base chemistry is essential in fields like biochemistry, environmental science, and pharmaceuticals, where the functionality and reactivity of molecules hinge on acid-base properties.

pH Calculations

pH calculations are essential in determining the acidity or basicity of a solution. The pH scale estimates how acidic or basic a substance is, with values ranging from 0 (most acidic) to 14 (most basic), where 7 is neutral.

The formula to determine pH in a solution is:

• pH = -log([H+])

This equation represents the negative logarithm of the hydrogen ion concentration \([H^+]\) in a solution.

In practice, pH calculations often involve using the Henderson-Hasselbalch equation for buffered solutions. This equation provides a direct link between pH, \( ext{pK}_a \), and the ratio of the concentration of the base form to the acid form of a compound:

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

This powerful equation allows us to calculate pH given the concentrations of acids and their conjugate bases, or solve for other variables, like \( ext{pK}_a \), making it a cornerstone of pH calculations in chemistry.

pKa Determination

Determining the \( ext{pK}_a \) value of an acid is an important concept in acid-base chemistry. The \( ext{pK}_a \) provides insight into the strength of an acid; it is the pH at which an acid is half dissociated. A lower \( ext{pK}_a \) indicates a stronger acid that dissociates more completely in solution.

To calculate \( ext{pK}_a \), you can rearrange and use the Henderson-Hasselbalch equation:

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

Given the pH, the concentrations of the acid \([HA]\) and its conjugate base \([A^-]\), you can solve for \( ext{pK}_a \). For example, if the pH is 6.0 and the concentrations of HA and \( ext{A}^- \) are 0.075 M and 0.025 M respectively, you would solve the equation:

• \( 6.0 = \text{pK}_a + \log \left( \frac{0.025}{0.075} \right) \)

Upon calculation, this results in \( \text{pK}_a = 6.48 \).

Knowing the \( ext{pK}_a \) of acids in a solution helps chemists understand reactivity, buffer capacity, and the shifting equilibrium of reactions. This \'roadmap\' is particularly useful for predicting how solutions behave in different chemical and biological contexts.