Problem 115 - Chemistry: The Central Science

Question

Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{pK}_{a}\) of butyric acid is \(4.84\). (a) Calculate the \(\mathrm{pK}_{b}\) for the butyrate ion. (b) Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of butyric acid. (c) Calculate the pH of a \(0.050 \mathrm{M}\) solution of sodium butyrate.

Step-by-Step Solution

Answer

(a) The pKb for the butyrate ion is approximately 9.16. (b) The pH of the 0.050 M solution of butyric acid is approximately 4.61. (c) The pH of the 0.050 M solution of sodium butyrate is approximately 4.67.

Step 1: (a) Calculate the pKb for the butyrate ion.

The relationship between the pKa and pKb of a conjugate acid-base pair is given by the following equation: \[pK_a + pK_b = 14\] To find the pKb, we can rearrange the equation as: \[pK_b = 14 - pK_a\] Now, plug in the given pKa value: \[pK_b = 14 - 4.84 = 9.16\] So, the pKb for the butyrate ion is 9.16.

Step 2: (b) Calculate the pH of a 0.050 M solution of butyric acid.

To determine the pH of the butyric acid solution, we can use the following equation, known as the Henderson-Hasselbalch equation: \[pH = pK_a + \log{\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}}\] For the butyric acid, \([\mathrm{A}^-]\) is the concentration of the butyrate ion, while \([\mathrm{HA}]\) is the concentration of the butyric acid. This equation tells us the pH of the solution given the acid dissociation constant (or pKa), and the concentrations of the acid and its conjugate base. Since we know the pKa of butyric acid (4.84) and the concentration of butyric acid (0.050 M), we can plug the values into the equation to find the pH: \[pH = 4.84 + \log{\frac{[\mathrm{A}^-]}{0.050}}\] However, we are not given the concentration of the butyrate ion, \([\mathrm{A}^-]\), directly. We can make a simplifying assumption that the concentration of butyrate ion will be equal to the decrease in the concentration of butyric acid, as it dissociates. Since the dissociation is in the form HA <=> H+ + A-, we can write it as: \[0.050 - x\] Where 'x' is the concentration of the butyrate ion, as well as hydrogen ions. Thus, the equation becomes: \[pH = 4.84 + \log{\frac{x}{0.050 - x}}\] We can solve for x by assuming that Ka (the acid dissociation constant) remains constant: \[K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}\] \[K_a = \frac{x^2}{0.050 - x}\] Using the pKa value, we can determine the Ka value: \[\mathrm{K}_{a} = 10^{-\mathrm{pKa}} = 10^{-4.84} = 1.445\times10^{-5}\] Now, substitute the Ka value into the equation and solve for x: \[1.445\times10^{-5} = \frac{x^2}{0.050 - x}\] We discard the larger root as it would be illogical for the concentration. Thus, x = 0.002018. Now, plugin the x value in the pH equation to find the pH value: \[pH = 4.84 + \log{\frac{0.002018}{0.050 - 0.002018}} \approx 4.61\] The pH of the 0.050 M solution of butyric acid is approximately 4.61.

Step 3: (c) Calculate the pH of a 0.050 M solution of sodium butyrate.

Sodium butyrate is a salt that dissociates to give sodium ions (Na+) and butyrate ions (A-). In such cases, we can use the Henderson-Hasselbalch equation to calculate the pH. We can rewrite the equation in terms of pKb and concentration of conjugate base, butyrate ion, and the conjugate acid, which is butyric acid (HA), as follows: \[pOH = pK_b + \log{\frac{[\mathrm{HA}]}{[\mathrm{A}^-]}}\] We know the pKb value (9.16) and the concentration of sodium butyrate (0.050 M). As sodium butyrate dissociates completely in solution, the concentration of butyrate ion will be equal to the initial concentration of sodium butyrate: \[[\mathrm{A}^-] = 0.050\] We can then use the Kb value to determine the concentration of butyric acid, HA: \[K_b = \frac{[\mathrm{HA}][\mathrm{OH}^-]}{[\mathrm{A}^-]}\] Using the pKb value, we can determine the Kb value: \[K_b = 10^{-\mathrm{pK}_{b}} = 10^{-9.16} = 7.24\times10^{-10}\] Now, substitute the Kb value and the concentration of butyrate ion: \[7.24\times10^{-10} = \frac{[\mathrm{HA}][\mathrm{OH}^-]}{0.050}\] We assume that the concentration of butyric acid ([HA]) is equal to the concentration of hydroxide ions ([OH-]), which is 'x'. Therefore, the equation becomes: \[7.24\times10^{-10} = \frac{x^2}{0.050}\] Solving for x, we find: \[x = 1.201\times10^{-5}\] We can now calculate the pOH using the given pKb value and the ratio of butyric acid to butyrate ion concentrations: \[pOH = 9.16 + \log{\frac{1.201\times10^{-5}}{0.050}} \approx 9.33\] Finally, we know that pH + pOH = 14. So, the pH of the 0.050 M solution of sodium butyrate is: \[pH = 14 - pOH = 14 - 9.33 = 4.67\] The pH of the 0.050 M solution of sodium butyrate is approximately 4.67.

Key Concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is crucial for anyone studying acid-base chemistry. This formula provides a direct way to calculate the pH of a buffer solution when the concentration of the acid and its conjugate base are known.

It is expressed as follows: \[ pH = pK_a + \log{\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}} \] Here, \( pK_a \) is the negative base-10 logarithm of the acid dissociation constant \( K_a \), \( [\mathrm{A}^-] \) represents the concentration of the conjugate base, and \( [\mathrm{HA}] \) is the concentration of the acid.

This equation comes from the rearrangement of the dissociation constant expression for weak acids and bases and assumes that the concentrations of the acid and its conjugate base do not change significantly compared to their initial values (which is a valid assumption for buffered solutions).

By using this equation, students can understand the buffering capacity of solutions and predict how the pH will change with varying amounts of acid and base.

pKa and pKb Relationship

The relationship between \( pK_a \) and \( pK_b \) provides insight into the strength of conjugate acid-base pairs. These values are inversely related, and their sum is always constant for a given solvent at a particular temperature. In water at 25°C, the sum is 14: \[ pK_a + pK_b = 14 \] To understand this relationship, students should recognize that when an acid is strong (low \( pK_a \)), its conjugate base will be weak (high \( pK_b \)), and vice versa. This is due to their respective abilities to donate or accept protons.

Kinetically, when an acid donates a proton easily, the reverse process for its conjugate base to take up a proton is less favorable, hence the higher \( pK_b \). This concept helps in predicting the behavior of substances in different pH environments and is essential in buffer calculations.

Acid Dissociation Constant

The acid dissociation constant, \( K_a \), is a quantitative measure of the strength of an acid in solution. It reflects the equilibrium concentration of a weak acid and its dissociated ions. A higher \( K_a \) value indicates a stronger acid because it reflects a greater tendency to lose its proton and form its conjugate base.

The mathematical expression for \( K_a \) from a general dissociation reaction: \[ HA \leftrightarrow H^+ + A^- \] is: \[ K_a = \frac{[H^+][A^-]}{[HA]} \] where \( [H^+] \), \( [A^-] \), and \( [HA] \) are the molar concentrations of hydrogen ions, conjugate base, and undissociated acid, respectively, in equilibrium. Since \( pK_a = -\log K_a \), a lower \( pK_a \) means a stronger acid. Understanding how to calculate and apply \( K_a \) helps predict the extent of dissociation and the resulting pH of a solution.

pH Calculation

pH calculation is a fundamental process in understanding the acidity or basicity of a solution, which is crucial in disciplines like chemistry, biology, and environmental science. The pH is calculated by taking the negative logarithm of the hydrogen ion concentration: \[ pH = -\log [H^+] \] For a weak acid or a weak base solution, the pH can be indirectly determined through various equilibrium expressions involving \( K_a \) or \( K_b \).

In the context of butyric acid and sodium butyrate solutions, knowing how to manipulate the dissociation equilibria and the corresponding constants is essential. Equilibrium calculations assume that initial concentrations are slightly adjusted due to the small amount of ionization. This assumption simplifies the calculations while providing a reasonably accurate estimate of the solution's pH. By mastering pH calculations, students can predict the behavior of different solutions during titrations, understand buffer systems, and handle various biochemical and industrial processes.