Problem 112 - Chemistry: The Central Science in SI Units, Global Edition

Question

In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(s)\) half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and \(0.050 \mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right)\). The measured cell voltage is \(1.030 \mathrm{~V}\). What is the \(\mathrm{p} K_{\mathrm{a}}\) of benzoic acid?

Step-by-Step Solution

Answer

The pKa of benzoic acid in this galvanic cell is approximately -7.48.

Step 1: Find the potential of the Ag+/Ag half-cell

The Nernst equation is used to calculate the potential of a half-cell at a given concentration. The equation is: \[ E_{cell} = E^\circ - \frac{0.05916}{n} \log Q \] where \(E_{cell}\) is the potential, \(E^\circ\) is the standard potential, \(n\) is the number of electrons transferred in the redox reaction, and \(Q\) is the reaction quotient. For the Ag+/Ag half-cell, the standard potential \(E^\circ = 0.7996\ V\), and the number of electrons transferred \(n = 1\), since the reduction reaction is: \[ \mathrm{Ag}^{+} + 1e^{-} \rightarrow \mathrm{Ag}(s) \] The reaction quotient, \(Q\), for the reduction reaction is: \[ Q = \frac{[\mathrm{Ag}^{+}]}{[\mathrm{Ag}(s)]} \] Since the activity of a solid is defined as 1, we can rewrite the reaction quotient as: \[ Q = [\mathrm{Ag}^{+}] \] We are given that the concentration of Ag+ ions is 1.00 M. Plugging the values into the Nernst equation, we get: \[ E_{\mathrm{Ag}^{+}/\mathrm{Ag}} = 0.7996 - \frac{0.05916}{1} \log(1) = 0.7996\ V \]

Step 2: Calculate the potential of the hydrogen half-cell

Since we know the overall cell voltage (\(1.030\ V\)), we can use the cell voltage equation: \[ E_{cell} = E_{cathode} - E_{anode} \] Plugging in the values we have found, we get: \[ 1.030 = 0.7996 - E_{anode} \] Solving for the anode potential, we get: \[ E_{anode} = 0.7996 - 1.030 = -0.2304\ V \]

Step 3: Calculate the pH of the solution

Now, we need to relate the potential of the hydrogen half-cell to the pH of the solution. For the hydrogen half-cell, the standard potential \(E^\circ = 0\ V\), and the number of electrons transferred \(n = 2\), since the reduction half-reaction is: \[ 2\mathrm{H}^{+} + 2e^{-} \rightarrow \mathrm{H}_{2}(g) \] Under standard conditions, \(\frac{[\mathrm{H}^{+}]}{P_{\mathrm{H}_{2}}}\) should be 1. However, we need to consider the non-standard conditions which can be done using the following formula at \(25\degree\)C: \[ E_{anode} = - \frac{0.05916}{2} \log([\mathrm{H}^{+}]) \] We have calculated the anode potential, so we can solve for the concentration of H+ ions: \[ -0.2304 = - \frac{0.05916}{2} \log([\mathrm{H}^{+}]) \] This gives us: \[ \log([\mathrm{H}^{+}]) = 7.78 \] and now we can calculate pH of the cell: \[ \mathrm{pH} = -\log([\mathrm{H}^{+}]) = -7.78 \]

Step 4: Calculate the pKa of benzoic acid

Using the Henderson-Hasselbalch equation, we can find the pKa of benzoic acid by plugging in the concentrations of the benzoate and the benzoic acid: \[ \mathrm{pH} = \mathrm{p} K_{a} + \log \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH}]} \] Substitute the pH value and concentrations of benzoate and benzoic acid in the equation: \[ -7.78 = \mathrm{p} K_{a} + \log \frac{0.050}{0.10} \] Now we can solve for the pKa of benzoic acid: \[ \mathrm{p} K_{a} = -7.78 - \log \frac{1}{2} \approx - 7.48 \] Therefore, the pKa of benzoic acid in this galvanic cell is approximately -7.48.

Key Concepts

Nernst equation

The Nernst equation is an essential tool in electrochemistry, used to determine the potential of an electrochemical cell under non-standard conditions. It connects to the cell's ability to perform work as the concentration of reactants and products changes. The equation is expressed as:
\[\ E_{cell} = E^\circ - \frac{0.05916}{n} \log Q\ \]where:

• \(E_{cell}\) is the cell potential under non-standard conditions.
• \(E^\circ\) is the standard cell potential.
• \(n\) is the number of electrons exchanged in the reaction.
• \(Q\) is the reaction quotient, representing the ratio of products to reactants at any point in time.

To apply the Nernst equation correctly, it is crucial to identify the correct reduction and oxidation reactions occurring in the cell. For a galvanic cell consisting of a silver cathode \(\mathrm{Ag}^+/\mathrm{Ag}(s)\), we specifically look at the reduction where silver ions gain an electron to become solid silver. The calculated potential under different concentrations allows us to understand how the reactions behave as conditions differ from the standard state.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is widely used in chemistry to estimate the pH of a buffer solution. It provides a simple way to connect the pH with the pKa of an acid and the concentrations of the acid and its conjugate base. The equation is given by:
\[\ \mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)\ \]where:

• \(\mathrm{pH}\) is the measure of acidity in the solution.
• \(\mathrm{p}K_a\) is the negative logarithm of the acid dissociation constant, reflecting the strength of the acid.
• \([\mathrm{A}^-]\) is the concentration of the conjugate base.
• \([\mathrm{HA}]\) is the concentration of the protonated acid form.

This equation is particularly useful for understanding how an acid-base equilibrium will respond to changes in concentration, pH, or pKa. In the context of the original exercise, knowing the pH and the ratio of salt to acid allowed a precise calculation of the pKa for benzoic acid. Such calculations are fundamental in chemistry, especially when preparing solutions with specific pH levels.

Galvanic cell potential

The galvanic cell potential is a measure of the voltage difference generated by a galvanic or voltaic cell. In these cells, spontaneous redox reactions occur between two half-cells to convert chemical energy into electrical energy. The overall cell potential \(E_{cell}\) is calculated as the difference between the cathode and anode potentials:
\[\ E_{cell} = E_{cathode} - E_{anode}\ \]Here, the cathode is where the reduction reaction takes place, while the anode is where oxidation occurs. Understanding the cell potential is crucial as it indicates the cell's ability to drive an electric current through an external circuit. Each half-cell's potential depends on the materials used and their respective standard potentials.
In the provided exercise, the silver half-cell serves as the cathode, while the hydrogen electrode solution acts as the anode. By knowing the standard potentials and calculating any deviations due to non-standard conditions using the Nernst equation, one can accurately determine the voltage available from the cell. Thus, the galvanic cell potential gives insights into the system's efficiency and effectiveness for generating electric current.