Problem 86 - General Chemistry: Principles and Modern Applications

Question

You prepare \(1.00 \mathrm{L}\) of a buffer solution that is \(1.00 \mathrm{M}\) \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) and \(1.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4} .\) The solution is divided in half between the two compartments of an electrolytic cell. Both electrodes used are Pt. Assume that the only electrolysis is that of water. If 1.25 A of current is passed for 212 min, what will be the \(\mathrm{pH}\) in each cell compartment at the end of the electrolysis?

Step-by-Step Solution

Answer

The pH in both the cathode and anode compartments will be 7.21 at the end of the electrolysis.

Step 1: Calculate the Total Charge

The total charge Q is obtained by the formula Q = I x t, where I = 1.25 A is current and t = 212 min x 60 s/min = 12720 s is time. Substituting those values into the formula gives Q = 1.25 A x 12720 s = 15900 C.

Step 2: Calculate the Moles of Electrons

We know that each mole of electrons carries a charge of 1 Faraday which is approximately 96500 C/mol. Therefore, the number of moles of electrons transferred can be calculated by dividing the total charge by the charge of one mole of electrons. So, moles of electrons = 15900 C / 96500 C/mol = 0.1648 mol.

Step 3: Calculate the Moles of Hydrogen and Hydroxide Ions

Under these conditions, each mole of electrons reduces one mole of water molecules, producing two moles of hydrogen ions \(H^{+}\) at the cathode and an equivalent amount of hydroxide ions \(OH^{-}\) at the anode. Therefore, we have produced 0.1648 mol of \(H^{+}\) and \(OH^{-}\) ions each.

Step 4: Adjust the Buffer Concentrations

As the \(OH^{-}\) ions are produced, they react with the \(NaH_{2}PO_{4}\) to form \(Na_{2}HPO_{4}\) and water in the one compartment. Similarly, as \(H^{+}\) ions are produced, they react with the \(Na_{2}HPO_{4}\) to form \(NaH_{2}PO_{4}\) and water in the other compartment. Since we originally had 1 M concentration of each buffer in 0.5 L, we had 0.5 mol of each buffer. After electrolysis, the buffer concentrations adjust making \(Na_{2}HPO_{4}\) into 0.5 mol - 0.1648 mol = 0.3352 mol and \(NaH_{2}PO_{4}\) into 0.5 mol - 0.1648 mol = 0.3352 mol respectively, in the corresponding compartments.

Step 5: Calculate the pH values

Given \(pKa_{2}\) for \(H_{2}PO_{4^{-}}/HPO_{4^{2-}}\) is 7.21, we can now calculate the pH in each compartment using the Henderson-Hasselbalch equation. For the first compartment: pH = \(pKa_{2}\) + log([\(HPO_{4^{2-}}\)] / [\(H_{2}PO_{4^{-}}\)]) = 7.21 + log(0.3352 / 0.3352) = 7.21. For the second compartment: pH = \(pKa_{2}\) + log([\(HPO_{4^{2-}}\)] / [\(H_{2}PO_{4^{-}}\)]) = 7.21 + log(0.3352 / 0.3352) = 7.21. So, the pH of both compartments remains the same.

Key Concepts

Buffer Solutions

Buffer solutions are essential in chemistry due to their ability to resist changes in pH upon the addition of small amounts of acid or base. They are made up of a weak acid and its conjugate base or a weak base and its conjugate acid. For instance, in the original exercise, you have a buffer composed of \(\mathrm{NaH}_{2}\mathrm{PO}_{4}\) (a weak acid) and \(\mathrm{Na}_{2}\mathrm{HPO}_{4}\) (its conjugate base). These substances work together to maintain a stable pH in a solution.

• The weak acid can neutralize added bases, and
• The conjugate base can neutralize added acids.

This dual action keeps the pH level steady even when small amounts of other substances are introduced.
In the exercise, the electrolysis process partially alters the concentrations of these components. However, thanks to the buffer system, the overall pH change across the compartments is minimized.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a valuable tool for estimating the pH of a buffer solution. It models the relationship between the pH, the pK extsubscript{a} (a measure of the strength of the acid), and the ratio of the concentrations of the conjugate base to the acid.The equation is as follows:\[pH = pK_a + \log\left( \frac{[\mathrm{Base}]}{[\mathrm{Acid}]} \right)\]This equation allows you to predict how the pH will shift as you add or remove components in the buffer system. In our exercise:

• pKa extsubscript{2} is given as 7.21 for the \(\mathrm{H}_{2}\mathrm{PO}_{4}^- / \mathrm{HPO}_{4}^{2-}\) system
• After electrolysis, both the acid and base concentrations were equal (0.3352 M each)

Using the Henderson-Hasselbalch equation, you find that \(\log(0.3352 / 0.3352) = 0\), so the overall pH remains unchanged at 7.21. This demonstrates the effectiveness of the buffer in maintaining pH.

Electrolysis of Water

Electrolysis is a process where electrical energy is used to drive a non-spontaneous chemical reaction. When it comes to water, electrolysis breaks down water into its base elements, hydrogen and oxygen gas, at separate electrodes.During the electrolysis of water:

• The cathode releases hydrogen gas by reducing water molecules, producing hydrogen ions \(H^{+}\).
• At the anode, water molecules are oxidized, releasing oxygen gas and producing hydroxide ions \(OH^{-}\).

The exercise uses platinum electrodes, which are inert and excellent conductors of electricity, to facilitate this reaction.
This process affects the buffer solution. The production of \(H^{+}\) and \(OH^{-}\) ions consequently modifies the concentrations of the acids and bases in the buffer system, as these ions react to form water with each other, or they alter the existing acid and base balance. Despite this, careful management of these reactions can keep the pH stable, as seen in the exercise, ensuring that the buffer performs its function effectively.