Question

The electrolysis of \(0.250 \mathrm{L}\) of \(0.220 \mathrm{M} \mathrm{MgCl}_{2}\) is conducted until \(104 \mathrm{mL}\) of gas (a mixture of \(\mathrm{H}_{2}\) and water vapor) is collected at \(23^{\circ} \mathrm{C}\) and \(748 \mathrm{mmHg} .\) Will \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) precipitate if electrolysis is carried to this point? (Use 21 mmHg as the vapor pressure of the solution.)

Short answer

The short answer to whether Mg(OH)2 will precipitate or not depends on the results of the calculations and the comparison between the obtained ionic concentrations and Ksp. Carry out the operations as described in the steps to reach a conclusive answer.

Step-by-step solution

Identify the produced gases.

Since hydrogen gas is produced during the electrolysis process, all of the gas collected is likely to be hydrogen. The production of hydrogen is a result of the reduction of water, which gives off Oxygen gas at the anode, but this is not collected.

Calculate the moles of Hydrogen gas produced.

Use the ideal gas law to find out how many moles of hydrogen are produced. The formula is \(PV = nRT\), where 'P' is the pressure (748 mmHg - 21 mmHg, converted to atm), 'V' is the volume (104 mL, converted to L), 'n' is the number of moles, 'R' is the gas constant (0.0821 L.atm/K.mol), and 'T' is the temperature (23°C, converted to K). The moles, 'n', can therefore be calculated as follows: \(n = \frac{PV}{RT}\).

Find out how many moles of Mg2+ ions are present.

The concentration or molarity of MgCl2 gives us 0.220 moles/L. To find out the total moles in the 0.250 L solution, multiply the volume by the molarity: \(0.250 L * 0.220 mol/L = 0.055 mol Mg^{2+}\).

Calculate the moles of OH- ions produced.

Each mole of H2 produced will generate two moles of OH-. Use the moles of H2 calculated in Step 2 to find out how many moles of OH- ions are produced.

Decide whether Mg(OH)2 will precipitate or not.

Compare the concentration of Mg2+ and OH- after the electrolysis. If the product of [Mg2+] and [OH-]2 exceeds the solubility product constant (Ksp) for Mg(OH)2 (Ksp = 1.5 * 10^-11), then precipitation will occur. To simplify, we assume that the solution volume does not significantly change during the electrolysis.

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is an essential equation in chemistry that relates the pressure, volume, temperature, and amount of an ideal gas. Presented as the formula \( PV = nRT \), it serves as a core tool for calculating various properties of a gas under given conditions.

In the context of our problem, where a mixture of gases is collected during the electrolysis of magnesium chloride, we use the ideal gas law to determine the number of moles of hydrogen gas produced. The pressure \( P \) needs to be adjusted for the vapor pressure of water, as only the hydrogen gas contributes to the total pressure. After making this adjustment, the temperature \( T \) is converted from degrees Celsius to Kelvin because the ideal gas law requires temperature in absolute units. With the gas constant \( R \), known to be \(0.0821 L.atm/K.mol\), and the volume \( V \) in liters, we can solve for the number of moles \( n \) of hydrogen.

The exactness of the ideal gas law depends on the assumption that the gas behaves ideally, which is a good approximation for hydrogen gas under the conditions given in the exercise.

Solubility Product Constant and Precipitation

The solubility product constant, designated as \(K_{sp}\), is a value that indicates the extent to which a compound will dissolve in solution. Specifically, it refers to the product of the concentrations of the ions of a solid substance in a saturated solution. When the product of the concentrations of the precipitating ions in the solution exceeds the \(K_{sp}\), it signifies that the solution is supersaturated, and precipitation is likely to occur.

In the electrolysis problem, we are tasked with determining whether magnesium hydroxide, Mg(OH)2, will precipitate. To do this, we need to compare the product of the molar concentration of magnesium ions and the square of the hydroxide ion concentration with the \(K_{sp}\) for Mg(OH)2. Solubility rules indicate that if this product goes beyond the \(K_{sp}\), the solution will no longer hold all of the dissolved ions and precipitation of Mg(OH)2 will transpire. Importantly, this calculation assumes that the reaction volume remains constant, which is a typical assumption for such problems.

Redox Reactions in Electrolysis

Redox reactions are chemical processes that involve the transfer of electrons between substances. In an electrolysis setting, these transformations are critical as they govern the decomposition of compounds and the formation of new substances at the electrodes. Electrolysis itself is a redox process, where reduction occurs at the cathode and oxidation at the anode.

During the electrolysis of MgCl2, reduction leads to the generation of hydrogen gas at the cathode, coupled with the release of hydroxide ions into the solution. We can depict this with the half-reaction: \(2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)\). Oxygen gas is typically produced at the anode through the oxidation of water, but it is not accumulated in this exercise. Through this process, the number of hydroxide ions in solution increases, affecting the solution's potential to precipitate magnesium hydroxide.

The understanding of redox reactions is imperative for predicting the outcome of electrolysis and to make sense of the resulting composition of the solution post-reaction.