Question

The solubility of MnSO_ \(\cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) is 70 \(\mathrm{g}\) per 100 \(\mathrm{mL}\) of water. (a) Is a 1.22 \(\mathrm{M}\) solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) saturated, supersaturated, or unsaturated? (b) Given a solution of MnSO_ \(\cdot \mathrm{H}_{2} \mathrm{O}\) of unknown concentration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

Short answer

The 1.22 M solution of MnS⁠O4·H₂O is unsaturated because its molarity is less than the solubility in molarity at \(20^{\circ} \mathrm{C}\) (4.094 M). To determine the saturation level of a MnS⁠O4·H₂O solution with an unknown concentration, you can perform an experiment by adding small amounts of solid MnS⁠O4·H₂O to the solution, filtering out undissolved solid, and evaporating the filtrate to half its volume before cooling it down to \(20^{\circ} \mathrm{C}\). Observe any precipitation or crystallization that occurs, and based on the amount of solid formed, you can determine if the original solution was unsaturated, saturated, or supersaturated.

Step-by-step solution

Convert the Solubility Information into Molarity

To compare the saturation of the given solution (1.22 M), we first need to convert the solubility of MnS⁠O4·H₂O, which is given in grams per 100 mL of water, into molarity. Use the molar mass of MnS⁠O4·H₂O to convert the solubility from grams to moles: Molar mass of MnS⁠O4·H₂O = 55.84 (Mn) + 32.07 (S) + 64.00 (4O) + 18.02 (H₂O) = 170.93 g/mol Now, convert the solubility into moles per 100 mL of water: Solubility in moles = (70 g / 170.93 g/mol) = 0.4094 mol per 100 mL of water Then, convert the solubility into molarity by using the volume of water: Solubility in Molarity = 0.4094 mol / 0.1 L = 4.094 M

Comparing the Given Solution with the Solubility Molarity

Now that we have the solubility of MnS⁠O4·H₂O in molarity (4.094 M), we can compare it to the given solution, which is 1.22 M. The saturation level of the given solution can be determined as follows: - Saturated: If the given solution molarity is equal to the solubility in molarity - Supersaturated: If the given solution molarity is greater than the solubility in molarity - Unsaturated: If the given solution molarity is less than the solubility in molarity Let's make the comparison: 1.22 M (given solution) < 4.094 M (solubility) Since the given solution molarity is less than the solubility in molarity, the given solution is unsaturated.

Proposing an Experiment to Determine the Saturation Level of an Unknown Solution

To determine the saturation level of a MnS⁠O4·H₂O solution with an unknown concentration, you can perform the following experiment: 1. Add small amounts of solid MnS⁠O4·H₂O to the unknown solution while stirring, until no more solid dissolves and a small amount of the solid remains at the bottom of the container. 2. Filter out the undissolved solid and measure the volume of the filtrate (solution). 3. Carefully evaporate the solution until it reaches half of its original volume, and then let it cool back to 20°C. 4. Observe any precipitation or crystallization that occurs as the solution cools: - If no solid forms, the original solution was unsaturated. - If a small amount of solid forms, the original solution was saturated. - If a large amount of solid forms, the original solution was supersaturated. With this experiment, you can determine whether the unknown MnS⁠O4·H₂O solution is saturated, supersaturated, or unsaturated.

Key concepts

Molarity Calculation

Understanding the concept of molarity is pivotal when discussing solution concentrations. Molarity, symbolized as M, is defined as the number of moles of a solute per liter of solution. To calculate molarity, one should divide the moles of the solute by the volume of the solution in liters. In a classroom scenario, this calculation helps students transition from theoretically knowing the molar concentration formula to applying it practically. For example, given the solubility of a substance in grams, we can calculate its molarity by obtaining the number of moles (using the substance's molar mass) and then translating these moles into molarity given the volume of the solution.

Using the exercise example, to determine if a 1.22 M solution of MnSO4·H2O is saturated, one needs first to convert the solubility form given in grams to molarity. The calculation involved finding the molar mass of MnSO4·H2O and then using the mass of the solute dissolved in 100 mL of water to find the moles. The resulting moles divided by the volume in liters provided the solubility in molarity, which was compared to the given solution's molarity.

Solution Saturation

Saturation of a solution is a key concept when it comes to understanding when a solution can no longer dissolve additional solute. A saturated solution holds the maximum amount of solute at a given temperature and pressure. If more solute can be dissolved, the solution is unsaturated. In contrast, a supersaturated solution contains more solute than it can theoretically hold at a given temperature and pressure – this state is achieved under specific conditions and is usually unstable.

To illustrate, if we consider the exercise’s 1.22 M solution of MnSO4·H2O and know the solubility molarity to be 4.094 M at 20°C, we can conclude that the solution is unsaturated because the concentration is less than what is required to achieve saturation at that temperature. This is an important distinction for students as it often leads to other discussions such as crystallization and solution dynamics under various conditions.

Solubility Experiments

Learning through experimentation enhances comprehension, and in the context of solubility, simple experiments can dramatically illustrate concepts like saturation levels. An experiment involves adding solute to a solvent until no more dissolves, indicating saturation. If a substance still dissolves, the solution is unsaturated; if precipitation occurs upon cooling or concentration, it was supersaturated.

In the context of our exercise, to identify the saturation point of an unknown MnSO4·H2O solution, you could gradually add solid MnSO4·H2O to the solution until it no longer dissolves, and then test for saturation by evaporating and cooling the solution. This hands-on approach gives students a tangible connection to the theoretical concepts they learn in textbooks and can be particularly engaging and revealing. It not only reinforces the theory but also encourages critical thinking and analysis as students predict and observe the outcomes of their experiments.