Question

The solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) 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 \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) of unknown concentration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

Short answer

The solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ}\mathrm{C}\) can be calculated by first determining its molecular weight and then converting the given solubility in grams to moles per liter. After comparing the solubility to the given concentration of \(1.22\,\mathrm{M}\), we can determine if the solution is saturated, supersaturated, or unsaturated. To determine the saturation state of an unknown solution experimentally, you can add a small amount of solid \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) to the solution and observe if it dissolves completely (unsaturated), partially (saturated), or if crystals re-form (supersaturated).

Step-by-step solution

Calculate the solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in moles per liter

First, we need to determine the solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in moles per liter. We know that \(70\,\mathrm{g}\) of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) dissolve in \(100\,\mathrm{mL}\) of water at \(20^{\circ} \mathrm{C}\). To convert this to moles per liter, we will first need to find the molecular weight of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\): Molecular weight = (Mass of Mn) + (Mass of S) + (4 * Mass of O) + (2 * Mass of H) + (Mass of O) = \(54.94 + 32.07 + 4\times 16 + 2\times 1.01 + 16\) Now, we can convert grams to moles: Moles of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) = \(\frac{70\,\mathrm{g}}{\text{Molecular weight}}\) Finally, since the amount of solute is given per \(100\,\mathrm{mL}\), we need to convert the volume to liters. To do so, divide the volume by \(1000\): Volume = \(\frac{100\,\mathrm{mL}}{1000}\,\mathrm{L}\) Now, we can find the solubility in moles per liter: Solubility = \(\frac{\text{Moles of }\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}}{\text{Volume in L}}\)

Compare solubility to the given concentration

Now, we will compare the solubility we calculated in moles per liter to the given concentration of \(1.22\,\mathrm{M}\) to determine whether the solution is saturated, supersaturated, or unsaturated. - If the solubility is equal to the concentration, the solution is saturated. - If the solubility is greater than the concentration, the solution is unsaturated. - If the solubility is less than the concentration, the solution is supersaturated.

Determine an experiment to test the saturation of an unknown solution

For a solution of unknown concentration, you can perform the following experiment to determine if the solution is saturated, supersaturated, or unsaturated: 1. Add a small amount of solid \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) to the solution and stir. 2. Observe whether the solid dissolves completely, partially, or not at all. - If the solid dissolves completely, the solution is unsaturated, as it can still dissolve more solute. - If the solid does not dissolve at all, the solution is saturated, as it cannot dissolve any more solute. - If the solid dissolves and then begins to form crystals again, the solution is supersaturated, as it is unstable and can only hold the excess solute temporarily.

Key concepts

Saturated solutions

In chemistry, a saturated solution is one in which no more solute can dissolve at a given temperature. Imagine you are making lemonade. There's a point where adding more sugar just leads to a pile settling at the bottom. That's saturation. In such a solution, the maximum amount of solute has dissolved in the solvent. If more solute is present, it will not dissolve because the solution has reached its capacity.
Understanding when a solution is saturated helps in predicting whether adding more solute will dissolve. In our exercise, identifying the saturation level helps ascertain if the solution contains the maximum dissolved solute (\(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\)) possible at 20°C. When solubility equals the concentration of the solution, the solution is considered saturated.

• Saturated solutions are balanced – the rates of dissolving and precipitating solute are equal.
• Determining this state is crucial for various applications like brewing, crystallization processes, and chemical reactions.

Supersaturated solutions

A supersaturated solution contains more dissolved solute than it normally would at the same temperature and can lead to crystallization. Picture a glass of sweet tea left to cool, and as it cools, sugar crystals form at the bottom. This is due to the solution being supersaturated.
Achieving a supersaturated state typically involves dissolving the solute at a high temperature before cooling the solution. The excess solute remains dissolved despite the lowered temperature, making the solution unstable. Small perturbations can cause the solute to precipitate out, creating crystals.
In practical terms, our exercise points out three scenarios when a solution is tested with added solute to classify it as supersaturated:

• If adding a small amount of solute results in rapid crystal formation, the solution is in a supersaturated state.
• Such solutions are beneficial for growing crystals and in certain chemical processes where excess solute needs removing.
• However, they must be handled carefully due to their potential to precipitate unexpectedly.

Molar concentration

Molar concentration, also known as molarity, is a way of expressing the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. Think of it as the amount of lemonade mix dissolved in water to make lemonade.
Calculating molarity is crucial in chemistry for solutions manufacture and when reacting chemical substances. It allows for precise control over reactants in chemical reactions. To find molarity:

• Convert the amount of solute from grams to moles using its molar mass.
• Measure the total solution volume in liters.
• Divide the moles of solute by the liters of solution to get molarity: \(\mathrm{M} = \frac{\text{moles of solute}}{\text{liters of solution}}\).

In our exercise, comparing the molarity of the given solution to the solubility helps classify it as saturated, unsaturated, or supersaturated. Recognizing these differences is essential for successful experimentation and solution preparation.