Problem 22 - Chemistry: The Central Science

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) \(\mathrm{ls}\) 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, \(w\) hat experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

Step-by-Step Solution

Answer

(a) The molar solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) is calculated to be \(4.14 \, M\). Since the given concentration (\(1.22 \, M\)) is lower than the molar solubility, the solution is unsaturated. (b) To determine the saturation state of a solution of unknown concentration, add a small amount of \(\mathrm{MnSO}_{4}\) crystals into the solution and observe if they dissolve. If the crystals dissolve entirely, the solution is unsaturated; if the crystals do not dissolve, the solution is saturated; if adding crystals causes more precipitate to form, the solution may be supersaturated.

Step 1: Find the molar mass of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\)

To calculate the molar solubility, first find the molar mass of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\). The molar mass can be calculated by adding the atomic weights of each element in the compound. \[M = (Mn) + (S) + 4(O) + 2(H) + (O)\] Using the atomic weights from the periodic table, calculate the molar mass: \[M = (54.94) + (32.07) + 4(16.00) + 2(1.01) + (16.00) = 169.02 \, g/mol\]

Step 2: Calculate the molar solubility

Next, calculate the molar solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\): \[Molar\, solubility = \frac{Solubility\, in\, g/100\, mL}{Molar\, mass}\times\frac{1000\, mL}{100\, mL}\] \[Molar\, solubility = \frac{70\, g/100\, mL}{169.02\, g/mol}\times\frac{1000\, mL}{100\, mL} = 4.14 \, M\]

Step 3: Determine the saturation state of the given solution

Compare the molarity of the given solution (\(1.22 \, M\)) to the calculated molar solubility (\(4.14 \, M\)). Since the given concentration is lower than the molar solubility, the solution is unsaturated.

Step 4: Suggest an experiment to determine the saturation state

To determine whether a solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) of unknown concentration is saturated, supersaturated, or unsaturated, add a small amount of \(\mathrm{MnSO}_{4}\) crystals into the solution and gently stir. Observe whether the crystals dissolve or not. - If the crystals dissolve entirely, the solution is unsaturated. - If the crystals do not dissolve and remain in the solution, the solution is saturated. - If adding the crystals and stirring cause more precipitate to form than expected, the solution might be supersaturated. In a supersaturated solution, the solute would typically precipitate until the concentration reaches the saturation level.

Key Concepts

Molar Solubility

Molar solubility refers to the number of moles of solute that can dissolve in a unit volume of solution to form a saturated solution. It is a crucial concept in understanding how much of a particular substance can dissolve in a solvent at a specific temperature.
For example, in the case of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\), the solubility given is 70 g per 100 mL of water at 20°C. To understand this in terms of molarity (M), we calculate the molar solubility as follows:

• First, find the molar mass of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\), which is 169.02 g/mol.
• Next, use this molar mass to convert grams per liter to moles per liter, accounting for volume changes from 100 mL to 1 L: \(\frac{70}{169.02} \times \frac{1000}{100} = 4.14\, M\).

This value, 4.14 M, represents the maximum concentration of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) that can dissolve in water at this temperature, which is its molar solubility.

Saturated Solution

A saturated solution occurs when no more solute can dissolve in the solvent at a given temperature. The solute's concentration has reached its maximum limit, and any additional solute will remain undissolved.
Consider \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water. If the molarity of this solution is equivalent to its molar solubility (4.14 M at 20°C), the solution is saturated.

• This means the solution is at equilibrium, and any extra solute will not dissolve.
• It's important to ensure conditions like temperature remain constant, as temperature changes can alter solubility.

If you're working with a saturated solution and attempt to add more solute, it will simply sit at the bottom of the container.

Supersaturated Solution

A supersaturated solution contains a higher concentration of solute than what is possible at equilibrium under normal conditions. Such a solution is an unstable state.
Supersaturation occurs when a solution is prepared by dissolving solute at an elevated temperature and then cooling it down slowly. If \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) is dissolved in water beyond its molar solubility at 20°C (4.14 M), the solution becomes supersaturated as it cools.

• In this state, the addition of a seed crystal or even slight perturbations can cause sudden precipitation of the excess solute.
• The presence of more solute than can permanently stay dissolved leads to a rapid crystallization process, as the solution seeks to reach a stable, saturated state.

Supersaturated solutions can be fascinating to watch but are difficult to maintain due to their instability.

Unsaturated Solution

An unsaturated solution is one in which more solute can be dissolved without reaching saturation. This means the concentration of solute is below its molar solubility level.
Let's take our \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) example. At 20°C, the molar solubility is 4.14 M. If the actual concentration is less than this, the solution is unsaturated, like a 1.22 M solution that is given.

• In this situation, you can continue adding solute, and it will dissolve until reaching the saturation point.
• Such solutions are more common than supersaturated ones and are stable under existing conditions unless more solute is added or temperature changes occur.

Unsaturated solutions allow for flexibility in experiments, as they can still accept additional solute until becoming saturated.

Molar Mass Calculation

Calculating the molar mass of a compound is a fundamental skill required for understanding and performing solubility calculations.
For molar mass calculation, you need to add up the atomic masses of all the atoms in a molecule. Consider \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\):

• The elements include Mn (Manganese), S (Sulfur), O (Oxygen), and H (Hydrogen). Their atomic masses are approximately 54.94, 32.07, 16.00, and 1.01 respectively.
• Calculate: \((54.94) + (32.07) + 4(16.00) + 2(1.01) + (16.00)\)
• The molar mass results in 169.02 g/mol, which is essential for converting between grams and moles in solution calculations.

This precise calculation allows you to determine the molar solubility or convert between mass-based and mole-based units, ensuring accurate experimentations and validations.