Question

Water is evaporated from \(125 \mathrm{mL}\) of \(0.198 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\) solution until the volume becomes \(105 \mathrm{mL}\). What is the molarity of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the remaining solution?

Short answer

The molarity of the remaining solution is \(0.236 \mathrm{M}\)

Step-by-step solution

Convert initial volume to litres

The initial volume is given in milliliters, to use it in the calculations, it should be converted to liters. We know that 1L = 1000mL. So, the initial volume \(V_{1}\) in litres is \(125 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.125 \mathrm{L}\)

Convert final volume to litres

Similarly, convert the final volume \(V_{2}\) to liters. So, \(105 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.105 \mathrm{L}\)

Use Molarity Conservation Principle

Since the amount of solute remains the same before and after evaporation of water, we can use the molarity conservation equation \(C_{1}V_{1} = C_{2}V_{2}\). Setting up the equation \((0.198 M)(0.125 L) = C_{2}(0.105 L)\)

Solve for final molarity

Rearranging the equation for \(C_{2}\), we get \(C_{2} = \frac{(0.198 \mathrm{M})(0.125 \mathrm{L})}{0.105 \mathrm{L}}\)

Calculate final molarity

On calculation, \(C_{2} = 0.236 \mathrm{M}\), which represents the molarity of \(K_2SO_4\) in the remaining solution after evaporation

Key concepts

Solution Concentration

When we talk about solution concentration, we refer to how much solute is dissolved in a given amount of solvent. In chemistry, molarity ( M ) is a common way to express solution concentration. It is defined as the number of moles of solute per liter of solution. A highly concentrated solution has more solute particles compared to a dilute solution.

In the context of the original exercise, we started with a solution of potassium sulfate ( K_2SO_4 ) that has a molarity of 0.198 M. This means each liter of the solution contains 0.198 moles of K_2SO_4 .

As we change the volume of the solution by evaporation, the concentration—or molarity—of the solute may change, depending on whether the amount of solute stays constant, which it does in this exercise.

Volume Conversion

Volume conversion is important when dealing with measurements in chemistry. Volumes are often measured in milliliters (mL) but need to be converted to liters (L) for molarity calculations, since molarity is based on liters.

For these calculations, use the conversion factor: 1 L = 1000 mL. This is a straightforward conversion:

• To convert from milliliters to liters, divide the volume in milliliters by 1000.
• For example, 125 mL becomes 0.125 L.
• Similarly, 105 mL becomes 0.105 L.

Converting units correctly is crucial for the accuracy of molarity calculations. Always double-check your conversion to ensure that your numbers make sense in the calculations.

Conservation of Mass

The principle of conservation of mass states that mass cannot be created or destroyed in a closed system. In chemical solutions, this means that unless solute is added or removed, the total mass of solute remains constant, even if the volume of the solvent changes. In the exercise, conservation of mass helps us to understand that while the water evaporates, the K_2SO_4 particles remain the same. The molarity calculation leverages this principle:

• The initial condition can be written as: Initial molarity C_1 times Initial volume V_1 equals the number of moles of solute, which remains unchanged.
• The unchanged number of moles equals the final molarity C_2 times the final volume V_2 .
• The equation C_1V_1 = C_2V_2 ensures that the amount of solute remains constant and allows calculation of the final concentration.

Chemical Solutions

Chemical solutions consist of a solute dissolved in a solvent. In this problem, potassium sulfate ( K_2SO_4 ) is the solute, and water is the solvent. Solutions can vary in concentration and volume, but the significant fact is that the solute particles are distributed evenly throughout the solvent.

Understanding how the solution changes with transformations like evaporation is critical.

• Evaporation reduces the volume of solvent (water), potentially increasing the concentration of solute, assuming no solute is lost.
• When water evaporates from the K_2SO_4 solution, the concentration increases because the same amount of solute is distributed over a smaller volume.

Recognizing these properties can help you manipulate solutions in the lab and solve various chemistry problems.