Question

If \(1.00 \mathrm{~L}\) of an \(\mathrm{LiOH}\) solution is boiled down to \(164 \mathrm{~mL}\) and its initial concentration is 0.00555 \(\mathrm{M}\), what is its final concentration?

Short answer

The final concentration is approximately 0.03384 M.

Step-by-step solution

Understand the Concepts

The initial concentration of the solution is given, as well as the initial and final volumes. The concentration changes because the solution volume changes when it is boiled down. We need to find the final concentration of the solution after boiling.

Set Up the Dilution Formula

The concentration and volume of a solution are related by the formula: \(C_1 V_1 = C_2 V_2\), where \(C_1\) and \(C_2\) are the initial and final concentrations, and \(V_1\) and \(V_2\) are the initial and final volumes, respectively. We need to solve for the final concentration \(C_2\).

Plug Values into the Formula

We know the initial concentration \(C_1 = 0.00555 \, \text{M}\), the initial volume \(V_1 = 1.00 \, \text{L} = 1000 \, \text{mL}\), and the final volume \(V_2 = 164 \, \text{mL}\). Substitute these values into the formula: \(0.00555 \, \text{M} \times 1000 \, \text{mL} = C_2 \times 164 \, \text{mL}\).

Solve for the Final Concentration

Rearrange the equation to find \(C_2\): \(C_2 = \frac{0.00555 \, \text{M} \times 1000 \, \text{mL}}{164 \, \text{mL}}\). Calculate \(C_2\) to find the final concentration.

Calculate the Numerical Value

Perform the calculation: \(C_2 = \frac{0.00555 \times 1000}{164}\). Simplifying this, we get \(C_2 = \frac{5.55}{164}\). Calculate \(C_2 \approx 0.03384 \, \text{M}\).

Key concepts

Dilution Formula

The dilution formula is a powerful tool to understand how the concentration of a solution changes with volume. Essentially, whenever you have a solution and you change its volume by either adding more solvent or evaporating some, its concentration will also change. The formula \[C_1 V_1 = C_2 V_2\]expresses this relationship. Here, \(C_1\) and \(C_2\) represent the initial and final concentrations, respectively, while \(V_1\) and \(V_2\) stand for the initial and final volumes. This formula relies on the idea that the total amount of solute (the substance dissolved) stays constant, even though the volume may change the concentration.

• \(C_1\) \(=\) initial concentration of the solution.
• \(V_1\) \(=\) initial volume of the solution.
• \(C_2\) \(=\) final concentration after dilution or evaporation.
• \(V_2\) \(=\) final volume after dilution or evaporation.

This straightforward formula greatly simplifies the process of finding the unknown concentration, as it only requires one to know three of the four variables. It's especially useful when dealing with tasks like boiling solutions down, like in our exercise.

Molarity Calculations

Molarity is a measure of the concentration of a solute in a solution. It's expressed in moles of solute per liter of solution, denoted by the symbol \(M\). This measurement provides a way to easily know how concentrated a solution is. Calculating molarity involves knowing how much of the solute you have and the total volume of the solution. In the context of dilution or concentration changes, this becomes vital in applying the dilution formula.For example, suppose you start with a solution of lithium hydroxide (\(\text{LiOH}\)) with a molarity (\(C_1\)) of 0.00555 \(M\). This means that in every liter of solution, there are 0.00555 moles of \(\text{LiOH}\). If you boil the solution down from 1.00 L (which is 1000 mL) to 164 mL, the amount of \(\text{LiOH}\) hasn't changed, but its concentration will increase because the same amount of solute is now in less liquid.Molarity is a critical concept because it directly links to how reactive or potent a solution might be, which is why accurate calculations are important in lab settings and chemical engineering.

Initial and Final Volume Changes

Understanding how volume changes affect a solution's concentration is key to grasping the results of any dilution or evaporation process. Volume changes can happen due to boiling (as in the given problem), addition of more solvent, or other physical means. Whenever the final volume of a solution differs from the initial volume, its concentration will be altered accordingly.In the given exercise, the initial volume of the \(\text{LiOH}\) solution is 1.00 L, which reduces significantly to 164 mL after boiling. It's important to convert units when necessary to maintain consistency, for example, converting liters to milliliters, since the formula used demands matching volume units. This ensures that unit discrepancies don't lead to calculation errors.Changes in volume directly affect the concentration:

• Decrease in volume increases concentration (less space for the same amount of solute).
• Increase in volume decreases concentration (more space for the same amount of solute).

By understanding initial and final volume changes, predicting concentration shifts becomes straightforward, helping to align expectations with experimental or practical outcomes.