Question

To prepare a solution that is \(0.50 \mathrm{M} \mathrm{KCl}\) starting with \(100.0 \mathrm{mL}\) of \(0.40 \mathrm{M} \mathrm{KCl},\) you should \((\mathrm{a})\) add \(20.0 \mathrm{mL}\) of water; (b) add 0.075 g KCl; (c) add 0.10 mol KCl; (d) evaporate \(20.0 \mathrm{mL}\) of water.

Short answer

The correct option to prepare a \(0.50 M\ \mathrm{KCl}\) solution starting with \(100.0 \mathrm{mL}\) of a \(0.40 M\ \mathrm{KCl}\) solution is to evaporate \(20.0 \mathrm{mL}\) of water.

Step-by-step solution

- Evaluate the first option

In the first option (a), we add \(20.0 \mathrm{mL}\) of water to the initial solution. This increases the volume of the solution, therefore the concentration decreases, which does not achieve the intended result.

- Evaluate the second option

In the second option (b), we add 0.075 g KCl to the initial solution. Potassium chloride (KCl) has a molar mass of \(74.55 \mathrm{g/mol}\). So, 0.075 g is approximately 0.001 mol. This increases the amount of solute, but not enough to reach a \(0.50 M\ \mathrm{KCl}\) concentration.

- Evaluate the third option

In the third option (c), we add 0.10 mol KCl to the initial solution. This significantly increases the amount of solute, leading to a concentration change. However, adding such a large amount of solute would far exceed the desired concentration.

- Evaluate the fourth option

In the fourth option (d), we evaporate \(20.0 \mathrm{mL}\) of water from the initial solution. This reduces the volume of the solution, leading to an increase in concentration. The new volume would be \(100.0 \mathrm{mL} - 20.0 \mathrm{mL} = 80.0 \mathrm{mL}\).\n\nBy substituting the numbers into the formula \(C = n/V\) where \(C\) is the molarity, \(n\) is the number of moles and \(V\) is the volume in liters, we get \n\n\(C = 0.40 \mathrm{mol/L} \times 0.100 \mathrm{L} / 0.080 \mathrm{L} = 0.50 \mathrm{M KCl}\).\n\nSo option (d) is the correct procedure to prepare a \(0.50 M\ \mathrm{KCl}\) solution starting with \(100.0 \mathrm{mL}\) of a \(0.40 M\ \mathrm{KCl}\) solution.

Key concepts

Molarity Calculations

Understanding molarity is crucial in chemistry, as it provides information on the concentration level in a solution. Essentially, molarity (\( M \)) is defined as the number of moles of solute per liter of solution. It's calculated using the formula: \[ M = \frac{n}{V} \] where \( n \) represents the amount of solute in moles, and \( V \) is the volume of the solution in liters.

To illustrate this, let's consider our textbook example where we want to achieve a solution of \(0.50 M \mathrm{KCl}\). Initially, we have \(100.0 mL\) of a \(0.40 M \mathrm{KCl}\) solution. To reach a molarity of \(0.50 M\), we need to manipulate the number of moles of solute or the volume of the solution. If we decide to add a certain mass of KCl, we first need to convert that mass to moles using the molar mass of KCl. This type of calculation can be tricky because you have to take the volume change into account after adding the solute.

However, by evaporating water, we decrease the volume and increase the concentration without changing the amount of solute—a clear illustration of how molarity can be affected.

Concentration of Solutions

Concentration of a solution is a measure of the amount of solute dissolved in a given quantity of solvent or solution. It's a broad term that includes molarity but also other measurements such as mass percent, volume percent, and molality. Concentration can determine the reactivity, boiling point, and many other properties of the solution.

It's important to remember that when adding more solute, you increase the concentration, assuming the volume remains constant. Alternatively, reducing the volume, as in our exercise through evaporation, also results in a higher molarity. Hence, the method of reaching the desired concentration depends on whether you are working with the solute amount, volume of solution, or both.

Dilution and Evaporation

Dilution and evaporation are two processes that can alter the concentration of a solution. Dilution involves adding more solvent, which increases the volume and lowers the concentration of the solution. In contrast, evaporation removes solvent, usually water, thereby reducing the volume and increasing the concentration of the solution.

These processes are applied in various scenarios in a laboratory or industrial setting. For instance, in our given exercise, to prepare a \(0.50 M \mathrm{KCl}\) solution from a \(0.40 M \mathrm{KCl}\) solution, we opt for evaporating water. Evaporating \(20.0 mL\) from our initial \(100.0 mL\) of solution increases the solution's concentration by decreasing its volume—an effective and practical approach to controlling concentration without the need for additional solute.