Question

Calculate the new molarity if each of the following dilutions is made. Assume the volumes are additive. a. \(55.0 \mathrm{~mL}\) of water is added to \(25.0 \mathrm{~mL}\) of \(0.119 \mathrm{M} \mathrm{NaCl}\) solution b. \(125 \mathrm{~mL}\) of water is added to \(45.3 \mathrm{~mL}\) of \(0.701 \mathrm{M} \mathrm{NaOH}\) solution c. \(550 . \mathrm{mL}\) of water is added to \(125 \mathrm{~mL}\) of \(3.01 \mathrm{M}\) KOH solution d. \(335 \mathrm{~mL}\) of water is added to \(75.3 \mathrm{~mL}\) of \(2.07 \mathrm{M} \mathrm{CaCl}_{2}\) solution

Short answer

The final molarity for each dilution is as follows: a. \(0.0372 \mathrm{M} \mathrm{NaCl}\) b. \(0.186 \mathrm{M} \mathrm{NaOH}\) c. \(0.556 \mathrm{M} \mathrm{KOH}\) d. \(0.378 \mathrm{M} \mathrm{CaCl}_{2}\)

Step-by-step solution

a. Dilution of NaCl solution

We are given the initial volume (\(V_1\)) of NaCl solution as: \(25.0 \mathrm{~mL}\) and the initial concentration (\(C_1\)) as: \(0.119 \mathrm{M}\). Also, we are given the added volume of water as: \(55.0 \mathrm{~mL}\). First, we need to calculate the total final volume (\(V_2\)): \(V_2 = V_1 + \mathrm{added\: volume\: of\: water}\) \(V_2 = 25.0\mathrm{~mL} + 55.0 \mathrm{~mL}\) \(V_2 = 80.0 \mathrm{~mL}\) Now, we will use the formula \(C_1V_1 = C_2V_2\) to find the final concentration (\(C_2\)) of the NaCl solution: \(C_2 = \dfrac{C_1V_1}{V_2}\) \(C_2 = \dfrac{(0.119 \mathrm{M})(25.0 \mathrm{~mL})}{80.0 \mathrm{~mL}}\) \(C_2 = 0.0372 \mathrm{M}\)

b. Dilution of NaOH solution

We are given the initial volume (\(V_1\)) of NaOH solution as: \(45.3 \mathrm{~mL}\) and the initial concentration (\(C_1\)) as: \(0.701 \mathrm{M}\). Also, we are given the added volume of water as: \(125 \mathrm{~mL}\). First, we need to calculate the total final volume (\(V_2\)): \(V_2 = V_1 + \mathrm{added\: volume\: of\: water}\) \(V_2 = 45.3\mathrm{~mL} + 125 \mathrm{~mL}\) \(V_2 = 170.3 \mathrm{~mL}\) Now, we will use the formula \(C_1V_1 = C_2V_2\) to find the final concentration (\(C_2\)) of the NaOH solution: \(C_2 = \dfrac{C_1V_1}{V_2}\) \(C_2 = \dfrac{(0.701 \mathrm{M})(45.3 \mathrm{~mL})}{170.3 \mathrm{~mL}}\) \(C_2 = 0.186 \mathrm{M}\)

c. Dilution of KOH solution

We are given the initial volume (\(V_1\)) of KOH solution as: \(125 \mathrm{~mL}\) and the initial concentration (\(C_1\)) as: \(3.01 \mathrm{M}\). Also, we are given the added volume of water as: \(550 \mathrm{~mL}\). First, we need to calculate the total final volume (\(V_2\)): \(V_2 = V_1 + \mathrm{added\: volume\: of\: water}\) \(V_2 = 125 \mathrm{~mL} + 550 \mathrm{~mL}\) \(V_2 = 675 \mathrm{~mL}\) Now, we will use the formula \(C_1V_1 = C_2V_2\) to find the final concentration (\(C_2\)) of the KOH solution: \(C_2 = \dfrac{C_1V_1}{V_2}\) \(C_2 = \dfrac{(3.01 \mathrm{M})(125 \mathrm{~mL})}{675 \mathrm{~mL}}\) \(C_2 = 0.556\mathrm{M}\)

d. Dilution of CaCl2 solution

We are given the initial volume (\(V_1\)) of CaCl2 solution as: \(75.3 \mathrm{~mL}\) and the initial concentration (\(C_1\)) as: \(2.07 \mathrm{M}\). Also, we are given the added volume of water as: \(335 \mathrm{~mL}\). First, we need to calculate the total final volume (\(V_2\)): \(V_2 = V_1 + \mathrm{added\: volume\: of\: water}\) \(V_2 = 75.3 \mathrm{~mL} + 335 \mathrm{~mL}\) \(V_2 = 410.3 \mathrm{~mL}\) Now, we will use the formula \(C_1V_1 = C_2V_2\) to find the final concentration (\(C_2\)) of the CaCl2 solution: \(C_2 = \dfrac{C_1V_1}{V_2}\) \(C_2 = \dfrac{(2.07 \mathrm{M})(75.3 \mathrm{~mL})}{410.3 \mathrm{~mL}}\) \(C_2 = 0.378\mathrm{M}\)

Key concepts

Molarity

Molarity is a term used to describe the concentration of a solution, representing the amount of solute dissolved in a unit volume of solution. It is typically expressed in moles per liter (\textbf{mol/L}) or molar (\textbf{M}). Knowing the molarity of a solution is critical for precise and accurate chemical reactions and experiments.

For instance, if we have a 0.119 M NaCl solution, this means there are 0.119 moles of sodium chloride in every liter of the solution. Molarity can be calculated using the formula: \[ Molarity (M) = \frac{moles\text{ }of\text{ }solute}{volume\text{ }of\text{ }solution\text{ }in\text{ }liters} \]
When it comes to dilutions, the molarity of a solution will change, and calculating this new molarity is crucial for tasks such as preparing a reaction mixture with the correct concentrations.

Concentration

Concentration generally refers to the amount of a substance in a defined space. When we talk about the concentration of a solution in chemistry, we are referring to how much solute is present relative to the amount of solvent or total solution.

There are various ways to express concentration, including percent composition by mass, mole fraction, molality, and, most commonly for liquid solutions, molarity. Concentration is a critical parameter because it determines how molecules interact in a solution, which can affect reaction rates and equilibrium states. The precise control of concentration is necessary to replicate experiments and to scale up from laboratory to production.

Dilution Formula

The dilution formula, sometimes referred to as the dilution equation, is a valuable tool for calculating the concentration of a solution after it has been diluted. It is based on the principle of conservation of mass, meaning the amount of solute remains constant before and after dilution, even as the volume changes.

The formula is often expressed as: \[ C_1V_1 = C_2V_2 \]
where \( C_1 \) is the initial concentration, \( V_1 \) is the initial volume, \( C_2 \) is the final concentration after dilution, and \( V_2 \) is the final volume. This formula assumes that the volumes are additive which means the final volume is the sum of the solvent and solution volumes.

Understanding and applying this formula is crucial when preparing solutions in a laboratory, such as diluting a stock solution to a lower concentration for use in an experiment.

Volume Addition

Volume addition is involved in the process of dilution where a solvent (typically water) is added to a solution to decrease its concentration. The key assumption is that the volumes of the solvent and the solution are additive, which is generally accurate for dilute solutions. However, for highly concentrated solutions or when working with solvents and solutes that might have significant interactions, volume changes upon mixing may need to be considered.

For the exercises provided, volume addition is straightforward: the total volume after dilution (\( V_2 \)) is the sum of the initial solution volume (\( V_1 \)) and the volume of water added. This calculation is fundamental in determining the new concentration of a solution after dilution using the dilution formula.