Question

Calculate the molarity of each aqueous solution: (a) \(25.5 \mathrm{~mL}\) of \(6.25 \mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{~L}\) with water (b) \(8.25 \mathrm{~mL}\) of \(2.00 \times 10^{-2} M\) KI diluted to \(12.0 \mathrm{~mL}\) with water

Short answer

In (a): 0.319 M HCl, and in (b): 1.38 \times 10^{-2} M KI

Step-by-step solution

Understand the Dilution Formula

The dilution of a solution can be calculated using the formula: \[ C_1 V_1 = C_2 V_2 \] where \( C_1 \) is the initial concentration, \( V_1 \) is the initial volume, \( C_2 \) is the final concentration, and \( V_2 \) is the final volume.

Apply the Formula for Solution (a)

For solution (a): - Initial concentration ( \( C_1 \) ) = 6.25 M - Initial volume ( \( V_1 \) ) = 25.5 mL - Final volume ( \( V_2 \) ) = 0.500 L = 500 mL Use the formula \( C_1 V_1 = C_2 V_2 \) to find \( C_2 \) : \[ (6.25 \, \mathrm{M}) \cdot (25.5 \, \mathrm{mL}) = C_2 \cdot (500 \, \mathrm{mL}) \] Solving for \( C_2 \), we get: \[ C_2 = \frac{6.25 \, \mathrm{M} \cdot 25.5 \, \mathrm{mL}}{500 \, \mathrm{mL}} \approx 0.31875 \, \mathrm{M} \approx 0.319 \, \mathrm{M} \]

Apply the Formula for Solution (b)

For solution (b): - Initial concentration ( \( C_1 \) ) = 2.00 \times 10^{-2} \, M - Initial volume ( \( V_1 \) ) = 8.25 mL - Final volume ( \( V_2 \) ) = 12.0 mL Use the formula \( C_1 V_1 = C_2 V_2 \) to find \( C_2 \) : \[ (2.00 \times 10^{-2} \, \mathrm{M}) \cdot (8.25 \, \mathrm{mL}) = C_2 \cdot (12.0 \, \mathrm{mL}) \] Solving for \( C_2 \), we get: \[ C_2 = \frac{2.00 \times 10^{-2} \, \mathrm{M} \cdot 8.25 \, \mathrm{mL}}{12.0 \, \mathrm{mL}} \approx 1.375 \times 10^{-2} \, \mathrm{M} \approx 1.38 \times 10^{-2} \, \mathrm{M} \]

Key concepts

headline of the respective core concept

To calculate the molarity of a diluted solution, we need to understand the concept of dilution. This process involves reducing the concentration of a solute in a solution, usually by adding more solvent. The Dilution Formula takes into account the volumes and concentrations of the solution before and after dilution.

The formula is: \[ C_1 V_1 = C_2 V_2 \]Here:

• \(C_1\) is the initial concentration.
• \(V_1\) is the initial volume.
• \(C_2\) is the final concentration.
• \(V_2\) is the final volume.

By rearranging this formula, we can solve for the new concentration after the solution has been diluted. This makes it easy to figure out how the molarity changes as we add more solvent to the solution.

headline of the respective core concept

Before we apply the dilution formula, we need to know the initial parameters of our solution: the initial concentration (\(C_1\)) and the initial volume (\(V_1\)).

For example, let's consider Solution (a):

• Initial concentration (\(C_1\)): 6.25 M. This is the concentration of the HCl solution before it is diluted.
• Initial volume (\(V_1\)): 25.5 mL. This is the volume of the HCl solution before dilution.

Likewise, for Solution (b):

• Initial concentration (\(C_1\)): 2.00 \times 10^{-2} M. This is the concentration of the KI solution before diluting.
• Initial volume (\(V_1\)): 8.25 mL. This is the volume of the KI solution before adding more solvent.

Having these parameters is crucial as they help us calculate how the concentration will change once we dilute the solution.

headline of the respective core concept

After determining the initial concentration and volume, we need to know the final concentration and volume, specifically after dilution.

Let's look at Solution (a) again:

• Final volume (\(V_2\)): 0.500 L. Since 0.500 L is equivalent to 500 mL, this is the volume after dilution.
• We use the dilution formula to calculate the final concentration (\(C_2\)). By plugging in the known values, we get\[ (6.25 \text{ M}) \cdot (25.5 \text{ mL}) = C_2 \cdot (500 \text{ mL}) \]Which simplifies to:\[ C_2 = \frac{6.25 \text{ M} \cdot 25.5 \text{ mL}}{500 \text{ mL}} \approx 0.319 \text{ M} \]

For Solution (b):

• Final volume (\(V_2\)): 12.0 mL. This is the total volume after adding more solvent.
• Using the dilution formula, we calculate\[ (2.00 \times 10^{-2} \text{ M}) \cdot (8.25 \text{ mL}) = C_2 \cdot (12.0 \text{ mL}) \]Which simplifies to:\[ C_2 = \frac{2.00 \times 10^{-2} \text{ M} \cdot 8.25 \text{ mL}}{12.0 \text{ mL}} \approx 1.38 \times 10^{-2} \text{ M} \]

Understanding these final values is essential to see how the concentration of a solution changes upon dilution.