Problem 29 - Chemistry Matter and Change

Question

Challenge Calculate pH and pOH for an aqueous solution containing \(1.0 \times 10^{-3}\) mol of HCl dissolved in 5.0 \(\mathrm{L}\) of solution.

Step-by-Step Solution

Answer

The pH of the solution is approximately 3.70, and the pOH is 10.30.

Step 1: Convert Given Information to Molarity

The problem provides the amount of HCl in moles and the volume of the solution. First, calculate the molarity of the HCl solution using the formula \( ext{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \). Here, the moles of HCl is \(1.0 \times 10^{-3}\) mol and the volume is 5.0 L. Thus, the molarity is \( \frac{1.0 \times 10^{-3}}{5.0} = 2.0 \times 10^{-4} \text{ M} \).

Step 2: Calculate pH from Molarity

Since HCl is a strong acid, it fully dissociates in water. Therefore, the concentration of \( \text{H}^+ \) ions is equal to the molarity of the HCl, which is \(2.0 \times 10^{-4} \text{ M} \). The \( \text{pH} \) is calculated using the formula \( \text{pH} = -\log[\text{H}^+] \). So, \( \text{pH} = -\log(2.0 \times 10^{-4}) \approx 3.70 \).

Step 3: Calculate pOH from pH

The relationship between \( \text{pH} \) and \( \text{pOH} \) is given by the equation \( \text{pH} + \text{pOH} = 14 \). Given that \( \text{pH} \approx 3.70 \), we can find \( \text{pOH} \) by rearranging the equation: \( \text{pOH} = 14 - 3.70 = 10.30 \).

Key Concepts

Molarity

Molarity is a key concept in chemistry that expresses the concentration of a solute in a solution. It tells us how much solute, in moles, is dissolved in a specific volume of solution. The formula to calculate molarity is simple:

• \( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \)

To solve problems involving molarity, ensure you have the moles of the solute and the volume of the solution in liters. In the given exercise, we used this formula to find the molarity of an HCl solution with 1.0 \( \times 10^{-3} \) moles of HCl in 5.0 liters of solution, which resulted in a molarity of 2.0 \( \times 10^{-4} \) M. This step is crucial for further pH and pOH calculations. Molarity simplifies understanding how concentrated a solution is, allowing for precise chemical calculations.

Strong Acid Dissociation

Strong acids, like HCl, are known for their complete dissociation in water. This means every molecule of HCl dissociates into ions. For HCl, it breaks down completely into \( \text{H}^+ \) and \( \text{Cl}^- \) ions.

• Complete dissociation implies that the concentration of \( \text{H}^+ \) ions is exactly equal to the molarity of the acid.

Understanding strong acid dissociation is crucial for calculating the concentration of \( \text{H}^+ \) in a solution. In our exercise, since 2.0 \( \times 10^{-4} \) M is the molarity of HCl, it is also the concentration of \( \text{H}^+ \) ions. This complete dissociation simplifies calculating pH from molarity, saving steps compared to weak acids where partial dissociation must be considered.

pH and pOH Relationship

The pH and pOH of a solution are intimately linked through a simple equation:

• \( \text{pH} + \text{pOH} = 14 \)

This relationship holds true for all aqueous solutions at 25°C. The pH measures the acidity, or the concentration of \( \text{H}^+ \) ions, while pOH measures the basicity, or the concentration of \( \text{OH}^- \) ions.
In the context of our exercise, once the pH was calculated (approximately 3.70), the pOH could be easily determined by rearranging the equation: \( \text{pOH} = 14 - 3.70 = 10.30 \). This provides a complete picture of the ionic balance in the solution, showcasing the simple elegance of chemical equilibrium concepts.

Logarithms in Chemistry

Logarithms are an essential mathematical tool in chemistry, especially for calculating pH and pOH. The formula for pH is:

• \( \text{pH} = -\log[\text{H}^+] \)

This formula shows how pH is a log scale, which means that each pH unit change represents a tenfold change in \( \text{H}^+ \) concentration. Knowing this, you can see why slight differences in pH actually correspond to significant differences in acidity or basicity.
In our problem, applying the logarithm \(-\log(2.0 \times 10^{-4})\) gave us a pH of approximately 3.70. Similarly, pOH is derived from \(-\log[\text{OH}^-]\). This use of logarithms allows chemists to work with the typically small and inconvenient numbers involved in concentration measurements, transforming them into a manageable scale.