Problem 14 - General Chemistry: Principles and Modern Applications

Question

What is the \(\mathrm{pH}\) of the solution obtained when \(125 \mathrm{mL}\) of \(0.606 \mathrm{M} \mathrm{NaOH}\) is diluted to \(15.0 \mathrm{L}\) with water?

Step-by-Step Solution

Answer

The pH of the solution, when \(125 mL\) of \(0.606 M NaOH\) is diluted to \(15.0 L\) with water, can be calculated using the steps outlined above.

Step 1: Calculate the new concentration of the NaOH solution

First, remember that when a solution is diluted, the amount of solute doesn't change, but the volume of the solvent increases. We can apply this principle in the formula \(M1V1 = M2V2\). Here, \(M1\) (original molarity) is \(0.606M\), \(V1\) (original volume) is \(125mL\) or \(0.125L\) and \(V2\) (new volume) is \(15.0L\). Solve for \(M2\) (new molarity), which represents the concentration of the NaOH solution after dilution.

Step 2: Determine the concentration of OH- ions

NaOH, being a strong base, will completely dissociate in water to form Na+ and OH-. Hence the concentration of OH- ions will be the same as the concentration of the NaOH solution found in Step 1.

Step 3: Calculate the pOH of the solution

The pOH of a solution can be calculated using the formula \(-\log[OH^-]\). Substitute the concentration of OH- ions found in Step 2 into this formula.

Step 4: Calculate the pH of the solution

The pH and the pOH of a solution are related by the formula \(pH + pOH = 14\). Substitute the pOH found in Step 3 into this formula and solve for pH, which refers to the pH of the solution after dilution.

Key Concepts

Dilution Formula

Dilution involves adding more solvent to a solution, thereby decreasing its concentration but not altering the amount of solute present. This process can be understood through the dilution formula:

• \(M_1V_1 = M_2V_2\)

Here, \(M_1\) represents the original molarity, \(V_1\) the original volume, \(M_2\) the final molarity (after dilution), and \(V_2\) the final volume. By rearranging this equation, you can solve for the new molarity \(M_2\), as shown below:

• \(M_2 = \frac{M_1 \times V_1}{V_2}\)

In this exercise, to find the new concentration of our NaOH solution, plug in:

• \(M_1 = 0.606 \text{ M}, V_1 = 0.125 \text{ L}, \text{ and } V_2 = 15.0 \text{ L}\)

The process allows us to determine exactly how the concentration changes when diluted.

NaOH Dissociation

Sodium hydroxide (NaOH) is known as a strong base, which means it dissociates completely in water. When dissolved, each NaOH unit splits into its constituent ions:

• \(\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-\)

Given this immediate and full dissociation, the concentration of the hydroxide ions (OH\(^-\)) in solution is equal to the initial concentration of the NaOH solution. During dilution, even though we decrease the concentration of NaOH, each NaOH molecule still splits completely upon dissolution. Therefore, when we calculate the concentration post-dilution, that will also represent the concentration of OH\(^-\) ions. This dissociation is crucial for further acidity or basicity calculations.

pOH Calculation

To understand how basic or alkaline a solution is, the pOH value is often used. It provides a measure of the concentration of hydroxide ions (OH\(^-\)) in a solution. The formula is straightforward:

• \(\text{pOH} = -\log[\text{OH}^-]\)

First, we determine the concentration of OH\(^-\) ions, which, in this case, is equivalent to the concentration of NaOH after dilution. Plug this concentration into the formula to calculate the pOH of the solution. Understanding pOH is a key step in evaluating the base strength of a solution, and subsequently, helps bridge the calculation for pH determination.

Strong Base

A strong base, such as NaOH, is characterized by its ability completely to dissociate in water. This means every molecule of NaOH that dissolves turns into free ions: Na\(^+\) and OH\(^-\). The complete dissociation implies that the amount of hydroxide ions truly represents the base's strength in solution.

When dealing with strong bases in pH calculations, this complete and predictable behavior simplifies the process.

• It allows for direct calculation of hydroxide ion concentration.
• Aids in accurately finding the pOH and converting it to pH.

Such a feature contrasts with weak bases, which do not fully dissociate. Recognizing whether a base is strong or weak helps in setting the right expectations for how it will behave in solution and impact pH and pOH calculations.