Problem 47 - Chemistry The Central Science In Si Units

Question

Calculate the concentration of an aqueous solution of \(\mathrm{NaOH}\) that has a pH of \(12.70 .\)

Step-by-Step Solution

Answer

The concentration of NaOH is approximately 0.0501 M.

Step 1: Understanding pH and pOH

The pH of a solution gives us the concentration of hydrogen ions. For a basic solution, which is what we have since the pH is greater than 7, we can find the concentration of hydroxide ions ( ext{OH}^- ext{ ions}) using the relation between pH and pOH. Recall that \( \text{pH} + \text{pOH} = 14 \).

Step 2: Calculating pOH

Subtract the given pH from 14 to find the pOH. This gives us \( \text{pOH} = 14 - 12.70 = 1.30 \).

Step 3: Relating pOH to [OH⁻] Concentration

The pOH can be used to find the concentration of hydroxide ions using the formula \( \text{pOH} = -\log[\text{OH}^-] \). Rearranging gives us \( [\text{OH}^-] = 10^{-\text{pOH}} \).

Step 4: Calculating [OH⁻] Concentration

Calculate \( [\text{OH}^-] \) by using \( [\text{OH}^-] = 10^{-1.30} = 5.01 \times 10^{-2} \text{ M} \). This is the concentration of \( \text{OH}^- \) in the solution and, since \( \text{NaOH} \) completely dissociates, this is also the concentration of \( \text{NaOH} \).

Key Concepts

pH and pOH relationship

The pH and pOH relationship is a core concept in understanding how acidic or basic an aqueous solution is. In any aqueous solution, the product of hydrogen ion concentrations \(\text{[H}^+\text{]}\) and hydroxide ion concentrations \(\text{[OH}^-\text{]}\) is a constant at a given temperature. This is known as the water dissociation constant (\(K_w\)), which is \(1.0 \times 10^{-14}\) at 25°C. Thus, the pH, which measures the acidity of a solution, and pOH, which measures the basicity, are interconnected through the equation:

• \( \text{pH} + \text{pOH} = 14 \)

If the pH of a solution is known, the pOH can easily be determined by subtracting the pH value from 14.In our given problem:

• The pH is 12.70, indicating a basic solution since it is above 7.
• To find the pOH, subtract the pH from 14: \(\text{pOH} = 14 - 12.70 = 1.30\).

Maintaining this relationship helps in determining the concentrations of hydrogen and hydroxide ions, important for predicting the behavior of solutions in various chemical reactions.

hydroxide ion concentration

Understanding hydroxide ion concentration is crucial for identifying how basic a solution is. In the context of our exercise, we first determined the pOH, which we then used to find the concentration of hydroxide ions (\([\text{OH}^-]\)). The relationship between pOH and hydroxide ion concentration is given by the formula:

• \(\text{pOH} = -\log[\text{OH}^-]\)

By rearranging this equation, we derive:

• \([\text{OH}^-] = 10^{-\text{pOH}}\)

Thus, once the pOH is calculated, you can find \(\text{[OH]}^−\) by raising 10 to the power of the negative pOH value. In our example:

• With a pOH of 1.30, the \(\text{[OH]}^−\) is calculated as \(10^{-1.30} = 5.01 \times 10^{-2} \text{ M}\).

This concentration tells us how many moles of hydroxide ions are present per liter of solution, a key piece of information when analyzing the basicity of the solution.

NaOH dissociation

NaOH, or sodium hydroxide, is a strong base that fully dissociates in water. This means that when NaOH is dissolved in water, it breaks apart completely into sodium (Na⁺) ions and hydroxide (OH⁻) ions. This complete dissociation is significant because it means that the concentration of \(\text{OH}^-\) ions directly corresponds to the concentration of the NaOH solution.Due to its strong basic nature:

• Every mole of NaOH will produce one mole of OH⁻ ions in solution.

Therefore:

• The concentration of OH⁻ ions is equal to the concentration of the NaOH solution.

In our calculation, the hydroxide ion concentration found as \(5.01 \times 10^{-2} \, \text{M}\) is the same as the concentration of the initial NaOH solution. This principle of full dissociation applies not just to NaOH but to all strong bases, making it easier to calculate the concentrations in such solutions.