Problem 12 - Chemistry and Chemical Reactivity

Question

What is the pH of a \(1.2 \times 10^{-4} \mathrm{M}\) solution of KOH? What is the hydronium ion concentration of the solution?

Step-by-Step Solution

Answer

The pH is 10.08 and the hydronium ion concentration is \(8.33 \times 10^{-11} \mathrm{M}\).

Step 1: Understanding the Problem

First, recognize that KOH is a strong base. In solution, KOH dissociates completely into K⁺ and OH⁻ ions. The concentration of OH⁻ is the same as the initial concentration of KOH, which is given as \(1.2 \times 10^{-4} \mathrm{M}\).

Step 2: Calculate Hydroxide Ion Concentration

As KOH is a strong base that dissociates completely, the concentration of hydroxide ions (OH⁻) in the solution is equal to the concentration of KOH: \([\text{OH}^-] = 1.2 \times 10^{-4} \mathrm{M}\).

Step 3: Calculate the pOH of the Solution

The formula to calculate pOH is \(\text{pOH} = -\log[\text{OH}^-]\). Substitute the concentration of hydroxide ions into the equation: \[\text{pOH} = -\log(1.2 \times 10^{-4}) \approx 3.92.\]

Step 4: Convert pOH to pH

Use the relation between pH and pOH, which is \(\text{pH} + \text{pOH} = 14\). Rearrange to solve for pH: \[\text{pH} = 14 - \text{pOH} = 14 - 3.92 = 10.08.\]

Step 5: Calculate the Hydronium Ion Concentration

The hydronium ion concentration \([\text{H}_3\text{O}^+]\) can be found using the relation \([\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14}\). Solve for \([\text{H}_3\text{O}^+]\) by substituting \([\text{OH}^-]\): \[[\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-4}} = 8.33 \times 10^{-11} \mathrm{M}.\]

Key Concepts

Strong Bases

Strong bases like KOH are substances that completely dissociate into ions when dissolved in water. This means that when you dissolve KOH, it breaks apart fully into potassium ions

• K⁺
• OH^-

This process ensures that every molecule of KOH contributes an hydroxide ion (OH^-) to the solution.

Knowing that KOH is a strong base helps us predict the concentration of ions present. It tells us that the number of OH^- ions equals the original concentration of KOH in the solution.
Understanding this concept is crucial for calculating the pH and pOH of basic solutions.

Hydroxide Ion Concentration

In a solution of a strong base like KOH, the hydroxide ion concentration

• [OH⁻]

is equal to the initial concentration of the base. This is because strong bases dissociate completely.

For example, in a KOH solution with a concentration of \(1.2 \times 10^{-4} \mathrm{M}\), the [OH⁻] is also \(1.2 \times 10^{-4} \mathrm{M}\). This is straightforward because each unit of KOH provides one OH⁻ ion.

To find the pOH, we use the formula:

• \( \text{pOH} = -\log[\text{OH}⁻] \)

Substitute the hydroxide ion concentration, and you'll get: \(\text{pOH} \approx 3.92\).

pOH is important because it helps us find the pH of the solution by using the relation \(\text{pH} + \text{pOH} = 14\).

Hydronium Ion Concentration

The hydronium ion concentration

• [H₃O⁺]

can be calculated from the hydroxide ion concentration using the water dissociation constant. Water maintains a balance in which the product of hydronium and hydroxide ion concentrations is equal to \(1.0 \times 10^{-14}\) at 25°C.

For a solution with an [OH⁻] of \(1.2 \times 10^{-4} \mathrm{M}\), use the formula:

• \([\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{[\text{OH}^-]}\)

Substituting the known [OH⁻] gives us: \([\text{H}₃\text{O}^+] \approx 8.33 \times 10^{-11} \mathrm{M}\).

This concentration tells us how acidic or basic a solution is. Although the pH of this KOH solution is above 7 (indicating a basic solution), knowing the hydronium ion concentration gives us deeper insight into the solution's properties.