Question

What is the \(\mathrm{pH}\) of a solution when \(\left[\mathrm{OH}^{-}\right]\) is \(0.0222 \mathrm{M}\) ?

Short answer

The \\( ext{pH}\\) of the solution is approximately 12.35.

Step-by-step solution

Understand the Concept

The concept of pH relates to the concentration of hydrogen ions \(\left[\mathrm{H}^{+}\right]\) in a solution. Since the concentration of hydroxide ions \(\left[\mathrm{OH}^{-}\right]\) is given, we will first find \(\left[\mathrm{H}^{+}\right]\) using the relation between \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\): \\[ \left[\mathrm{H}^{+}\right] \times \left[\mathrm{OH}^{-}\right] = 1.0 \times 10^{-14} \]

Solve for \\(\left[\mathrm{H}^{+}\right]\\)

Given \(\left[\mathrm{OH}^{-}\right] = 0.0222 \, \mathrm{M}\), we use the formula for water's ion product: \[ \left[\mathrm{H}^{+}\right] = \frac{1.0 \times 10^{-14}}{0.0222} \approx 4.50 \times 10^{-13} \; \mathrm{M} \]

Calculate \\( ext{pH}\\)

To find the \(\mathrm{pH}\), use the relation: \[ \mathrm{pH} = -\log_{10} \left(\left[\mathrm{H}^{+}\right]\right) \]\Substitute \(\left[\mathrm{H}^{+}\right] = 4.50 \times 10^{-13} \mathrm{M}\): \[ \mathrm{pH} = -\log_{10}(4.50 \times 10^{-13}) \approx 12.35 \]

Key concepts

Hydrogen Ion Concentration

In the realm of chemistry, understanding hydrogen ion concentration is key to determining the acidity or basicity of a solution. Hydrogen ions, denoted as \([\mathrm{H}^+]\), play a crucial role in various chemical reactions.

In water, a small fraction of water molecules dissociate to form hydrogen ions \([\mathrm{H}^+]\) and hydroxide ions \([\mathrm{OH}^-]\). This occurs naturally at a constant rate, leading to very small concentrations of \([\mathrm{H}^+]\) in neutral solutions. To calculate the pH of a solution, it is necessary to first determine the concentration of these hydrogen ions by using the water ion product equilibrium formula.

The formula \( \left[\mathrm{H}^+\right] \times \left[\mathrm{OH}^-\right] = 1.0 \times 10^{-14} \ \text{(at 25°C)} \) provides us with a straightforward way to find \([\mathrm{H}^+]\) when the concentration of \([\mathrm{OH}^-]\) is known. This relationship is vital because it enables us to understand how changes in one ion concentration affect the other and thus determine the solution's pH.

Hydroxide Ion Concentration

Hydroxide ions \([\mathrm{OH}^-]\) are negatively charged ions that are critical in the characterization of basic solutions. When a solution has a high concentration of hydroxide ions, it is considered basic or alkaline.

In the context of pH calculations and water chemistry, the concentration of hydroxide ions is directly linked to the concentration of hydrogen ions. The more \([\mathrm{OH}^-]\) ions in a solution, the fewer \([\mathrm{H}^+]\) ions are present, which increases the solution's pH. This inversely proportional relationship is essential for students to grasp because it explains why, as the concentration of \([\mathrm{OH}^-]\) increases, the pH value rises, indicating a more basic solution.

For example, in our exercise, given \(\left[\mathrm{OH}^-\right] = 0.0222 \ \text{M}\), students are tasked to find \([\mathrm{H}^+]\) in order to compute pH—hence understanding and manipulating \([\mathrm{OH}^-]\) is crucial for pH calculation.

Water Ion Product

The water ion product, often expressed as \(K_w\), is a fundamental concept in understanding the chemistry of aqueous solutions. It refers to the constant product of the concentrations of hydrogen ions and hydroxide ions in water.

At 25°C, the value of \(K_w\) is \(1.0 \times 10^{-14}\). This small but significant number is extremely useful because it allows chemists to determine the concentrations of either ion if the other is known. This equilibrium expression is a cornerstone of acid-base chemistry.

• The constant value \(K_w\) implies that if one changes, the other must adjust to maintain the equilibrium.
• In neutral water, \([\mathrm{H}^+]\) and \([\mathrm{OH}^-]\) are equal, hence each is \(1.0 \times 10^{-7} \ \text{M}\).
• For solutions other than pure water, \(\left[\mathrm{H}^+\right] \times \left[\mathrm{OH}^-\right] \) still equals \(1.0 \times 10^{-14}\).

Understanding \(K_w\) is vital because it explains how acids and bases interact in water, serving as the gateway to calculating pH and understanding the nature of different solutions.