Problem 31 - Chemistry The Central Science In Si Units

Question

At the boiling point of water \(\left(100^{\circ} \mathrm{C}\right), K_{w}=5.6 \times 10^{-13} .\) Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Step-by-Step Solution

Answer

\([\mathrm{H}^+] = [\mathrm{OH}^-] = 7.48 \times 10^{-7}\, \text{M}.\)

Step 1: Understand the Relationship

The ionic product of water, denoted as \(K_w\), represents the equilibrium constant for the self-ionization of water: \( K_w = [\mathrm{H}^+][\mathrm{OH}^-] \). For a neutral solution, the concentrations of \([\mathrm{H}^+]\) and \([\mathrm{OH}^-]\) are equal.

Step 2: Set Up the Equation

Since \([\mathrm{H}^+] = [\mathrm{OH}^-]\) in a neutral solution, we can replace \([\mathrm{H}^+]\) with \(x\) and set \([\mathrm{OH}^-] = x\). Therefore, the equation for the ionic product becomes: \( x^2 = K_w \).

Step 3: Solve for \(x\)

Substitute the given value of \(K_w = 5.6 \times 10^{-13}\) into the equation: \( x^2 = 5.6 \times 10^{-13} \). Solve for \(x\) by taking the square root of both sides: \( x = \sqrt{5.6 \times 10^{-13}} \).

Step 4: Calculate the Concentration

Compute the square root: \[ x = \sqrt{5.6 \times 10^{-13}} \approx 7.48 \times 10^{-7} \]. Hence, \([\mathrm{H}^+] = [\mathrm{OH}^-] = 7.48 \times 10^{-7}\, \text{M}.\)

Key Concepts

Self-Ionization of Water

Water is a fascinating substance, not just because it sustains life, but also due to its unique property of self-ionization. This process involves water molecules interacting with each other to produce hydrogen ions (\([\text{H}^+]\)) and hydroxide ions (\([\text{OH}^-]\)). Despite sounding complex, it's a natural and constant process.

• Each water molecule can donate a hydrogen ion to another nearby water molecule.
• This exchange results in the production of one \([\text{H}^+]\) and one \([\text{OH}^-]\) ion simultaneously.

This autoprotolysis, or self-ionization, is crucial as it sets the stage for water's neutral nature, encouraging further reactions and contributing to its solvent abilities.
At any given time, the concentrations of \([\text{H}^+]\) and \([\text{OH}^-]\) in pure water are tiny but measurable, demonstrating water's ability to dissociate into ions constantly.

Neutral Solution

A neutral solution is something to frequently encounter in chemistry. The concept is straightforward.
In a neutral solution, the concentrations of hydrogen ions (\([\text{H}^+]\)) and hydroxide ions (\([\text{OH}^-]\)) are exactly equal.

• This balance means that the solution is neither acidic nor basic.
• For pure water at 25°C, this is typically around \(1.0 \times 10^{-7}\, \text{M}\).

However, this balance can shift with temperature.
For example, as demonstrated in the original exercise at 100°C, though the ion concentrations are different due to increased molecular activity at boiling temperatures, neutrality is still achieved by having equal \([\text{H}^+]\) and \([\text{OH}^-]\) ion concentrations, preserving the essential characteristic of a neutral solution.

Equilibrium Constant

The equilibrium constant is a key concept when discussing reactions, including the self-ionization of water. This constant, denoted as \(K_w\) for water, quantifies the ratio at which the reactants and products of a reversible reaction reach equilibrium.

• It's a fixed numerical value at a given temperature.
• For water, the formula is \(K_w = [\text{H}^+][\text{OH}^-]\).

This means that regardless of whether the system starts with mostly reactants or products, it will adjust concentration of ions to this equilibrium constant.
As temperature increases, the value of \(K_w\) changes, reflecting increased molecular energy and frequency of water molecule collisions.
This dynamic nature of \(K_w\) ensures that scientists can predict ion concentrations at various temperatures, understanding how this crucial constant influences the behavior of aqueous solutions.