Explain the relationship between the self-ionization of water and \(K_{w}\) .

Short Answer

Expert verified
The self-ionization of water describes how water molecules can split into hydronium and hydroxide ions. The water ion product, \(K_{w}\), is the equilibrium constant for this reaction that measures the concentrations of these ions in the solution at equilibrium. They are related in such a way that any changes in the concentration of one ion due to shifts in the ionization equilibrium will affect the other ion's concentration to maintain \(K_{w}\) constant.

Step by step solution

01

Understand the Self-Ionization of Water

Water can behave as both an acid (proton donor) and a base (proton receiver) - this dual nature enables water to self-ionize. In the self-ionization process, one water molecule (H2O) donates a proton (H+) to another water molecule, resulting in the formation of a hydronium ion (H3O+) and a hydroxide ion (OH-). This can be represented by the equation: 2H2O(l) ↔ H3O+(aq) + OH⁻(aq).
02

Understand the Water Ion Product (Kw)

The water ion product, denoted as \(K_{w}\), is the equilibrium constant for this self-ionization reaction. It describes the product of the molar concentration of hydrogen ions (H3O+) and hydroxide ions (OH-) in water at a specific temperature. Mathematically, it's represented as: \(K_{w} = [H3O+][OH-]\). At 25°C, Kw is equal to \(1.0 x 10^{-14}\).
03

Relate Self-Ionization and Kw

In a neutral solution at 25°C, due to the equal concentrations of H3O+ and OH-, each is equal to \(\sqrt{K_{w}} = 1.0 x 10^{-7}\) M. If the concentrations of these ions change, the solution may become acidic (if [H3O+] > [OH-]) or basic (if [OH-] > [H3O+]). However, at a given temperature, the product of their concentrations, \(K_{w}\), remains constant. This clearly shows the relationship between the self-ionization of water and \(K_{w}\), as any changes in one of these ions' concentration due to a shift in ionization equilibrium will influence the other to maintain a constant \(K_{w}\).

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