Question

• Draw/Insert a picture of water molecule.

• Explain and illustrate what happens when water dissociates/ionizes.

Answer 1

Water dissociates/ionizes into positively charged hydrogen ions (H+) and negatively charged hydroxide ions (OH-) when its molecules split apart.

Explanation

Water, composed of two hydrogen atoms and one oxygen atom, undergoes dissociation or ionization when energy is introduced, typically through an electrical current or other catalysts. This process breaks the hydrogen-oxygen bonds within water molecules, leading to the separation of H+ (hydrogen ions) and OH- (hydroxide ions).

The hydrogen atom loses an electron, becoming a positively charged ion (H+), while the remaining hydroxide group gains the electron, acquiring a negative charge (OH-). This dissociation creates an equilibrium between ions, maintaining the balance between the concentrations of H+ and OH- in a neutral solution.

This phenomenon is essential in various chemical reactions and biological processes. For instance, in acids, higher concentrations of H+ are present due to a greater dissociation, making the solution more acidic. Conversely, bases have more OH- ions, leading to higher pH levels.

The dissociation of water is critical in maintaining homeostasis in living organisms as it influences the pH levels in cells and bodily fluids. Additionally, this ionization enables water to act as a solvent, facilitating the dissolution of various substances due to its polar nature, essential for biological functions like nutrient transport and metabolic reactions.

Answer 2

Water undergoes self-ionization, producing equal amounts of hydroxide ions (OH⁻) and hydronium ions (H₃O⁺).

Step-by-step explanation:

When water undergoes dissociation or ionization, it forms ions through the transfer of a proton (H⁺). The chemical equation for this process is:

`\[ \text{H}_2\text{O} \rightleftharpoons \text{H}⁺ + \text{OH}⁻ \]`

In this reaction, a water molecule (H₂O) donates a proton to another water molecule, forming a hydroxide ion (OH⁻) and a hydronium ion (H₃O⁺). The concentration of each ion in pure water at 25°C is
`\(1.0 * 10^(-7) \, \text{mol/L}\)` due to the equilibrium constant for water, known as the ion product of water
`(\(K_w\)), which is \(1.0 * 10^(-14)\)` at this temperature. The mathematical expression for
`\(K_w\)` is given by:

`\[ K_w = [\text{H⁺}][\text{OH⁻}] \]`

At 25°C, if we let
`\([ \text{H⁺} ]\) be \(x\), then \([ \text{OH⁻} ]\)` is also x, and the equation becomes:

`\[ x * x = 1.0 * 10^(-14) \]`

Solving for x gives the concentrations of both
`\([ \text{H⁺} ]\) and \([ \text{OH⁻} ]\) as \(1.0 * 10^(-7) \, \text{mol/L}\).` This demonstrates the dynamic equilibrium between the production and recombination of hydronium and hydroxide ions in water.

In summary, water dissociation leads to the simultaneous generation of hydronium ions and hydroxide ions, resulting in a delicate balance represented by the ion product of water at a given temperature.