Write the balanced equation for the dissociation of solid KOH in water. This reaction is not in equilibrium and goes to completion. (Be sure to include all states of matter in the equation.)
The correct answer and explanation is:
Balanced Equation for the Dissociation of Solid KOH in Water: KOH (s)→H2OK+(aq)+OH−(aq)\text{KOH (s)} \xrightarrow{\text{H}_2\text{O}} \text{K}^+ (aq) + \text{OH}^- (aq)
Explanation:
Potassium hydroxide (KOH) is a strong base and an ionic compound that dissociates completely when dissolved in water. The solid form, KOH (s), consists of potassium ions (K⁺) and hydroxide ions (OH⁻) held together by ionic bonds in a crystal lattice.
When KOH is added to water, it dissolves and dissociates into its constituent ions. This dissociation is a physical change driven by the strong ion-dipole interactions between water molecules and the ions. Water molecules surround and stabilize the individual K⁺ and OH⁻ ions through hydration (solvation), which helps pull the ions apart from the solid lattice.
The reaction can be represented as: KOH (s)→K+(aq)+OH−(aq)\text{KOH (s)} \rightarrow \text{K}^+ (aq) + \text{OH}^- (aq)
- KOH (s) = solid potassium hydroxide
- K⁺ (aq) = potassium ion solvated in water (aqueous)
- OH⁻ (aq) = hydroxide ion solvated in water (aqueous)
Since KOH is a strong base, it dissociates completely in water; thus, the reaction goes to completion and is not an equilibrium process. This means that essentially all the KOH molecules separate into ions, and no significant amount remains as intact KOH in solution.
This dissociation is important in many chemical and biological processes because the hydroxide ions increase the solution’s pH, making it basic (alkaline). This is why KOH solutions are commonly used in laboratories and industry to provide a strongly basic environment.
Summary:
- KOH dissociates fully into K⁺ and OH⁻ ions in water.
- The process is not reversible (no equilibrium).
- The ions are hydrated by water, stabilizing them in solution.
- This makes the solution strongly basic due to the OH⁻ ions.