What is the Kb of NaOH?
The correct answer and explanation is:
Correct Answer:
The Kb of NaOH (sodium hydroxide) is not applicable or undefined, because NaOH is a strong base.
Explanation (300 words):
In acid-base chemistry, the base dissociation constant (Kb) is used to measure the strength of a weak base in water. It represents the equilibrium constant for the reaction in which a base reacts with water to produce hydroxide ions (OH⁻). For example: B+H2O⇌BH++OH−\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-
The Kb value quantifies how much the base dissociates to form OH⁻ at equilibrium. A higher Kb indicates a stronger base (more dissociation), while a lower Kb indicates a weaker base (less dissociation).
However, NaOH (sodium hydroxide) is classified as a strong base, meaning it dissociates completely in water: NaOH→Na++OH−\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-
Since the reaction goes to completion and there is no equilibrium, we do not define a Kb value for strong bases like NaOH. The concept of Kb is only useful for weak bases where an equilibrium is established.
In contrast, if we were discussing a weak base like ammonia (NH₃), which partially reacts with water: NH3+H2O⇌NH4++OH−\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-
We can calculate a meaningful Kb (for NH₃, Kb ≈ 1.8 × 10⁻⁵ at 25°C).
Because NaOH dissociates 100% into Na⁺ and OH⁻, we simply say that its basicity is maximal, and a Kb value does not exist or is approaching infinity in practical terms. Therefore, in problems involving NaOH, you typically don’t use Kb—instead, you use stoichiometry and concentration directly to find [OH⁻] and pOH.