Final answer:
The Kb for the conjugate base of a weak acid with Ka of 2.0 × 10^-6 is 5.0 × 10^-9, calculated using the relationship Kw = Ka × Kb where Kw is the ion product of water.
Step-by-step explanation:
The subject of your question falls under the field of chemistry, specifically dealing with acid-base equilibrium. In order to find the Kb value of the conjugate base of a weak acid, we use the ion product of water (Kw), which is 1.0 × 10^-14 at 25°C.
The relationship between Ka, Kb, and Kw is given by the equation Kw = Ka × Kb. In your case, Ka is given as 2.0 × 10^-6. By rearranging the equation to solve for Kb, we get Kb = Kw / Ka.
So for the conjugate base of a weak acid with Ka of 2.0 × 10^-6, Kb would be calculated as Kb = (1.0 × 10^-14) / (2.0 × 10^-6) = 5.0 × 10^-9.
Learn more about Acid-Base Equilibrium