An amine, R–NH2, accepts a proton from water as shown below. R-NH2 +…

The correct answer is Option (4) → \( K_b = \frac{[\ce{R-NH3^+}][\ce{OH^-}]}{[\ce{R-NH2}]} \)

• The equilibrium constant K of the reaction has the products on top and the reactants below, water included:

      \( \displaystyle K = \frac{[\ce{R-NH3^+}][\ce{OH^-}]}{[\ce{R-NH2}][\ce{H2O}]} \)

• Water is present in large excess, so its concentration stays practically constant. It is taken to the left side and joined with K; the product K[H₂O] is the base dissociation constant, K_b:

      \( \displaystyle K[\ce{H2O}] = K_b = \frac{[\ce{R-NH3^+}][\ce{OH^-}]}{[\ce{R-NH2}]} \)

      \( \displaystyle \mathrm{p}K_b = -\log K_b \)

• A larger K_b, or a smaller pK_b, means a stronger base.

Option (1) \( K_b = \frac{[\ce{R-NH3^+}][\ce{OH^-}]}{[\ce{R-NH2}][\ce{H2O}]} \) — Incorrect. This is K, not K_b; K_b does not carry the [H₂O] term.

Option (2) \( K_b = \frac{[\ce{R-NH2}]}{[\ce{R-NH3^+}][\ce{OH^-}]} \) — Incorrect. This is the expression upside down, that is 1/K_b.

Option (3) \( K_b = \frac{[\ce{R-NH2}][\ce{H^+}]}{[\ce{R-NH3^+}]} \) — Incorrect. This describes R–NH₃⁺ giving up a proton, which is the reverse kind of process, not the amine accepting a proton from water.

NCERT (2026-27 edition):
9.6 Chemical Reactions – “Basic character of amines can be better understood in terms of their K_b and pK_b values as explained below:”

“Larger the value of K_b or smaller the value of pK_b, stronger is the base.”