MHT CET202522 Apr 2025Evening ShiftChemistryIonic EquilibriumActual
A weak monoacidic base dissociates to 1.5 % in 0.001 M solution at 298 K . Calculate the dissociation constant of weak base.
Options
- A2.25 10⁻⁷
- B3.05 10⁻⁷
- C2.5 10⁻⁵
- D3.725 10⁻⁶
Correct answer
A. 2.25 10⁻⁷
Step-by-step solution
The weak monoacidic base BOH dissociates as BOH B ^+ + OH ^- with initial concentration C = 0.001 , M = 10⁻³ , M and percentage dissociation 1.5 % . The degree of dissociation is = 1.5 / 100 = 0.015 . At equilibrium, [ B ^+] = [ OH ^-] = C and [ BOH ] = C(1 - ) . The base dissociation constant is given by K_b = [ B ^+][ OH ^-] [ BOH ] = C^2 ^2 C(1 - ) = C ^2 1 - . Since = 0.015 is small, we approximate 1 - 1 , yielding K_b C ^2 = (10⁻³) (0.015)^2 = (10⁻³) (2.25 10⁻⁴) = 2.25 10⁻⁷ . The exact value would be K_b = 2.2