NEETChemistryIonic Equilibrium
The pH of a 0.05 ~M aqueous solution of a weak monoprotic acid is 3 . The acid dissociation constant (K_ a ) of this weak acid is :
Options
- A2 10⁻⁵
- B2 10⁻²
- C1 10⁻⁶
- D1 10⁻³
Correct answer
A. 2 10⁻⁵
Step-by-step solution
Given, concentration of the weak acid, C = 0.05 ~M pH = 3 We know that, pH = - [ H ^+] [ H ^+] = 10^ - pH = 10⁻³ ~M For a weak monoprotic acid, the acid dissociation constant K_ a is approximated by: K_ a = [ H ^+]^2 C K_ a = (10⁻³)^2 0.05 = 10⁻⁶ 5 10⁻² K_ a = 0.2 10⁻⁴ = 2 10⁻⁵ Answer: 2 10⁻⁵