An aqueous solution of a weak monoprotic acid (HX) has a molality of 0.1 m . The observed freezing point of this solution is -0.2232^ C . The acid dissociation constant ( K_a ) of the weak acid is : [Given : Freezing point depression constant of water, K_f = 1.86 K kg mol ⁻¹ ]

Step-by-step solution

The depression in freezing point is given by: T_f = i K_f m Substituting the given values: 0.2232 = i 1.86 0.1 i = 0.2232 0.186 = 1.2 For a weak monoprotic acid HX, the dissociation is HX H ^+ + X ^- . The van 't Hoff factor is i = 1 + . 1 + = 1.2 = 0.2 The acid dissociation constant K_a is given by: K_a = m ^2 1 - K_a = 0.1 (0.2)^2 1 - 0.2 = 0.1 0.04 0.8 = 0.004 0.8 = 0.005 = 5.0 10⁻³ . Answer: 5.0 10⁻³