The resistance of a 0.05 M solution of a weak acid HA in a conductivity cell is 50 . The cell constant of the conductivity cell is 0.2 cm ⁻¹ . The limiting molar conductivities of H ^+ and A ^- are 350 S cm ^2 mol ⁻¹ and 50 S cm ^2 mol ⁻¹ respectively. The acid dissociation constant ( K_a ) of the weak acid is x 10⁻⁴ . The value of x is ________.

Step-by-step solution

Specific conductivity ( ) of the solution is given by: = G^* R = 0.2 50 = 4 10⁻³ S cm ⁻¹ Molar conductivity ( _m ) is: _m = 1000 C = 4 10⁻³ 1000 0.05 = 80 S cm ^2 mol ⁻¹ Limiting molar conductivity ( _m^ ) of the weak acid HA is: _m^ = ^ _ H ^+ + ^ _ A ^- = 350 + 50 = 400 S cm ^2 mol ⁻¹ Degree of dissociation ( ) is: = _m _m^ = 80 400 = 0.2 The acid dissociation constant ( K_a ) is: K_a = C ^2 1 - = 0.05 (0.2)^2 1 - 0.2 = 0.05 0.04 0.8 = 0.002 0.8 = 0.0025 = 25 10⁻⁴ Thus, the value of x is 25 . Answer: 25