The limiting molar conductivities ( _m^ ) of three strong electrolytes are given below: Electrolyte _m^ ( S cm ^2 mol ⁻¹ ) HNO ₃ 421 KNO ₃ 145 KA 104 The molar conductivity of a 0.02 M solution of the weak acid HA is 19 S cm ^2 mol ⁻¹ . What is the dissociation constant ( K_a ) of the weak acid HA ? (Assume 1 - 1 )

Step-by-step solution

According to Kohlrausch's law of independent migration of ions: _m^ ( HA ) = _m^ ( HNO ₃) + _m^ ( KA ) - _m^ ( KNO ₃) _m^ ( HA ) = 421 + 104 - 145 = 380 S cm ^2 mol ⁻¹ The degree of dissociation ( ) is given by: = _m _m^ = 19 380 = 0.05 The dissociation constant ( K_a ) is calculated as: K_a = c ^2 K_a = 0.02 (0.05)^2 K_a = 0.02 0.0025 = 5.0 10⁻⁵ mol L ⁻¹ Answer: 5.0 10⁻⁵ mol L ⁻¹