The molar conductivity of a 10⁻³ M solution of a weak monoprotic acid HA is 3.9 S cm ^2 mol ⁻¹ . The limiting molar conductivities ( ^ _m ) of HCl, NaCl, and NaA are 426 S cm ^2 mol ⁻¹ , 126 S cm ^2 mol ⁻¹ , and 90 S cm ^2 mol ⁻¹ respectively. The pK_a of the acid HA is ____. [Assume degree of dissociation 1 ]

Step-by-step solution

According to Kohlrausch's law of independent migration of ions: ^ _m( HA ) = ^ _m( HCl ) + ^ _m( NaA ) - ^ _m( NaCl ) ^ _m( HA ) = 426 + 90 - 126 = 390 S cm ^2 mol ⁻¹ The degree of dissociation is: = _m ^ _m = 3.9 390 = 0.01 = 10⁻² For a weak acid (assuming 1 ), the dissociation constant K_a is: K_a = C ^2 = 10⁻³ (10⁻²)^2 = 10⁻⁷ The pK_a of the acid is: pK_a = - (K_a) = - (10⁻⁷) = 7 Answer: 7