The conductivity of 0.01 M solution of CH ₃ COOH at 298 K is 1.65 10⁻⁴ Scm ⁻¹ What is the pK _ a value of the acid if ^0 ( H ⁺ ) and ^0 ( CH ₃ COO )⁻¹ are 349.1 Scm ^2 ~mol ⁻¹ and 40.9 Scm ^2 ~mol ⁻¹ respectively?

Step-by-step solution

The molar conductivity _m is given by the formula _m = 1000 C , where = 1.65 10⁻⁴ S cm ⁻¹ and C = 0.01 M . _m = 1.65 10⁻⁴ 1000 0.01 = 16.5 S cm ^2 mol ⁻¹ . The limiting molar conductivity _m^0 is ^0( H ^+) + ^0( CH ₃ COO ^-) = 349.1 + 40.9 = 390.0 S cm ^2 mol ⁻¹ . The degree of dissociation is given by = _m _m^0 = 16.5 390.0 0.0423 . The dissociation constant K_a is given by K_a = C ^2 1 - . Since is small, 1 - 1 . K_a C ^2 = 0.01 (0.0423)^2 = 0.01 0.001789 = 1.789 10⁻⁵ . The pK _a value is - ₁₀(K_a) = - ₁₀(1.789 1