Question

Calculate the solution conductivities of 0.1 M KOH, 0.1 M NaOH and 0.1 M HNO3. Assume that they are fully dissociated

Answer 1

L_KOH = l_K⁺ + l_OH^-         [L=molar conductivity, l_K⁺ and l_OH^- molar ionic conductivity of K⁺                             and OH^- respectively]

  = 73.5+198.6 Sm²mol^-1

  = 272.1 Sm²mol^-1

L = (1000×K)/C     K = conductivity

Conductivity of 0.1 M KOH is

 K = 272.1× 0.1/1000 Sm^-1

      = 0.027 Sm^-1 Answer

L_NaOH = l_Na⁺ + l_OH^-

      = 50.1+198.6 Sm²mol^-1

      = 248.7 Sm²mol^-1

Conductivity of 0.1 M NaOH is

K = 248.7× 0.1/1000 Sm^-1

  = 0.024 Sm^-1 Answer

L_HNO3 = l_H⁺ + l_NO3^-

        = 349.8+71.4 Sm²mol^-1

       = 421.2 Sm²mol^-1

Conductivity of 0.1 M HNO₃ is

K = 421.2× 0.1/1000 Sm^-1

  = 0.042 Sm^-1 Answer