Answer to Question #123800 in General Chemistry for resom

Question #123800
Calculate the solution conductivities of 0.1 M KOH, 0.1 M NaOH and 0.1 M HNO3. Assume that they are fully dissociated
1
Expert's answer
2020-06-25T07:45:18-0400

LKOH = lK+ + lOH-         [L=molar conductivity, lK+ and lOH- 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

LNaOH = lNa+ + lOH-

      = 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

LHNO3 = lH+ + lNO3-

        = 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



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