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The conductivity of a 0.01 M solution of a 1:1 weak electrolyte at 298 K is 1.5 times 10^(-4)" S " cm^(-1). i) molar conductivity of the solution ii) degree of dissociation and the dissociation constant of the weak electrolyte Given that lambda_("cation")^(@)=248.2" S "cm^(2)mol^(-1) lambda_("anion")^(@)=51.8" S "cm^(2)mol^(-1) |
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Answer» SOLUTION :Given C = 0.01 M `""lambda_("cation")^(@)=248.2" S "cm^(2)mol^(-1)` `K=1.5 times 10^(-4)" S " cm^(-1) " "lambda_("anion")^(@)=51.8" S "cm^(2)mol^(-1)` (i) Molar conductivity `wedge_(m)^(@)=(""^(@)(SM^(-1)) times 10^(-3))/(C("in M"))mol^(-1)m^(3)` `" "^(@)=1.5 times 10^(-4)" S " cm^(-1)` `""=(1.5 times 10^(2) times 10^(-3))/(0.01)S" "mol^(-1)m^(2)` `1cm^(-1)=10^(2)m^(-1)` `""=1.5 times 10^(-3)" S "m^(2)mol^(-1)` `""=1.5 times 10^(2)` (ii) Degree of DISSOCIATION `alpha=WEDGE^(@)/wedge_(infty)^(@)` `wedge_(infty)^(@)=lambda_("cation")^(@)+lambda_("anion")^(@)` `" "=(248.2+51.8)" S "cm^(2)mol^(-1)` `" "=300" S " cm^(2)mol^(-1)` `" "=300 times 10^(-14)" S "m^(2)mol^(-1)` `alpha=(1.5 times 10^(-3)" S"m^(2)mol^(-1))/(300 times 10^(-4)Sm^(2)mol^(-1))` `alpha=0.05` |
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