1.

At 300 K specific conductivity of ethanol is 4xx10^(-10)mhocm^(-1). The ionic conductances of H^(+),C_(2)H_(5)O^(-) at his temperature is 300 and 100 mhocm^(2)" equivalent"^(-1) respectively. Then the negative logarithm of ionic product of alcohol will be 18.

Answer»


Solution :`lamda^(infty)=lamda_(H^(+))^(infty)+lamda_(C_(2)H_(5)O^(-))^(infty)=400`
`thereforelamda=kxx(1000)/(C)""C=[H^(+)]=[.^(-)OC_(2)H_(5)]`
`thereforeC=(4xx10^(-10)xx1000)/(400)=10^(-9)M`
`thereforeK_("ALCOHOL")=[H^(+)][OC_(2)H_(5)]=(10^(-9))^(2)`
`thereforepK_("alcohol")=-log(10^(-18))=18`


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