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

Question

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

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.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - T
`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`

4 views Answered 2023-02-05 15:29