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Molar conductivity of aqueous solution o...

Molar conductivity of aqueous solution of `HA` is `200Scm^(2)mol^(-1),pH` of this solution is `4`
Calculate the value of `pK_(a)(HA)` at `25^(@)C`.
Given `^^_(M)^(oo)(NaA)=100scm^(2)mol^(-1),`
`^^_(M)^(oo)(HCl)=425Scm^(2)mol^(-1),`
`^^_(M)^(oo)(NaCl)=125 Scm^(2)mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
4
`^^_(M)^(oo)(HA) =^^_(M)^(oo)(HCl)+^^_(M)^(oo)(NaA)-^^_(M)^(oo)(NaCl)`
`=425+100-125=400cm^(2)mol^(-1)`
`pH=4,[H^(+)]=10^(-4)=alphaC`
`alpha=(^^m)/(^^_(m)^(oo))=200/400=0.5`
`K_(a)=((Calpha)alpha)/((1-alpha))=(10^(-4)(0.5))/((1-alpha))=10^(-4), pK_(a)=4`
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