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If the molar conductivities at infinite dilution of NaCl, HCl and CH_(3)COONa(NaAc) are 126.4, 425.9 and 91.0 S cm^(2)mol^(-1) respectively, what will be that of acetic acid (Hac)? |
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Answer» Solution :According to KOHLRAUSCH's law, `wedge^(@)` for `CH_(3)COOH=lamda_(CH_(3)COO^(-))^(@)+lamda_(H^(+))^(@)` `wedge^(@)` for NaCl=`lamda_(Na^(+))^(@)+lamda_(Cl^(-))^(@)=126.4" S "cm^(2)mol^(-1)`. . . (i) `wedge^(@)` for `HCl=lamda_(H^(+))^(@)+lamda_(Cl^(-))^(@)=425.9" S "cm^(2)mol^(-1)`. . . (ii) `wedge^(@)` for `CH_(3)COONa=lamda_(CH_(3)COO^(-))^(@)+lamda_(Na^(+))^(@)=91.0" S "cm^(2)mol^(-1)`. . . (III) ADDING eqns. (ii) and (iii) and subtracting (i), we get `lamda_(H^(+))^(@)+lamda_(Cl^(-))^(@)+lamda_(CH_(3)COO^(-))^(@)-lamda_(Na^(+))^(@)-lamda_(Na^(+))^(@)-lamda_(Cl^(-))^(@)=425.9+91.0-126.4" S "cm^(2)mol^(-1)` or `lamda_(CH_(3)COO^(-))^(@)+lamda_(H^(+))^(@)=390.5" S "cm^(2)mol^(-1)` i.e., `wedge^(@)` for `CH_(3)COOH=390.5" S "cm^(2)mol^(-1)`. |
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