Question

(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is `8.0 xx 10^(-5)" S "cm^(-1)`.
Given . `lambda_(H^(+))^(@) = 349.6" S " cm^(2)" mol"^(-1), lamda_(CH_(3)COO^(-))^(@)= 40.9" S " cm^(2) " mol "^(-1)`
(b) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases to a smaller extent while that of ‘A’ increases to a much larger extent comparatively. Which of the two is a strong electrolyte? Justify your answer.

Answer 1

(a) `^^ ^(@)(CH_(3)COOH)=lamda_(H+)^(@)+lamda_(CH_(3)COO^(-))^(@)`
`=349.6+40.9=390.5" S cm"^(2)" mol "^(-1)`
`^^_(m)=(kxx1000)/(c)`
`^^_(m)(8.0xx10^(-5)" S "cm^(-1)L^(-1))/(0.0024" mol "L^(-1))=33.33" S cm"^(2)" mol"^(-1)`
`alpha=(^^_(m))/(^^_(m)^(@))`
`alpha=(33.33" S cm"^(2)" mol"^(-1))/(390.5" S cm"^(2)" mol"^(-1))=0.085`
Electrolyte B is a strong electrolyte.
Limiting molar conductivity increases only to a smaller extent for a dtrong electrolyts, as on dilution the interionic interactions are overcome.
Limiting molar conductivity the degree of dissociation increases, therefore the number of ions in total volume of solution increases.