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InterviewSolution
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The specific conductance `(k)`of a `0.1M NaOH` solution is `0.0221 S cm^(-1)`. On addition of an equal volume `(V)` of `0.10 M HCI` solution, the value of specific conductance `(k)` falls to `0.0056 S cm^(-1)`. On further addition of same volume `(V)` of `0.10 M HCI`, the value of k rises to `0.017 S cm^(-1)`. calculate equivalent conductivity `(^^)` for `HCI` in `S cm^(2) mol^(-1)`. |
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Answer» Correct Answer - 9 On further adding `V` volume `0.10 M HCI`, concentration of `NaCI` and `HCI` in the final solution are: `[NaCI] = (0.05 xx2)/(3) = (1)/(30)M` `[HCI] = (0.1xx1)/(3) = (1)/(30)M` Also, since specific conductivity is additive: `k = K_(NaCI) +K_(HCI)` `rArr K_(HCI) = 0.017 - 0.0056` `= 0.0114 S cm^(-1)` `rArr ^^_(HCI) = (0.0114 xx 1000)/((1)/(30)) = 342 S cm^(2) mol^(-1)` So ans is `3 +4 +2 = 9` |
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