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)`.

Answer

9

Explanation

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`