The specific conductance (k)of a 0.1M NaOH solution is 0.0221 S cm^(-1). On addition of an equal volume (V) of0.10 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).

The specific conductance (k)of a 0.1M NaOH solution is 0.0221 S cm^(-1

1.	The specific conductance (k)of a 0.1M NaOH solution is 0.0221 S cm^(-1). On addition of an equal volume (V) of0.10 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» Solution :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)` Soans is `3 +4 +2 = 9`