The conductance of 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross-section of `1 cm^(2)`. The conductance of this solution was found to be `5xx10^(-7) S`. The pH of the solution is 4. Calculate the value of limiting molar conductivity.

Explanation

Step I. Calculation of specific conductance
Specific conductance (k) `="conductance"xx"cell constant"`
`=(5xx10^(-7)"ohm"^(-1))xx((120 cm))/((1 cm^(2)))=6xx10^(-5)"ohm"^(-1)cm^(-1)`
Step II. Calculation of molar conductance
Molar concentration (C )`=1.5xx10^(-3)" mol "L^(-1)`
`=(1.5xx10^(-3)" mol")/(1"L")=(1.5xx10^(-6)" mol")/(10^(3)cm^(3))=1.5xx10^(-6)" mol "cm^(-3)`
`"Molar conductance" (Lambda_(m)^(c ))=(k)/(C )=((6xx10^(-5)"ohm"^(-1)cm^(-1)))/((1.5xx10^(-6)" mol "cm^(-3)))=40 " ohm"^(-1)cm^(2)" mol "L^(-1)`
Step III. Calculation of molar conductance at infinite dilution `(Lambda_(m)^(0))`
pH of solution` =4,[H^(+)]=10^(-4)" mol "L^(-1)`
`[H^(+)]=Ca" or "alpha=([H^(+)])/(C )=((10^(-4)" mol "L^(-1)))/((1.5xx10^(-1)" mol "L^(-1)))=(1)/(15)`
Now `alpha=(Lambda_(m)^(C ))/(Lambda_(m)^(0))" or "Lambda_(m)^(0)=(Lambda_(m)^(C ))/(alpha)=(40"ohm"^(-1)cm^(2)" mol"^(-1))/((1)/(15))=600"ohm"^(-1)cm^(2)mol^(-1)`