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

`k=Gxx1/a=5xx10^(-7)xx120/1=6xx10^(-) S cm^(-1)`
`A_m^(C)=(kxx1000)/("Molarity")=(6xx10^(-5)xx1000)/(0.0015)=40 S Cm^(2) mol ^(-1)`
`pH=4=-log[H^(+)] therefore [H^(+)]=10^(-4)M`
Ineternal Conc : 0/.0015 0
Equillibrium Conc. 0.0015-alpha 0.0015 alpha o.oo15 alpha
Thus,`[H^(++)]=0.0015alpha=10^(-4)implies alpha (10^(-4))/0.0015`
Also ,`alpha (A_m^(C))/A_m^(0)therefore (10^(-))/0.0015 =40/(A_m^(@))`
`A_m^(@) =(40xx0.0015)/(10^(-4))=600=6xx10^(2) S Cm^(2) m ol ^(-1)`
On comparing it wZ=6ith `Z xx10 ^(2) S Cm ^(2) mo l^(-1)` , we get : `z=6`