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The conductance of a 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. The value of limiting molar conductivity (wedge_(m)^(0)) of this weak monobasic acid is aqueous solution is Zxx10^(2)S" "cm^(-1)mol^(-1). the value of Z is:- |
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Answer» Solution :`C=0.0015M""l=120cm` `G=5xx10^(-7)s""a=1cm^(2)` `G=KXX(a)/(l)` `5xx10^(-7)=Kxx(1)/(120)` `K=6xx10^(-5)S" "cm^-1` `(^^)_(m)^(c)=(Kxx1000)/(M)=(6xx10^(-5)xx1000)/(0.0015)` `pH=4implies[H^(+)]=10^(-4)=calpha=0.0015alpha` `alpha=(10^(-4))/(0.0015)""alpha=((^^)_(m)^(c))/((^^)_(m)^(0))impleis(10^(-4))/(0.0015)=((6xx10^(-5)xx1000)/(0.0015))/((^^)_(m)^(0))` `(^^)_(m)^(0)=6xx10^(2)s" "cm^(2)"mole"-1` |
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