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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 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 the limiting molar conductivity (wedge_(m)^(@)) of the weak monobasic acid is aqueous solution is Zxx10^(2).S" "cm^(2)mol^(-1). the value of Z is |
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Answer» Solution :CONC `C=0.0015M`, conductance, `G=5xx10^(-7)S` `l=120cm,a=1cm^(2)` `wedge_(m)^(c)=(Kxx1000)/("Molarity")=(6xx10^(-5)xx1000)/(0.0015)` `=40" S "cm^(2)mol^(-1)` `{:(HA,HARR,H^(+),+,A^(-)),(C,,0,,0),(C-Calpha,,Calpha,,Calpha):}` `[H^(+)]=Calpha=10^(-4)M""(becausepH=4)` `theeforealpha=(10^(-4))/(C)=(10^(-4))/(0.0015)` `alpha=(wedge_(m)^(c))/(wedge_(m)^(@)) therefore(10^(-4))/(0.0015)=(40)/(wedge_(m)^(@))` or `wedge_(m)^(@)=(40xx0.0015)/(10^(-4))=6xx10^(2)" S "cm^(2)mol^(-1)` |
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