The conductivity of 0.001028 mol L–1 acetic acid is 4.95 × 10–5 S cm–1. Calculate its dissociation constant if Ëm ° for acetic acid is 390.5 S cm2 mol–1.
prasanjeet kumar
12 Years agoGrade 8
1 Answer
Sunil Kumar FP
11 Years ago
conductivity of .001028 mol/l acetic acid =4.95*10^-5Scm^-1 therefore molar conductivity=4.95*10^-5/.001028 =4815.17/1000=48.15Scm^3mol^-1 degree of dissociation,d=48.15/390.5=.1233 now dissociation constane=cd^2/1-d=.001028*.1233^2/1-.1233=1.78*10^-5molL^-1