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The molar conductivity of 0.025 mol L^(-1) methanoic acid (HCOOH) is 46.1 Scm^2 mol^(-1). Calculate its degree of dissociation and dissociation constant. Given lambda^@ (H^+)=349.6 S cm^2 mol^-1 and lambda (HCOO^-)= 54.6 S cm^2 mol^-1 |
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Answer» Solution :`lambda^@_m (HCOOH)= lambda^@(H^+) + lambda^@(HCOO^-)=349.6 + 54.6 = 404.2 S cm^2 mol^-1` Degree of DISSOCIATION of HCOOH = `Lambda_m/Lambda_m^@= 46.1/404.2=0.114` for WEAK electrolytes, `K_a = (Calpha^2)/(1-alpha)THEREFORE K_a=(0.025xx(0.114)^2)/(1-0.0114)=3.67xx10^-4` |
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