Conductivity of 0.00241 M acetic acid is 7.896 x 10^(-5) S cm^(-1). Calculate its molar conductivity. If `A°_m` for acetic acid is `390.5 S cm^2` `mol^(-1)`, what is its dissociation constant?

Explanation

Step 1 : Calculation of molar conductance `(wedge^(C)m)`
`K=7.896xx10^(-5)Scm^(-1)`
`C=0.00241 mol L^(-1)`
`M=(0.00241mol)/(10^(3)cm^(3))=241xx10^(-8)mol cm^(-3)`
`(wedge^(C)m)=(K)/(C)=((7.896xx10^(-5)Scm^(-1)))/((241xx10^(-8)mol cm^(-3)))`
`=32.76Scm^(2)mol^(-1)`
Step II : Calculation of degree of dissociation of acetic acid
`a=(wedge^(C)m)/(wedge^(@)m)=((32.765cm^(2)mol^(-1)))/((390.55cm^(2)mol^(-1)))`
`=0.084=8.4xx10^(-2)`
Step III : Calculation of dissociation constant
`K_(c)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])=(CaxxCa)/((1-a))=(ca^(2))/(1-a)`
`=((0.0024mol L^(-1))xx(0.084)^(2))/(1-0.084)`
`=0.0000185mol L^(-1)=1.85xx10^(-5)mol L^(-1)`