conductivity of 0.00241 M acetic acid solution if 7.896 x 10^-5. calculate the molar conductivity of this solution . if lamda zero m for acetic acid be 390.5 S.Cm^2mol^-1. what will be its dissociation constant?
Given :
k= 7.896 *10-5 S cm-1
M=0.00241 mol L-1
Molar Conductivity λm =( k*1000 )/M
=(7.896*10-5 *1000 )/0.00241 =32.76 Scm2 mol-1
Degree of dissociation (α)=λm/λ0m =32.76/390.5 =0.084
Dissociation constant (K) =Cα/1-α
K= 0.00241*0.084/(1-.084) =1.86 *10-5 mol L-1
k= 7.896 *10-5 S cm-1
M=0.00241 mol L-1
Molar Conductivity λm =( k*1000 )/M
=(7.896*10-5 *1000 )/0.00241 =32.76 Scm2 mol-1
Degree of dissociation (α)=λm/λ0m =32.76/390.5 =0.084
Dissociation constant (K) =Cα/1-α
K= 0.00241*0.084/(1-.084) =1.86 *10-5 mol L-1