#K = 1.79 × 10^"-5"#; pH = 2.9

For a weak electrolyte like acetic acid, the degree of dissociation α is given by

#color(blue)(|bar(ul(color(white)(a/a)α = Λ/Λ_∞color(white)(a/a)|)))" "#

where

#Λ color(white)(m)=# the molar conductance of the solution

#Λ_∞ = #the molar conductance at infinite dilution

In this problem,

#Λ color(white)(m)= "0.000 52 Ω"^"-1"·"m"^2"mol"^"-1"#

#Λ_∞ = "0.039 07 Ω"^"-1"·"m"^2"mol"^"-1"#

∴ #α = ("0.000 52" color(red)(cancel(color(black)(Ω^"-1"·"m"^2"mol"^"-1"))))/("0.039 07" color(red)(cancel(color(black)(Ω^"-1"·"m"^2"mol"^"-1")))) = 0.0133#

The apparent dissociation constant is given by the formula

#color(blue)(|bar(ul(color(white)(a/a)K = (α^2C) /(1 – α)color(white)(a/a)|)))" "#

∴ #K = (0.0133^2 × 0.1)/("1 – 0.0133") = 1.79 × 10^"-5"#

#["H"_3"O"^"+" ] = αC = 0.0133 × "0.1 mol/L" = "0.001 33 mol/L"#

#"pH" = "-log(0.001 33") = 2.9#