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