Question #66439

1 Answer
Apr 16, 2017

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