Question

How do you prove that the equilibrium concentrations of `"H"^(+)` and `"OH"^(-)` are `1.00 xx 10^(-7) "M"` at neutral pH using molar conductivity data at `25^@ "C"`?

Answer 1

Suppose we have `"1 L"` of water. Note the following molar conductivities:

• `barlambda_("H"^(+)) = 0.34982 ("S"cdot"L")/("mol"cdot"cm")`
• `barlambda_("OH"^(-)) = 0.1986 ("S"cdot"L")/("mol"cdot"cm")`

Electrical conductivity is often measured in `mu"S/cm"`, and the electrical conductivity of chemically pure water is `0.055 mu"S/cm"`. Since it must be that ions are in solution to give a nonzero electrical conductivity, let us show that it is due to `"OH"^(-)` and `"H"^(+)`.

If we represent electrical conductivity as `lambda_"soln"` and molar conductivity as `barlambda_i`, then:

`bb(lambda_"soln" = 10^6 sum_"strong electrolytes"[El ectrolyte]barlambda_E)`

`= 10^6 sum_(i) nu_i [X_i]barlambda_i`

`= color(green)(10^6(stackrel("H"^(+))overbrace(nu_(+)["H"^(+)]barlambda_("H"^(+)) )+ stackrel("OH"^(-))overbrace(nu_(-)["OH"^(-)]barlambda_("OH"^(-)))))`

where:

• The units are `(mu"S")/("cm") = "mol"/"L" xx ("S"cdot"L")/("mol"cdot"cm")`, and the `10^6` converts `"S"` into `mu"S"`.
• `nu_(+)` is the stoichiometric coefficient of the cation and `nu_(-)` is that of the anion.
• `"[X]"` means concentration of `"X"` in `"mol/L"`.

To prove that both concentrations are `10^(-7) "mol/L"`, I will work backwards to solve for those.

`(lambda_"soln")/(10^6 mu"S/S") = nu_(+)["H"^(+)]*barlambda_("H"^(+)) + nu_(-)["OH"^(-)]*barlambda_("OH"^(-))`

Since we know that the components `"H"^(+)` and `"OH"^(-)` add together in an equimolar amount to give `"H"_2"O"` via conservation of mass and charge, we know that `["OH"^(-)] = ["H"^(+)]` and `nu_(+) = nu_(-)`. i.e.,

`"H"_2"O" rightleftharpoons "H"^(+) + "OH"^(-)`,

and so, we can simplify this to become:

`(lambda_"soln")/(10^6 mu"S/S") = ["H"^(+)]cdot(barlambda_("H"^(+)) + barlambda_("OH"^(-)))`

Solving for one of them,

`["H"^(+)] = ["OH"^(-)]`

`= (lambda_"soln")/((10^6 mu"S")/"S" * (barlambda_("H"^(+)) + barlambda_("OH"^(-)))`

`= (0.055 cancel(mu"S")"/"cancel"cm")/((10^6 cancel(mu"S"))/cancel("S") * (0.34982 (cancel("S")cdot"L")/("mol"cdotcancel"cm") + 0.1986 (cancel("S")cdot"L")/("mol"cdotcancel"cm"))`

`= 1.00288xx10^(-7)` `"mol/L"`

`~~ color(blue)(1.00xx10^(-7))` `color(blue)("mol/L")`

Therefore, the concentrations of `"OH"^(-)` and `"H"^(+)` are nonzero, and are in fact, `1.00xx10^(-7)` `"M"`. That means water does perform an autoionization. QED