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"#?

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