# 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"?

Mar 27, 2016

Suppose we have $\text{1 L}$ of water. Note the following molar conductivities:

• ${\overline{\lambda}}_{\text{H"^(+)) = 0.34982 ("S"cdot"L")/("mol"cdot"cm}}$
• ${\overline{\lambda}}_{\text{OH"^(-)) = 0.1986 ("S"cdot"L")/("mol"cdot"cm}}$

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

If we represent electrical conductivity as ${\lambda}_{\text{soln}}$ and molar conductivity as ${\overline{\lambda}}_{i}$, then:

$\boldsymbol{{\lambda}_{\text{soln" = 10^6 sum_"strong electrolytes}} \left[E l e c t r o l y t e\right] {\overline{\lambda}}_{E}}$

$= {10}^{6} {\sum}_{i} {\nu}_{i} \left[{X}_{i}\right] {\overline{\lambda}}_{i}$

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

where:

• The units are $\left(\mu \text{S")/("cm") = "mol"/"L" xx ("S"cdot"L")/("mol"cdot"cm}\right)$, and the ${10}^{6}$ converts $\text{S}$ into $\mu \text{S}$.
• ${\nu}_{+}$ is the stoichiometric coefficient of the cation and ${\nu}_{-}$ is that of the anion.
• $\text{[X]}$ means concentration of $\text{X}$ in $\text{mol/L}$.

To prove that both concentrations are ${10}^{- 7} \text{mol/L}$, I will work backwards to solve for those.

$\left({\lambda}_{\text{soln")/(10^6 mu"S/S") = nu_(+)["H"^(+)]*barlambda_("H"^(+)) + nu_(-)["OH"^(-)]*barlambda_("OH}}^{-}\right)$

Since we know that the components ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ add together in an equimolar amount to give $\text{H"_2"O}$ via conservation of mass and charge, we know that $\left[{\text{OH"^(-)] = ["H}}^{+}\right]$ and ${\nu}_{+} = {\nu}_{-}$. i.e.,

${\text{H"_2"O" rightleftharpoons "H"^(+) + "OH}}^{-}$,

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,

$\left[{\text{H"^(+)] = ["OH}}^{-}\right]$

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

$= \left(0.055 \cancel{\mu \text{S")"/"cancel"cm")/((10^6 cancel(mu"S"))/cancel("S") * (0.34982 (cancel("S")cdot"L")/("mol"cdotcancel"cm") + 0.1986 (cancel("S")cdot"L")/("mol"cdotcancel"cm}}\right)$

$= 1.00288 \times {10}^{- 7}$ $\text{mol/L}$

$\approx \textcolor{b l u e}{1.00 \times {10}^{- 7}}$ $\textcolor{b l u e}{\text{mol/L}}$

Therefore, the concentrations of ${\text{OH}}^{-}$ and ${\text{H}}^{+}$ are nonzero, and are in fact, $1.00 \times {10}^{- 7}$ $\text{M}$. That means water does perform an autoionization. QED