Past exam of the biological physics course of the University of Oxford / 2015 / 3 / b / Solution

Full dissociation gives bulk concentrations in $\mathrm{mM}$ of $155$ for sodium, $5$ for potassium, $2$ for calcium, $114$ for chloride, and $25$ for hydrogen phosphate. The positive and negative charge concentrations are both $164\mathrm{mM}$, so the bulk is electrically neutral. At $T=298.15\mathrm{K}$,The Boltzmann distribution then givesFor completeness the coion concentrations are $114e^{-\nu }=6.65196\mathrm{mM}$ and $25e^{-2\nu }=0.0851195\mathrm{mM}$. They are depleted, not zero. Using $1\mathrm{mM}=1\mathrm{mol}{\mathrm{m}}^{-3}$ and multiplying molar concentrations by the Avogadro constant in the Grahame equation givesThis calculation retains every species. Since $|{\varphi }_0|/V_T=2.84$, a linear Debye–Hückel approximation is not justified. The large accumulation, especially of divalent counterions, follows from the valence in the Boltzmann factor. However, multi-molar local concentrations also warn that point-ion mean-field theory neglects finite ion size, hydration, specific adsorption, correlations, and changes in local permittivity. These are predictions of the stated idealized model, not precise measured interfacial concentrations.