= 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 $T=298.15\,\mathrm K$,
$$
V_T=\frac{k_BT}{e}=25.6926\,\mathrm{mV},\qquad
u=-\frac{\phi_0}{V_T}=2.84129.
$$
The <Boltzmann distribution> then gives
$$
\boxed{c_{1+}(0)=160e^\nu=2742.05\,\mathrm{mM},\qquad
c_{2+}(0)=2e^{2\nu}=587.409\,\mathrm{mM}.}
$$
For 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 $1\,\mathrm{mM}=1\,\mathrm{mol\,m^{-3}}$ and multiplying molar <concentrations> by the <Avogadro constant> in the <Grahame equation> gives
$$
\boxed{\sigma=-0.103243\,\mathrm{C\,m^{-2}}.}
$$
This calculation retains every species. Since $|\phi_0|/V_T=2.84$, 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>.
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