Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 2 / A7 / d

Let $\alpha$ be the degree of ionization. Conservation of nuclei and charge neutrality, with no pre-existing electrons or other ions, giveSubstitution into the definition of $K_c$ givesThe physical root lies between zero and one:The second form avoids subtracting nearly equal terms at large $y$. The limits are $\alpha \simeq \sqrt{y}$ when $y\ll 1$ and $\alpha \simeq 1-1/y$ when $y\gg 1$. The dependence on initial number density matters: $K_c$ alone, without $c_{\mathrm{tot}}$ and the standard-state convention, does not specify the ionized fraction.

For hydrogen, the electronic ground state has $g_{\mathrm{H}}=2$, while the proton has no electronic degeneracy, $g_{{\mathrm{H}}^{+}}=1$. For sodium, the neutral $3s{}^2{S}_{1/2}$ ground state has $g_{\mathrm{Na}}=2$ and the closed-shell ion has $g_{\mathrm{N}{\mathrm{a}}^{+}}=1$. Ignoring hyperfine structure and consistently cancelling nuclear-spin factors, $2g_{{\mathrm{M}}^{+}}/g_{\mathrm{M}}=1$ for both.

At $5000\mathrm{K}$,It is convenient to calculate the density $D=K_c{c}^{\circ }$, which avoids choosing an unnecessary numerical value of the standard density. The results areThus hydrogen is about $0.00409\%$ ionized, whereas sodium is about $53.6\%$ ionized in the stated ground-state ideal-gas model. The large difference is dominated by the smaller sodium ionization energy in the exponential. The weak-ionization approximation is excellent for hydrogen but would be inaccurate for sodium, so the quadratic solution is needed there.