How to recast Henderson–Hasselbalch when fitting data on a chemical shift vs. pH graph for NMR I've just been getting into organic and been reading about H-H related to $\mathrm{NMR}$. I've been seeing articles state how recasting of H-H is necessary and want to know how equations like below are derived.

$$σ_\mathrm{peak} = σ_\ce{HA} + \frac{Δσ}{1 + 10^{(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}}\tag{1}$$
$$y = y_\mathrm{min} + \frac{y_\mathrm{max} - y_\mathrm{min}}{1 + 10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})}}\tag{2}$$

Eq. 1: https://www.nature.com/articles/srep43748
&quot;In the equation, $\sigma_\mathrm{peak}$ is the measured $\mathrm{NMR}$ chemical shift of the peak of interest, $\sigma_\ce{HA}$ is the chemical shift of the protonated form and $\Delta\sigma$ is the difference between $\sigma_\ce{HA}$ and the chemical shift of the deprotonated form of $\ce{HA}$ ($\sigma_\ce{A}$).&quot;
Eq. 2: https://pubs.acs.org/doi/full/10.1021/ed3000028?mobileUi=0
&quot;$y$ is the experimentally measured parameter (chemical shift, in this case), and $(y_{max} - y)$ and $(y - y_{min})$ are proportional to the concentrations of the deprotonated and protonated forms of the acid, respectively.&quot;
The derivation of these expressions begins from the premise that the observed value of the NMR parameter (chemical shift) is an average over values for two states weighted by their mole fractions: $$\sigma = \sigma_1 \chi_1 + \sigma_2 \chi_2= \sigma_1 \chi_1 + \sigma_2 (1-\chi_1)$$ This is a reasonable assumption in NMR experiments if there are two states (protonated and deprotonated) with distinct shifts and exchange between the two states is fast compared to the chemical shift difference such that the observable signal is a single time average. This allows us to write that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma_2)} = \chi_1 $$ Similarly we can write that $$\frac{(\sigma-\sigma_1)}{(\sigma_2-\sigma_1)} = \chi_2 $$ such that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{\chi_1}{\chi_2} $$ or, in terms of concentrations, that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{c_1}{c_2} $$ Introducing this last expression into the Henderson-Hasselbalch equation $$pK_a=pH+\log_{10}\left(\frac{c_{HA}}{c_{A}}\right)$$ gives upon rearrangement
$$10^{(pK_a-pH)}=\left(\frac{c_{HA}}{c_A}\right)=\frac{(\sigma-\sigma_A)}{(\sigma_{HA}-\sigma)}$$ which can be further rearranged into $$1+10^{(pK_a-pH)}=\frac{(\sigma_{HA}-\sigma_A)}{(\sigma_{HA}-\sigma)}=\frac{\Delta \sigma}{(\sigma-\sigma_{HA})}$$ and then further into the expressions in the articles (note the sign convention for $\Delta \sigma$).