Interpreting spectral data. The graph shows the H-nuclear magnetic resonance chemical shift of the${\mathbf{H}}_{\mathbf{2}}$ proton on pyridine as a function of $\mathbf{pH}$Chemical shift is related to the environment of a proton in a molecule. If the environment changes, the chemical shift changes. Suggest an explanation for why the chemical shift changesbetween low and high Estimate ${\mathbf{pK}}_{\mathbf{2}}$ for pyridinium ion$\left(C_5{H}_5{\mathrm{NH}}^{+}\right)$

By using the Henderson-Hasselbalch equation and/or utilizing the titration curve to establish the equivalency point (i.e., where $\left[B\right]\mathbf{}\mathbf{=}\left[\mathrm{BH}+\right]\mathbf{}\mathbf{and}\mathbf{}\mathbf{pKa}\mathbf{}\mathbf{=}\mathbf{}\mathbf{pH}$the experimental pKa of pyridine can be easily obtained.

The $p{K}_a$value of pyridinium ionlocalid="1654775985807" $\approx 5.2$

For$\approx 5.2$chemical shift at low is $8.67$and at high pH is localid="1654776794757" $8.67$At $\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}$there should be equal quantity of both species and chemical shift will belocalid="1654776260122" $1/2\left(8.87+7.89\right)=8.28\mathrm{ppm}.\mathrm{At}\mathrm{pH}=5.2$At this chemical shift interrupts the curve via data point.

Here the ${\mathrm{pK}}_{\mathrm{a}}$is around$5.2$

Chemical shift at low$\mathrm{pH}\left(8.67\mathrm{ppm}\right)$is given forlocalid="1654776395452" ${\mathrm{C}}_5{\mathrm{H}}_5{\mathrm{NH}}^{+}$

Chemical shift at high $pH\left(7.89ppm\right)$is for ${\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N}$.

When the$\mathrm{pH}=\mathrm{pKa}$

Both species are present in equal amounts, and the chemical shift would be:

$\frac{\left(8.87+7.89\right)}{2}=8.28\mathrm{ppM}$

Thus when the pH is approximately 5.2, this chemical shift would intercept the curve through the data points.