Slope ≠ 1 in Henderson-Hasselbalch graph for determination of pKa
OP's question: From the papers I've read, most if not all of them use the intercept with the $\mathrm{pH}$ axis instead of the $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$ axis in this situation, why is this?

Answer for this question is easy. According to the graphs plotted using the equation $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = \mathrm{pH} - \mathrm{p}K_\mathrm{a}$ (Henderson-Hasslebalch equation), when $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = 0$ (meaning $[\ce{In-}] = [\ce{HIn}]$), then $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $. Therefore, the $x$-value where the straight-line crosses the $x$-axis is the best value for $\mathrm{p}K_\mathrm{a} $.