Make the plot $10^{\rm{pH}}(A_{L^-}-A)$$10^{\mathrm{pH}}(A_{\ce{L-}} - A)$ versus A
the$A$, the adjusted equation will be of the form $$ 10^{\rm{pH}}(A_{L^-}-A) = 10^{\rm{pKa}}A-10^{\rm{pKa}}A_{\rm{HL}} $$$$ 10^{\mathrm{pH}}(A_{\ce{L-}} - A) = 10^{\mathrm{p}K_\mathrm{a}}A - 10^{\mathrm{p}K_\mathrm{a}}A_{\ce{HL}}, $$ the slope is $10^{\rm pKa}$$10^{\mathrm{p}K_\mathrm{a}}$, that is, $\rm pKa = log_{10}(slope)$$\mathrm{p}K_\mathrm{a} = \log_{10}(\text{slope})$.
