The ${\mathrm{P}}^{\mathrm{H}}$ of 0.1 M monobasic acid is 4.5. The ${\mathrm{K}}_{\mathrm{a}} \mathrm{and} {\mathrm{P}}^{{\mathrm{K}}_{\mathrm{a}}}$ values, respectively, are

${\mathrm{P}}^{\mathrm{H}}=-\log \left[{\mathrm{H}}^{+}\right]$

$\left[{\mathrm{H}}^{+}\right]=\mathrm{Antilog} \left(-{\mathrm{P}}^{\mathrm{H}}\right)$

$=\mathrm{Antilog} (-4.5)$

$=\mathrm{Antilog} \left(\overline{5}.5\right)=3.16\times {10}^{-5}$

$\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{A}}^{-}\right]=3.16\times {10}^{-5}$

$\left[\mathrm{HA}\right]=\mathrm{C}-\mathrm{C\alpha }=\mathrm{C}-\left[{\mathrm{H}}^{+}\right]$

         $=0.1-3.16\times {10}^{-5}=0.1$

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}=\frac{{\left(3.16\times {10}^{-5}\right)}^2}{0.1}=1.0\times {10}^{-8}$

${\mathrm{P}}^{{\mathrm{K}}_{\mathrm{a}}}=-\log {\mathrm{K}}_{\mathrm{a}}=-\log {10}^{-8}=+8$