Dissociation constants of two acids HA and HB are respectively $4 \times {10}^{-10}$ and $4 \times {10}^{-5}$, whose pH value will be higher for a given molarity

$\mathrm{HA} \rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-}$

$\mathrm{HB} \rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{B}}^{-}$

For HA

${\mathrm{K}}_{\mathrm{a}} = {\mathrm{\alpha }}^2\mathrm{C}$

${\mathrm{\alpha }}_1^{ } = \sqrt{\frac{{\mathrm{K}}_{{\mathrm{a}}_1}}{\mathrm{C}}} = \sqrt{\frac{4 \times {10}^{-10}}{1}}$

${\alpha }_1 = 2 \times {10}^{-5}$ for C = 1 M

Similarly  $\Rightarrow \left[{\mathrm{H}}^{+}\right] = \mathrm{C\alpha } = 2 \times {10}^{-5} ; \mathrm{pH} =-\log \left[{\mathrm{H}}^{+}\right] = 4.7$

For HB:

  ${\mathrm{\alpha }}_2^{ } = \sqrt{\frac{{\mathrm{K}}_{{\mathrm{a}}_2}}{\mathrm{C}}} = \sqrt{\frac{1.8 \times {10}^{-5}}{1}}$ for  C = 1 M

${\mathrm{\alpha }}_2 = 4.2 \times {10}^{-3} \Rightarrow \left[{\mathrm{H}}^{+}\right]=C\mathrm{\alpha }$

                                               $= 4.2 \times {10}^{-3}$

$pH =-\log \left(4.2 \times {10}^{-3}\right) = 2.37$;   ${\mathrm{P}}^{\mathrm{H}} \mathrm{of} \mathrm{HA}>\mathrm{HB}$