A $50.0 mL$ sample of a $1.00 M$ solution of a diprotic acid ${\mathrm{H}}_2\mathrm{A}\left(K_{a_1}=1.0\times {10}^{-6}\text{ and }{K}_{a_2}=1.0\times {10}^{-10}\right.)$ is titrated with $2.00 \mathrm{M} \mathrm{NaOH}$. What is the minimum volume of $2.00 M NaOH$needed to reach a $pH$ of $10.00$? 

${\mathrm{H}}_2\mathrm{A} + {\mathrm{OH}}^{-} ⟶ {\mathrm{HA}}^{-} + {\mathrm{H}}_2\mathrm{O}$

50                 2V                  -

-                   2V-50             50

${\mathrm{HA}}^{-} +{ \mathrm{OH}}^{-} ⟶ {\mathrm{A}}^{2-}+ {\mathrm{H}}_2\mathrm{O}$

50                2V -50

100 - 2 V     -                    2 V - 50

$\mathrm{pH}=p{K}_2+\log \frac{\left[{\mathrm{A}}^{2-}\right]}{\left[{\mathrm{HA}}^{-}\right]}=10$

$100-2V=2V-50$

$150=4\mathrm{V}$

$V=37.5 \mathrm{mL}$

Hence, option B is correct.