At $90^{\circ} \mathrm{C}$, the hydronium ion concentration in pure water is $10^{-6} \mathrm{M}$. If $100 \mathrm{ml}$ of $0.5 \mathrm{M}-\mathrm{NaOH}$ solution is mixed with $250 \mathrm{ml}$ of $0.2 \mathrm{M}-\mathrm{HNO}_{3}$ solution at $90^{\circ} \mathrm{C}$,
$\mathrm{pH}$ of the resulting solution will be
(a) $7.0$
(b) $6.0$
(c) $8.0$
(d) $0.85$