${\left[{\mathrm{H}}^{+}\right]}_{{\mathrm{H}}_2{\mathrm{SO}}_4}$ = 2 × 10^–10 M
${\left[{\mathrm{H}}^{+}\right]}_{{\mathrm{H}}_2\mathrm{O}}$ = 10^–7 M (slightly less due to common ion effect.)
[H⁺]_Total = 10^–7 + 2 × 10^–10 M (approximate)
pH = 6.9 (due to slightly acidic so less than 7 but closer to 7]