For very dilute strong acid solutions, the H+ from the autoionization of water must be considered. Total $[H^+] = 10^{-8} (	ext{from HCl}) + 10^{-7} (	ext{from water}) = 1.1 \times 10^{-7}$ M. The pH = -log($1.1 \times 10^{-7}$) ≈ 6.98.