The pH of a 10^{-8} M HCl aqueous solution at 298 K is closest to:

For an extremely dilute acid solution like $10^{-8}$ M HCl, the contribution of $H^+$ ions from the autoionization of water cannot be neglected. Using the charge neutrality condition $[H^+] = [Cl^-] + [OH^-]$, let $x = [H^+]$. Then $x = 10^{-8} + \frac{10^{-14}}{x}$. Solving this quadratic equation gives $x \approx 1.051 \times 10^{-7}$ M, so $\text{pH} = -\log_{10}(1.051 \times 10^{-7}) \approx 6.98$.