What is the pH of a \( 1.0 \times 10^{-8}\text{ M} \) aqueous \( \text{HCl} \) solution at 298 K? (Take \( \log 1.05 = 0.02 \))

Accounting for \( [\text{OH}^-] = x \), \( [\text{H}_3\text{O}^+] = 10^{-8} + x \), solving \( x(10^{-8}+x) = 10^{-14} \) gives \( x = 9.5 \times 10^{-8} \), so \( \text{pOH} = 7.02 \) and \( \text{pH} = 14 - 7.02 = 6.98 \).