Calculate the exact pH of a aqueous solution at . — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For extremely dilute acids, water's auto-ionization cannot be ignored. Total $[H^+] = x + 10^{-8}$ and $[OH^-] = x$. Solving $x(x + 10^{-8}) = 10^{-14}$ yields $[H^+] \approx 1.05 \times 10^{-7} \text{ M}$, giving $pH \approx 6.98$.