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

💡 Solution & Explanation

Since the concentration of HCl is very low ($< 10^{-6} \text{ M}$), the $H^+$ contribution from the auto-ionization of water cannot be neglected. Let $x$ be the $[H^+]$ from water. Then $[H^+]_{total} = x + 10^{-8}$. Solving the equilibrium $x(x + 10^{-8}) = 10^{-14}$ yields $x \approx 0.95 \times 10^{-7}$. Total $[H^+] = 1.05 \times 10^{-7} \text{ M}$, yielding $pH = -\log(1.05 \times 10^{-7}) \approx 6.98$.