For very dilute acidic solutions like $10^{-8}\text{ M}$ HCl, the contribution of $H^+$ from the auto-ionization of water cannot be neglected. Let the $[H^+]$ contributed by the dissociation of water be $x$. Since water produces $H^+$ and $OH^-$ in equal amounts, $[OH^-] = x$. The total concentration of hydrogen ions is the sum of those from HCl and water: Total $[H^+] = [H^+]_{\text{HCl}} + [H^+]_{\text{water}} = 10^{-8} + x$ We know the ionic product of water is $K_w = [H^+][OH^-] = 10^{-14}$ at 298 K. Substituting the values: $(10^{-8} + x)(x) = 10^{-14}$ $x^2 + 10^{-8}x - 10^{-14} = 0$ Solving this quadratic equation for $x$: $x = \frac{-10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2}$ $x = \frac{-10^{-8} + \sqrt{10^{-16} + 400 \times 10^{-16}}}{2}$ $x = \frac{-10^{-8} + \sqrt{401 \times 10^{-16}}}{2}$ $x = \frac{-10^{-8} + 20.025 \times 10^{-8}}{2} = 9.5125 \times 10^{-8}\text{ M}$ Now, calculating the total $[H^+]$: Total $[H^+] = 10^{-8} + 9.5125 \times 10^{-8} = 10.5125 \times 10^{-8}\text{ M} \approx 1.0525 \times 10^{-7}\text{ M}$.