Dilute HCl pH Calculation

To determine the pH of a $10^{-8} \text{ M HCl}$ solution, we must consider both the dissociation of the strong acid HCl and the autoionization of water ($H_2O \rightleftharpoons H^+ + OH^-$). The autoionization constant of water is $K_w = 10^{-14}$ (since $pK_w = 14$).

In pure water, $[H^+] = [OH^-] = 10^{-7} \text{ M}$. Since the concentration of HCl ($10^{-8} \text{ M}$) is lower than that of $H^+$ from water, water's contribution is significant.

Hydrogen Ion Concentration Analysis

Let the total hydrogen ion concentration be $[H^+] = x$. The contribution from HCl is $10^{-8} \text{ M}$. The autoionization of water produces equal amounts of $H^+$ and $OH^-$. Let this amount be $s$. Then, the total concentrations are:

• $[H^+]_{total} = [H^+]_{HCl} + [H^+]_{water} = 10^{-8} + s = x$
• $[OH^-]_{total} = [OH^-]_{water} = s

Using the $K_w$ expression, $[H^+]_{total} \times [OH^-]_{total} = K_w$: $x \times s = 10^{-14}$ From the first equation, $s = x - 10^{-8}$. Substitute this into the $K_w$ expression:

$x(x - 10^{-8}) = 10^{-14}$ $x^2 - 10^{-8}x - 10^{-14} = 0$

Solving for $[H^+]$

This is a quadratic equation for $x = [H^+]$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-10^{-8}$, $c=-10^{-14}$:

$x = \frac{-(-10^{-8}) \pm \sqrt{(-10^{-8})^2 - 4(1)(-10^{-14})}}{2(1)}$ $x = \frac{10^{-8} \pm \sqrt{10^{-16} + 4 \times 10^{-14}}}{2}$ $x = \frac{10^{-8} \pm \sqrt{10^{-16} + 400 \times 10^{-16}}}{2}$ $x = \frac{10^{-8} \pm \sqrt{401 \times 10^{-16}}}{2}$ $x = \frac{10^{-8} \pm \sqrt{401} \times 10^{-8}}{2}$

We take the positive root since concentration cannot be negative:

$x = \frac{10^{-8}(1 + \sqrt{401})}{2}$ Using $\sqrt{401} \approx 20.025$: $x \approx \frac{10^{-8}(1 + 20.025)}{2}$ $x \approx \frac{21.025 \times 10^{-8}}{2}$ $x \approx 10.5125 \times 10^{-8} \text{ M}$ $x \approx 1.05125 \times 10^{-7} \text{ M}$

pH Determination

Calculate the pH using the formula $pH = -\log_{10}[H^+]$:

$pH = -\log_{10}(1.05125 \times 10^{-7})$ $pH = 7 - \log_{10}(1.05125)$ Using $\log_{10}(1.05125) \approx 0.0217$: $pH \approx 7 - 0.0217$ $pH \approx 6.9783$

Final pH Value

Rounding the pH to two decimal places gives $6.98$.