The hydrogen ion concentration of a `10^(-8)` M HCl aqueous solution at 298 K `(K_(w)=10^(-14))` is

Options

A. `9.525 xx 10^(-8)` M
B. `1.0 xx 10^(-8)` m
C. `1.0 xx 10^(-6)` M
D. `1.0525 xx 10^(-7) M `

Explanation

To find the hydrogen ion concentration of a `10^(-8)` M HCl aqueous solution at 298 K, we need to consider both the contribution of HCl and the autoionization of water. Here’s a step-by-step solution:

### Step 1: Understand the dissociation of HCl
HCl is a strong acid and completely dissociates in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Given the concentration of HCl is `10^(-8)` M, the concentration of hydrogen ions from HCl is also `10^(-8)` M.

### Step 2: Consider the contribution from water
Water also contributes to the hydrogen ion concentration due to its autoionization:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
At 298 K, the ion product of water \( K_w \) is given as:
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]
In pure water, the concentration of hydrogen ions and hydroxide ions is equal:
\[ [\text{H}^+] = [\text{OH}^-] = 10^{-7} \, \text{M} \]

### Step 3: Calculate the total hydrogen ion concentration
The total hydrogen ion concentration in the solution will be the sum of the hydrogen ions from HCl and from water:
\[ [\text{H}^+]_{\text{total}} = [\text{H}^+]_{\text{HCl}} + [\text{H}^+]_{\text{water}} \]
Substituting the values:
\[ [\text{H}^+]_{\text{total}} = 10^{-8} + 10^{-7} \]

### Step 4: Simplify the expression
To add these two concentrations, we can express \( 10^{-8} \) in terms of \( 10^{-7} \):
\[ [\text{H}^+]_{\text{total}} = 10^{-8} + 10^{-7} = 10^{-8}(1 + 10) = 10^{-8} \times 11 \]

### Step 5: Final calculation
Now, calculating the total concentration:
\[ [\text{H}^+]_{\text{total}} = 11 \times 10^{-8} = 1.1 \times 10^{-7} \, \text{M} \]

### Conclusion
Thus, the hydrogen ion concentration of the `10^(-8)` M HCl aqueous solution at 298 K is approximately:
\[ \boxed{1.1 \times 10^{-7} \, \text{M}} \]