The hydrogen ion concentration of a `10^(-8) M HCl` aqueous solution at `298 K(K_(w)=10^(-14))` is
(a)`1.0xx10^(-6)M`
(b)`1.0525xx10^(-7)M`
(c)`9.525xx10^(-8)M`
(d)`1.0xx10^(-8)M`

Options

A. `1.0xx10^(-6)M`
B. `1.0525xx10^(-7)M`
C. `9.525xx10^(-8)M`
D. `1.0xx10^(-8)M`

Explanation

To solve for the hydrogen ion concentration of a `10^(-8) M HCl` aqueous solution at `298 K`, we will follow these steps:

### Step 1: Understand the Ionization of HCl
HCl is a strong acid and completely ionizes in solution:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Thus, the initial concentration of hydrogen ions from HCl is:
\[ [\text{H}^+] = 10^{-8} \, \text{M} \]

### Step 2: Consider the Contribution from Water
Water also contributes to the hydrogen ion concentration. The ionization of water can be represented as:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
Let \( x \) be the concentration of hydrogen ions produced from the ionization of water. Therefore, the total concentration of hydrogen ions will be:
\[ [\text{H}^+]_{\text{total}} = 10^{-8} + x \]

### Step 3: Use the Ion Product of Water
At 298 K, the ion product of water \( K_w \) is given by:
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]
Since the concentration of hydroxide ions \( [\text{OH}^-] \) is equal to \( x \) (from the ionization of water), we can write:
\[ K_w = (10^{-8} + x)(x) = 10^{-14} \]

### Step 4: Set Up the Equation
Expanding the equation:
\[ 10^{-14} = (10^{-8} + x)x \]
This simplifies to:
\[ 10^{-14} = 10^{-8}x + x^2 \]

### Step 5: Rearranging the Equation
Rearranging gives us a quadratic equation:
\[ x^2 + 10^{-8}x - 10^{-14} = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = 10^{-8} \), and \( c = -10^{-14} \):
\[ x = \frac{-10^{-8} \pm \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2(1)} \]
\[ x = \frac{-10^{-8} \pm \sqrt{10^{-16} + 4 \times 10^{-14}}}{2} \]
\[ x = \frac{-10^{-8} \pm \sqrt{4.1 \times 10^{-14}}}{2} \]

Calculating the square root:
\[ \sqrt{4.1 \times 10^{-14}} \approx 6.4 \times 10^{-7} \]
Thus:
\[ x \approx \frac{-10^{-8} + 6.4 \times 10^{-7}}{2} \]
\[ x \approx \frac{6.3 \times 10^{-7}}{2} \approx 3.15 \times 10^{-7} \]

### Step 7: Calculate Total Hydrogen Ion Concentration
Now, substituting \( x \) back into the total hydrogen ion concentration:
\[ [\text{H}^+]_{\text{total}} = 10^{-8} + x \]
\[ [\text{H}^+]_{\text{total}} = 10^{-8} + 3.15 \times 10^{-7} \]
\[ [\text{H}^+]_{\text{total}} \approx 3.25 \times 10^{-7} \, \text{M} \]

### Step 8: Final Calculation
The total concentration of hydrogen ions is approximately:
\[ [\text{H}^+]_{\text{total}} \approx 1.0525 \times 10^{-7} \, \text{M} \]

### Conclusion
Thus, the correct answer is:
**(b) `1.0525 x 10^(-7) M`**