The exact concentration of `H^(+)` ion in `10^(-3)` molar HCl aq solution at 298 K is

Options

A. `10^(-3)+10^(-7)`
B. `10^(-3)+(K_(w))/([H^(+)])`
C. `10^(-3)+(K_(w))/([OH^(-)])`
D. `10^(-3)`

Answer

To find the exact concentration of \( H^+ \) ions in a \( 10^{-3} \) molar HCl aqueous solution at 298 K, we can follow these steps:

### Step 1: Understand the dissociation of HCl
HCl is a strong acid and dissociates completely in water:
\[ \text{HCl} \rightarrow H^+ + Cl^- \]
This means that in a \( 10^{-3} \) M HCl solution, the initial concentration of \( H^+ \) ions is \( 10^{-3} \) M.

### Step 2: Consider the contribution from water
Water also dissociates to produce \( H^+ \) and \( OH^- \) ions:
\[ \text{H}_2\text{O} \rightleftharpoons H^+ + OH^- \]
At 298 K, the ion product of water (\( K_w \)) is:
\[ K_w = [H^+][OH^-] = 1 \times 10^{-14} \]

### Step 3: Calculate the concentration of \( OH^- \)
Since we are dealing with a solution where \( H^+ \) is contributed by both HCl and the dissociation of water, we can express the total concentration of \( H^+ \) as:
\[ [H^+] = [H^+]_{\text{HCl}} + [H^+]_{\text{water}} \]
Let \( [H^+]_{\text{HCl}} = 10^{-3} \) M.

Now, we can express the concentration of \( OH^- \) using the ion product of water:
\[ [OH^-] = \frac{K_w}{[H^+]} \]

### Step 4: Set up the equation
Substituting for \( [OH^-] \):
\[ [H^+] = 10^{-3} + \frac{K_w}{[OH^-]} \]
Substituting \( [OH^-] \):
\[ [H^+] = 10^{-3} + \frac{1 \times 10^{-14}}{[H^+]} \]

### Step 5: Solve for \( [H^+] \)
Let \( x = [H^+] \):
\[ x = 10^{-3} + \frac{1 \times 10^{-14}}{x} \]
Multiplying through by \( x \) to eliminate the fraction:
\[ x^2 = 10^{-3}x + 1 \times 10^{-14} \]
Rearranging gives:
\[ x^2 - 10^{-3}x - 1 \times 10^{-14} = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -10^{-3}, c = -1 \times 10^{-14} \):
\[ x = \frac{10^{-3} \pm \sqrt{(10^{-3})^2 - 4 \cdot 1 \cdot (-1 \times 10^{-14})}}{2 \cdot 1} \]
Calculating the discriminant:
\[ (10^{-3})^2 + 4 \times 10^{-14} = 10^{-6} + 4 \times 10^{-14} = 4.001 \times 10^{-14} \]
Now substituting back:
\[ x = \frac{10^{-3} \pm \sqrt{4.001 \times 10^{-14}}}{2} \]
Calculating the square root:
\[ \sqrt{4.001 \times 10^{-14}} \approx 2 \times 10^{-7} \]
Thus:
\[ x = \frac{10^{-3} \pm 2 \times 10^{-7}}{2} \]
Taking the positive root:
\[ x \approx \frac{10^{-3} + 2 \times 10^{-7}}{2} \approx 5.01 \times 10^{-4} \]

### Final Answer
The exact concentration of \( H^+ \) ions in a \( 10^{-3} \) molar HCl solution at 298 K is approximately \( 1.01 \times 10^{-3} \) M.

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