pH is the negative logarithm of `H^(+)`
` pH=log[H^(+)]`
`HCllt H^(+)+Cl^(-)`
`H_(2)OhArrH^(+)+OH^(-)`
` K_(W)=[H^(+)][OH^(-)]`
`K_(W)` depend on the temperatue. With rise in temperature `K_(W)` increases.
At 298 K, pH of pure water = 7
The exact concentration of `H^(+)` in `10^(-6)` M HCl given by

Options

A. `10^(-6)+10^(-8)`
B. `10^(-6)+10^(-7)`
C. `10^(-6)`
D. `10^(-6)-10^(-7)`

Answer

To solve the problem, we need to determine the total concentration of \( H^+ \) ions in a solution of \( 10^{-6} \) M HCl.

### Step-by-Step Solution:

1. **Understanding pH and \( H^+ \) Concentration**:
- The pH of a solution is defined as:
\[ \text{pH} = -\log[H^+] \]
- At 298 K, pure water has a pH of 7, which means:
\[ [H^+] = 10^{-7} \text{ M} \]

2. **Dissociation of HCl**:
- When \( HCl \) is dissolved in water, it dissociates completely:
\[ HCl \rightarrow H^+ + Cl^- \]
- Therefore, a \( 10^{-6} \) M solution of \( HCl \) will contribute \( 10^{-6} \) M of \( H^+ \) ions.

3. **Total Concentration of \( H^+ \)**:
- The total concentration of \( H^+ \) ions in the solution will be the sum of the \( H^+ \) ions contributed by \( HCl \) and those from the ionization of water:
\[ [H^+]_{\text{total}} = [H^+]_{\text{HCl}} + [H^+]_{\text{water}} \]
- Substituting the known values:
\[ [H^+]_{\text{total}} = 10^{-6} \text{ M} + 10^{-7} \text{ M} \]

4. **Calculating the Total Concentration**:
- To add these two concentrations, we can express \( 10^{-6} \) M as \( 1.0 \times 10^{-6} \) M and \( 10^{-7} \) M as \( 0.1 \times 10^{-6} \) M:
\[ [H^+]_{\text{total}} = 1.0 \times 10^{-6} + 0.1 \times 10^{-6} = 1.1 \times 10^{-6} \text{ M} \]

5. **Conclusion**:
- The total concentration of \( H^+ \) ions in the \( 10^{-6} \) M HCl solution is:
\[ [H^+]_{\text{total}} = 1.1 \times 10^{-6} \text{ M} \]

### Final Answer:
The exact concentration of \( H^+ \) in \( 10^{-6} \) M HCl is \( 1.1 \times 10^{-6} \) M.

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