The hydrogen ion concentration and pH of the solution made by mixing 100 mL of 1.0 M`HNO_(3)` with 100 mL of 0.8 M KOH, are

Options

A. `[H^(+)] = 0.1, pH = 1`
B. `[H^(+)] = 0.01, pH = 2`
C. `[H^(+)] = 1 xx 10^(-12), pH = 12`
D. `[H^(+)] = 1 xx 10^(-7), pH = 7`

Answer

To solve the problem of finding the hydrogen ion concentration and pH of the solution made by mixing 100 mL of 1.0 M HNO₃ with 100 mL of 0.8 M KOH, we will follow these steps:

### Step-by-Step Solution:

1. **Calculate the number of millimoles of HNO₃:**
\[ \text{Millimoles of HNO₃} = \text{Molarity} \times \text{Volume (mL)} = 1.0 \, \text{M} \times 100 \, \text{mL} = 100 \, \text{mmol} \]

2. **Calculate the number of millimoles of KOH:**
\[ \text{Millimoles of KOH} = \text{Molarity} \times \text{Volume (mL)} = 0.8 \, \text{M} \times 100 \, \text{mL} = 80 \, \text{mmol} \]

3. **Determine the limiting reactant and the amount left after the reaction:**
- HNO₃ reacts with KOH in a 1:1 ratio.
- Since we have 100 mmol of HNO₃ and 80 mmol of KOH, KOH is the limiting reactant.
- After the reaction:
- HNO₃ remaining = 100 mmol - 80 mmol = 20 mmol
- KOH is completely consumed.

4. **Calculate the concentration of H⁺ ions:**
- Since HNO₃ is a strong acid, it dissociates completely:
\[ \text{HNO₃} \rightarrow \text{H}^+ + \text{NO}_3^- \]
- Therefore, the remaining 20 mmol of HNO₃ will produce 20 mmol of H⁺ ions.

5. **Calculate the total volume of the solution:**
\[ \text{Total volume} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]

6. **Calculate the concentration of H⁺ ions:**
\[ \text{Concentration of H}^+ = \frac{\text{Millimoles of H}^+}{\text{Total volume (L)}} = \frac{20 \, \text{mmol}}{200 \, \text{mL}} = \frac{20 \times 10^{-3} \, \text{mol}}{0.2 \, \text{L}} = 0.1 \, \text{M} \]

7. **Calculate the pH of the solution:**
\[ \text{pH} = -\log[\text{H}^+] = -\log(0.1) = 1 \]

### Final Results:
- **Hydrogen ion concentration**: 0.1 M
- **pH of the solution**: 1