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 find 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 can follow these steps:

### Step 1: Calculate the moles of HNO₃ and KOH
- **Moles of HNO₃**:
\[ \text{Moles of HNO}_3 = \text{Concentration} \times \text{Volume} = 1.0 \, \text{M} \times 0.1 \, \text{L} = 0.1 \, \text{mol} \, (100 \, \text{mL}) \]

- **Moles of KOH**:
\[ \text{Moles of KOH} = 0.8 \, \text{M} \times 0.1 \, \text{L} = 0.08 \, \text{mol} \, (100 \, \text{mL}) \]

### Step 2: Determine the limiting reactant
In the neutralization reaction:
\[ \text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O} \]
- Here, 1 mole of HNO₃ reacts with 1 mole of KOH.
- We have 0.1 moles of HNO₃ and 0.08 moles of KOH. Since KOH is the limiting reactant, it will completely react with HNO₃.

### Step 3: Calculate the remaining moles of HNO₃ after the reaction
- Moles of HNO₃ remaining:
\[ \text{Remaining HNO}_3 = 0.1 \, \text{mol} - 0.08 \, \text{mol} = 0.02 \, \text{mol} \]

### Step 4: Calculate the total volume of the solution
- Total volume after mixing:
\[ \text{Total Volume} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]

### Step 5: Calculate the concentration of H⁺ ions
- Since HNO₃ is a strong acid, it fully dissociates:
\[ \text{Concentration of H}^+ = \frac{\text{Remaining moles of HNO}_3}{\text{Total Volume}} = \frac{0.02 \, \text{mol}}{0.2 \, \text{L}} = 0.1 \, \text{M} \]

### Step 6: Calculate the pH of the solution
- pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] = -\log(0.1) = 1 \]

### Final Answer:
- The hydrogen ion concentration is **0.1 M** and the pH of the solution is **1**.

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