`0.1mol` of `RNH_(2)(K_(b) = 5 xx 10^(-4))` is mixed with `0.08mol` of `HC1` and diluted to `1L`. Calculate the `[H^(o+)]` in the solution.

Options

A. `8 xx 10^(-11)M`
B. `1.6 xx 10^(-11)M`
C. `8 xx 10^(-5)M`
D. `8 xx 10^(-2)M`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the reaction
When `RNH2` (a weak base) reacts with `HCl` (a strong acid), it forms `RNH3Cl`. The reaction can be written as:
\[ RNH_2 + HCl \rightarrow RNH_3^+ + Cl^- \]

### Step 2: Determine the initial moles
- Moles of `RNH2` = 0.1 mol
- Moles of `HCl` = 0.08 mol

### Step 3: Determine the limiting reagent
Since `HCl` is a strong acid and will completely react with the weak base `RNH2`, we can see that `HCl` is the limiting reagent.

### Step 4: Calculate the moles after the reaction
- Moles of `RNH2` remaining after reaction:
\[ 0.1 \, \text{mol} - 0.08 \, \text{mol} = 0.02 \, \text{mol} \]
- Moles of `RNH3^+` formed:
\[ 0.08 \, \text{mol} \]

### Step 5: Calculate the concentrations in 1 L solution
- Concentration of `RNH2`:
\[ [RNH_2] = \frac{0.02 \, \text{mol}}{1 \, \text{L}} = 0.02 \, \text{M} \]
- Concentration of `RNH3^+`:
\[ [RNH_3^+] = \frac{0.08 \, \text{mol}}{1 \, \text{L}} = 0.08 \, \text{M} \]

### Step 6: Use the buffer equation
For a basic buffer, we can use the following equation:
\[ pOH = pK_b + \log \left( \frac{[RNH_3^+]}{[RNH_2]} \right) \]

### Step 7: Calculate \( pK_b \)
Given \( K_b = 5 \times 10^{-4} \):
\[ pK_b = -\log(5 \times 10^{-4}) \approx 3.3 \]

### Step 8: Substitute values into the equation
Now substituting the values into the buffer equation:
\[ pOH = 3.3 + \log \left( \frac{0.08}{0.02} \right) \]
\[ pOH = 3.3 + \log(4) \]
Since \( \log(4) \approx 0.6 \):
\[ pOH = 3.3 + 0.6 = 3.9 \]

### Step 9: Calculate pH
Now, we can find the pH:
\[ pH = 14 - pOH = 14 - 3.9 = 10.1 \]

### Step 10: Calculate the concentration of \( [H^+] \)
Using the relationship:
\[ [H^+] = 10^{-pH} = 10^{-10.1} \]
Calculating this gives:
\[ [H^+] \approx 7.94 \times 10^{-11} \, \text{M} \]

### Final Answer
The concentration of \( [H^+] \) in the solution is approximately \( 8 \times 10^{-11} \, \text{M} \).

---