`0.1` mole of `CH_(3)CH (K_(b) = 5 xx 10^(-4))` is mixed with `0.08` mole of HCl and diluted to one litre. What will be the `H^(+)` concentration in the solution

Options

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

Explanation

To solve the problem, we need to find the concentration of \( H^+ \) ions in a solution where \( 0.1 \) moles of \( CH_3NH_2 \) (a weak base) is mixed with \( 0.08 \) moles of \( HCl \) (a strong acid) and diluted to one liter.

### Step-by-Step Solution:

1. **Identify the Reaction:**
When \( CH_3NH_2 \) reacts with \( HCl \), it forms the salt \( CH_3NH_3^+ \) and \( Cl^- \):
\[ CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^- \]

2. **Calculate Initial Moles:**
- Moles of \( CH_3NH_2 \) = \( 0.1 \) moles
- Moles of \( HCl \) = \( 0.08 \) moles

3. **Determine Remaining Moles After Reaction:**
Since \( HCl \) is a strong acid, it will completely react with the weak base \( CH_3NH_2 \):
- Moles of \( CH_3NH_2 \) after reaction = \( 0.1 - 0.08 = 0.02 \) moles
- Moles of \( CH_3NH_3^+ \) formed = \( 0.08 \) moles

4. **Calculate Concentrations:**
Since the total volume of the solution is \( 1 \) liter:
- Concentration of \( CH_3NH_2 \) (base) = \( \frac{0.02 \text{ moles}}{1 \text{ L}} = 0.02 \, \text{M} \)
- Concentration of \( CH_3NH_3^+ \) (salt) = \( \frac{0.08 \text{ moles}}{1 \text{ L}} = 0.08 \, \text{M} \)

5. **Use Henderson-Hasselbalch Equation:**
The Henderson-Hasselbalch equation for a basic buffer is:
\[ pOH = pK_b + \log\left(\frac{[Salt]}{[Base]}\right) \]
Here, \( K_b = 5 \times 10^{-4} \), so:
\[ pK_b = -\log(5 \times 10^{-4}) \approx 3.301 \]
Substituting the values:
\[ pOH = 3.301 + \log\left(\frac{0.08}{0.02}\right) = 3.301 + \log(4) \approx 3.301 + 0.602 = 3.903 \]

6. **Calculate \( pH \) from \( pOH \):**
Since \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 3.903 = 10.097 \]

7. **Calculate \( H^+ \) Concentration:**
The concentration of \( H^+ \) ions can be calculated using:
\[ [H^+] = 10^{-pH} = 10^{-10.097} \approx 8.0 \times 10^{-11} \, \text{M} \]

### Final Answer:
The concentration of \( H^+ \) in the solution is approximately \( 8.0 \times 10^{-11} \, \text{M} \).