`CH_(3)NH_(2)(0.1"mole",K_(b)=5xx10^(-4))` is added to 0.08 moles of HCI and the solution is diluted to one litre. The resulting hydrogen ion concentration is

Options

A. `1.6xx10^(-11)`
B. `8xx10^(-11)`
C. `5xx10^(-5)`
D. `8xx10^(-2)`

Explanation

To solve the problem, we need to determine the resulting hydrogen ion concentration when 0.1 moles of CH₃NH₂ (methylamine) is added to 0.08 moles of HCl and diluted to one liter.

### Step 1: Identify the Reaction
When methylamine (a weak base) reacts with hydrochloric acid (a strong acid), it forms the salt CH₃NH₃Cl (methylammonium chloride):
\[ \text{CH}_3\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{Cl}^- \]

### Step 2: Determine Initial Moles
- Moles of CH₃NH₂ = 0.1 moles
- Moles of HCl = 0.08 moles

### Step 3: Calculate Moles After Reaction
Since HCl is a strong acid, it will completely react with methylamine. The reaction will consume 0.08 moles of CH₃NH₂:
- Moles of CH₃NH₂ remaining = 0.1 - 0.08 = 0.02 moles
- Moles of CH₃NH₃Cl formed = 0.08 moles

### Step 4: Concentration of Species in 1 Litre Solution
After dilution to 1 liter:
- Concentration of CH₃NH₂ = \( \frac{0.02 \text{ moles}}{1 \text{ L}} = 0.02 \, \text{M} \)
- Concentration of CH₃NH₃Cl = \( \frac{0.08 \text{ moles}}{1 \text{ L}} = 0.08 \, \text{M} \)

### Step 5: Calculate pOH of the Buffer Solution
Using the Henderson-Hasselbalch equation for a basic buffer:
\[ \text{pOH} = -\log K_b + \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \]

Where:
- \( K_b = 5 \times 10^{-4} \)
- \([\text{Salt}] = 0.08 \, \text{M}\)
- \([\text{Base}] = 0.02 \, \text{M}\)

Calculating:
1. \( -\log K_b = -\log(5 \times 10^{-4}) \)
- \( \log(5) \approx 0.699 \)
- Therefore, \( -\log(5 \times 10^{-4}) = 4 - 0.699 = 3.301 \)

2. Calculate the ratio:
\[ \log \left( \frac{0.08}{0.02} \right) = \log(4) \approx 0.602 \]

3. Combine:
\[ \text{pOH} = 3.301 + 0.602 = 3.903 \]

### Step 6: Calculate pH
Using the relationship \( \text{pH} + \text{pOH} = 14 \):
\[ \text{pH} = 14 - 3.903 = 10.097 \]

### Step 7: Calculate Hydrogen Ion Concentration
Using the formula:
\[ [\text{H}^+] = 10^{-\text{pH}} \]
\[ [\text{H}^+] = 10^{-10.097} \]

Calculating:
\[ [\text{H}^+] \approx 8 \times 10^{-11} \, \text{M} \]

### Final Result
The resulting hydrogen ion concentration is:
\[ [\text{H}^+] \approx 8 \times 10^{-11} \, \text{M} \]

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