`0.1` mole of `CH_(3)NH_(2) (K_(b)=5xx10^(-4))` is mixed with `0.08` mole of `HCl` and diluted to one liter. The `[H^(+)]` in solution is

Options

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

Explanation

To solve the problem of finding the concentration of \([H^+]\) in the solution after mixing \(0.1\) mole of \(CH_3NH_2\) (methylamine) with \(0.08\) mole of \(HCl\) and diluting to one liter, we can follow these steps:

### Step 1: Identify the Reaction
When \(CH_3NH_2\) (a weak base) reacts with \(HCl\) (a strong acid), it forms \(CH_3NH_3^+\) (the conjugate acid) and \(Cl^-\). The reaction can be written as:
\[ CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^- \]

### Step 2: Determine Initial Moles
Before the reaction occurs, we have:
- Moles of \(CH_3NH_2 = 0.1\) moles
- Moles of \(HCl = 0.08\) moles

### Step 3: Determine Moles After Reaction
Since \(HCl\) is the limiting reagent, it will react completely with \(0.08\) moles of \(CH_3NH_2\):
- Moles of \(CH_3NH_2\) remaining = \(0.1 - 0.08 = 0.02\) moles
- Moles of \(CH_3NH_3^+\) formed = \(0.08\) moles
- Moles of \(Cl^-\) formed = \(0.08\) moles

### Step 4: Calculate Concentrations
After the reaction, the total volume of the solution is \(1\) liter. Therefore, the concentrations are:
- \([CH_3NH_2] = \frac{0.02 \text{ moles}}{1 \text{ L}} = 0.02 \text{ M}\)
- \([CH_3NH_3^+] = \frac{0.08 \text{ moles}}{1 \text{ L}} = 0.08 \text{ M}\)

### Step 5: Calculate \(pK_b\) and \(pK_a\)
Given \(K_b = 5 \times 10^{-4}\):
\[ pK_b = -\log(5 \times 10^{-4}) \approx 3.30 \]
Using the relationship \(pK_a + pK_b = 14\):
\[ pK_a = 14 - pK_b = 14 - 3.30 = 10.70 \]

### Step 6: Use the Henderson-Hasselbalch Equation
To find \(pH\), we can use the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[base]}{[acid]}\right) \]
Here, the base is \(CH_3NH_2\) and the acid is \(CH_3NH_3^+\):
\[ pH = 10.70 + \log\left(\frac{0.02}{0.08}\right) \]
Calculating the logarithm:
\[ \log\left(\frac{0.02}{0.08}\right) = \log(0.25) \approx -0.60 \]
Thus,
\[ pH = 10.70 - 0.60 = 10.10 \]

### Step 7: Calculate \([H^+]\)
Now, we can find \([H^+]\) from \(pH\):
\[ [H^+] = 10^{-pH} = 10^{-10.10} \approx 7.94 \times 10^{-11} \text{ M} \]

### Final Answer
The concentration of \([H^+]\) in the solution is approximately \(7.94 \times 10^{-11} \text{ M}\).

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