0.1 mole of `CH_3NH_2` `(K_b=5xx10^(-4))` is mixed with 0.08 mole of HCl and diluted to 10 litre, what will be the `OH^-` concentration in the solution ?

Options

A. `4.25xx10^(-4) M`
B. `1.25xx10^(-4) M`
C. `7.9xx10^(-4) M`
D. `2.5xx10^(-4) M`

Answer

To find the concentration of \( OH^- \) in the solution after mixing 0.1 moles of \( CH_3NH_2 \) (methylamine) with 0.08 moles of HCl and diluting to 10 liters, we can follow these steps:

### Step 1: Identify the Reaction
When \( CH_3NH_2 \) (a weak base) is mixed with HCl (a strong acid), they react to form \( CH_3NH_3Cl \) (the salt of the weak base and strong acid):
\[ CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^- \]

### Step 2: Determine Moles After Reaction
- Initial moles of \( CH_3NH_2 \) = 0.1 moles
- Initial moles of HCl = 0.08 moles

Since HCl is a strong acid, it will completely react with \( CH_3NH_2 \):
- Moles of \( CH_3NH_2 \) remaining after reaction:
\[ 0.1 - 0.08 = 0.02 \text{ moles} \]
- Moles of \( CH_3NH_3^+ \) formed:
\[ 0.08 \text{ moles} \]

### Step 3: Calculate Concentrations in 10 L Solution
The total volume of the solution is 10 L. We can now find the concentrations:
- Concentration of \( CH_3NH_2 \):
\[ \text{Concentration of } CH_3NH_2 = \frac{0.02 \text{ moles}}{10 \text{ L}} = 0.002 \text{ M} \]
- Concentration of \( CH_3NH_3^+ \):
\[ \text{Concentration of } CH_3NH_3^+ = \frac{0.08 \text{ moles}}{10 \text{ L}} = 0.008 \text{ M} \]

### Step 4: Use the Henderson-Hasselbalch Equation
Since we have a weak base and its salt, we can use the Henderson-Hasselbalch equation for a basic buffer:
\[ pOH = pK_b + \log\left(\frac{[CH_3NH_3^+]}{[CH_3NH_2]}\right) \]

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

### Step 6: Substitute Values into the Equation
Substituting the concentrations into the equation:
\[ pOH = 3.3 + \log\left(\frac{0.008}{0.002}\right) \]
Calculating the log term:
\[ \frac{0.008}{0.002} = 4 \quad \Rightarrow \quad \log(4) \approx 0.602 \]
Thus,
\[ pOH = 3.3 + 0.602 = 3.902 \]

### Step 7: Calculate \( OH^- \) Concentration
Using the relationship between \( pOH \) and \( OH^- \) concentration:
\[ pOH = -\log[OH^-] \quad \Rightarrow \quad [OH^-] = 10^{-pOH} \]
Calculating \( [OH^-] \):
\[ [OH^-] = 10^{-3.902} \approx 1.25 \times 10^{-4} \text{ M} \]

### Final Answer
The concentration of \( OH^- \) in the solution is approximately \( 1.25 \times 10^{-4} \text{ M} \).

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