An aqueous solution contains 0.01 M `RNH_2 (K_b=2 xx 10^(-6)) & 10^(-4) M NaOH.` The concentration of `OH^(-)` is nearly:

Options

A. `2.414 xx 10^(-4)M`
B. `10^(-4)M`
C. `1.414 xx 10^(-4)M`
D. `2 xx 10^(-4)M`

Answer

To solve the problem, we need to find the concentration of hydroxide ions (OH⁻) in a solution containing both a weak base (RNH₂) and a strong base (NaOH). Let's break down the steps:

### Step 1: Determine the pKb of RNH₂
Given:
- Kb of RNH₂ = 2 × 10⁻⁶

To find pKb:
\[ \text{pKb} = -\log(K_b) = -\log(2 \times 10^{-6}) \]
Calculating this:
\[ \text{pKb} \approx 6 - \log(2) \approx 6 - 0.301 = 5.699 \]

### Step 2: Calculate the pH from RNH₂
Using the formula for pH of a weak base:
\[ \text{pH} = 14 - \frac{1}{2} \left( \text{pKb} + \log[C] \right) \]
Where C is the concentration of RNH₂ (0.01 M):
\[ \text{pH} = 14 - \frac{1}{2} \left( 5.699 + \log(0.01) \right) \]
Calculating:
\[ \log(0.01) = -2 \]
So,
\[ \text{pH} = 14 - \frac{1}{2} \left( 5.699 - 2 \right) = 14 - \frac{1}{2} \times 3.699 = 14 - 1.8495 \approx 12.1505 \]

### Step 3: Calculate the pOH
Using the relationship \( \text{pOH} = 14 - \text{pH} \):
\[ \text{pOH} = 14 - 12.1505 \approx 1.8495 \]

### Step 4: Calculate the concentration of OH⁻ from RNH₂
Using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Thus,
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-1.8495} \approx 0.0141 \, \text{M} \, \text{(or } 1.41 \times 10^{-2} \text{ M)} \]

### Step 5: Calculate the contribution of NaOH
Given:
- Concentration of NaOH = 10⁻⁴ M

Since NaOH is a strong base, it completely dissociates:
\[ [\text{OH}^-]_{\text{NaOH}} = 10^{-4} \, \text{M} \]

### Step 6: Calculate the total concentration of OH⁻
Now, we add the contributions from RNH₂ and NaOH:
\[ [\text{OH}^-]_{\text{total}} = [\text{OH}^-]_{\text{RNH2}} + [\text{OH}^-]_{\text{NaOH}} = 1.41 \times 10^{-2} + 1.0 \times 10^{-4} \]
Since \(1.41 \times 10^{-2}\) is much larger than \(1.0 \times 10^{-4}\), we can approximate:
\[ [\text{OH}^-]_{\text{total}} \approx 1.41 \times 10^{-2} \, \text{M} \]

### Final Answer
The concentration of OH⁻ in the solution is approximately:
\[ \text{Concentration of } OH^- \approx 1.41 \times 10^{-2} \, \text{M} \]