On addition of 100mL 0.01 M of NaOH solultion to 100mL 0.01M triethyl amine solution .The concentration of Triethyl ammonium ion `[C_(6)NH_(16)]^(+)` in resulting solution will be :

Options

A. `100K_(b)`
B. `200K_(b)`
C. `10K_(b)`
D. `K_(b)`

Explanation

To find the concentration of triethylammonium ion \([C_6H_{16}N]^+\) in the resulting solution after mixing 100 mL of 0.01 M NaOH with 100 mL of 0.01 M triethylamine, we can follow these steps:

### Step 1: Determine the moles of NaOH and triethylamine
- **Moles of NaOH**:
\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.01 \, \text{M} \times 0.1 \, \text{L} = 0.001 \, \text{mol} \]

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