`Hin` is an acidic indicator `(K_(Ind) =10^(-7))` which dissociates into aqueous acidic solution of `30mL` of `0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13))`
If this solution is treated with `30mL`of `NaOH` solution, then what molarity of `NaOH` is needed to reach the equivalence point with indicator?

Options

A. `0.1M`
B. `0.2M`
C. `0.3M`
D. `0.4M`

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Determine the N-factor of H3PO4
H3PO4 (phosphoric acid) can dissociate in three steps:
1. H3PO4 → H2PO4⁻ + H⁺ (N-factor = 1)
2. H2PO4⁻ → HPO4²⁻ + H⁺ (N-factor = 1)
3. HPO4²⁻ → PO4³⁻ + H⁺ (N-factor = 1)

When we consider the complete neutralization of H3PO4, it can donate up to three protons. However, for the first neutralization, it will convert to H2PO4⁻, and for the second neutralization, it will convert to HPO4²⁻. Therefore, for the first equivalence point, the N-factor is 1, and for the second equivalence point, the N-factor is 2.

### Step 2: Calculate the mEq of H3PO4
Given:
- Volume of H3PO4 solution = 30 mL = 0.030 L
- Molarity of H3PO4 = 0.05 M
- N-factor for the first neutralization = 1

The mEq (milliequivalents) of H3PO4 can be calculated as:
\[ \text{mEq of H3PO4} = \text{Molarity} \times \text{Volume (L)} \times \text{N-factor} \]
\[ \text{mEq of H3PO4} = 0.05 \, \text{mol/L} \times 0.030 \, \text{L} \times 1 = 0.0015 \, \text{mol} \]

### Step 3: Set up the equation for NaOH
Since we are treating the H3PO4 solution with NaOH, we can set up the equation:
\[ \text{mEq of NaOH} = \text{mEq of H3PO4} \]
Given that the N-factor for NaOH is 1, we can express the mEq of NaOH as:
\[ \text{mEq of NaOH} = \text{Molarity of NaOH} \times \text{Volume of NaOH (L)} \times \text{N-factor} \]
Let the molarity of NaOH be \( M \) and the volume of NaOH = 30 mL = 0.030 L. Thus:
\[ M \times 0.030 \times 1 = 0.0015 \]

### Step 4: Solve for the molarity of NaOH
Rearranging the equation gives:
\[ M = \frac{0.0015}{0.030} = 0.05 \, \text{mol/L} \]

### Step 5: Calculate the final molarity of NaOH needed to reach the equivalence point
Since we need to reach the second equivalence point (where the N-factor is 2), we need to double the amount of NaOH:
\[ \text{Molarity of NaOH} = \frac{0.0015 \times 2}{0.030} = 0.1 \, \text{mol/L} \]

Thus, the molarity of NaOH needed to reach the equivalence point with the indicator is **0.1 M**.