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))`
Calculate the `[(Ind^(Theta))/(Hin)]`

Options

A. `1.413 xx 10^(-5)`
B. `1.413 xx 10^(-4)`
C. `3.128xx10^(-5)`
D. `3.128xx10^(-14)`

Explanation

To solve the problem, we need to calculate the ratio of the concentration of the dissociated form of the indicator \([Ind^-]\) to the concentration of the undissociated form \([HIn]\) in a solution of phosphoric acid.

### Step-by-Step Solution:

1. **Identify the dissociation of the indicator**:
The indicator \(HIn\) dissociates as follows:
\[ HIn \rightleftharpoons H^+ + Ind^- \]
The equilibrium constant for this dissociation is given by:
\[ K_{Ind} = \frac{[H^+][Ind^-]}{[HIn]} \]

2. **Write the expression for the ratio**:
Rearranging the equation gives us:
\[ \frac{[Ind^-]}{[HIn]} = \frac{K_{Ind}}{[H^+]} \]

3. **Determine the concentration of \(H^+\)**:
We need to find the concentration of \(H^+\) ions in the solution. Since we are given a 0.05 M solution of \(H_3PO_4\), we will consider its first dissociation:
\[ H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \]
The equilibrium constant \(K_1\) for this dissociation is \(10^{-3}\).

The concentration of \(H^+\) can be approximated using the formula:
\[ [H^+] \approx \sqrt{K_1 \cdot C} \]
where \(C\) is the concentration of the acid (0.05 M).

Substituting the values:
\[ [H^+] \approx \sqrt{10^{-3} \cdot 0.05} = \sqrt{5 \times 10^{-5}} = 7.07 \times 10^{-3} \text{ M} \]

4. **Substitute \(K_{Ind}\) and \([H^+]\) into the ratio**:
Now we can substitute \(K_{Ind} = 10^{-7}\) and \([H^+] = 7.07 \times 10^{-3}\) into the ratio:
\[ \frac{[Ind^-]}{[HIn]} = \frac{10^{-7}}{7.07 \times 10^{-3}} \]

5. **Calculate the ratio**:
Performing the calculation:
\[ \frac{[Ind^-]}{[HIn]} = \frac{10^{-7}}{7.07 \times 10^{-3}} \approx 1.41 \times 10^{-5} \]

### Final Answer:
The ratio \(\frac{[Ind^-]}{[HIn]}\) is approximately \(1.41 \times 10^{-5}\).