During neutralisation of an acid by a base, the end point refers for the completion of reaction. The detection of end point in acid -base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthlein) or a weak base (Mrthyl orange). At about `50%` ionisation which depends on the medium, the anion furnished by an indicator (acid) or cation furnished by indicator (basic) imparts its characteristic colour to solution at point. For example phenolphthalein, the dissociation is
`underset("Colourless")(H In hArr H^(+))+underset("Pink")(In^(-)), K_(H In)= ([H^(+)][In^(-)])/([H In])`
favoured in presence of alkali and pink colour of phenolphalein ion is noticed as soon as the medium changes to alkaline nature. The end point of acid-base neutralisation not necessarily coincides with equivalent point but it is closer and closer to equivalence point. Also at equivalence point of acid-base neutralisation pH is not necessarliy equal to 7.
The dissociation constant of an acid-base indicator which furnished coloured anion is `1xx10^(-5)`. The pH of solution at which indicator is `80%` in dissociated form is:

Options

A. `5.2310`
B. `5.6020`
C. `8.3980`
D. `8.4820`

Explanation

To solve the problem, we need to determine the pH of a solution at which an acid-base indicator is 80% dissociated. The dissociation constant (K) for the indicator is given as \(1 \times 10^{-5}\).

### Step-by-Step Solution:

1. **Understand the dissociation of the indicator**:
The dissociation of the indicator can be represented as:
\[ \text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^- \]
The equilibrium expression for this dissociation is:
\[ K = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]} \]

2. **Define the initial concentration**:
Let's assume the initial concentration of the indicator (HIn) is \(C\). When the indicator is 80% dissociated, we can express the concentrations as follows:
- \([\text{In}^-] = 0.8C\) (80% dissociated)
- \([\text{HIn}] = 0.2C\) (20% remains undissociated)

3. **Substitute into the equilibrium expression**:
Now we can substitute these values into the equilibrium expression:
\[ K = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]} = \frac{[\text{H}^+](0.8C)}{(0.2C)} \]
Simplifying this gives:
\[ K = \frac{4[\text{H}^+]}{1} \]
Therefore:
\[ K = 4[\text{H}^+] \]

4. **Substitute the value of K**:
We know \(K = 1 \times 10^{-5}\), so:
\[ 1 \times 10^{-5} = 4[\text{H}^+] \]
Solving for \([\text{H}^+]\):
\[ [\text{H}^+] = \frac{1 \times 10^{-5}}{4} = 2.5 \times 10^{-6} \]

5. **Calculate the pH**:
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the value of \([\text{H}^+]\):
\[ \text{pH} = -\log(2.5 \times 10^{-6}) \]
Using a calculator, we find:
\[ \text{pH} \approx 5.60 \]

### Final Answer:
The pH of the solution at which the indicator is 80% dissociated is approximately **5.60**.