An acid type indicator, H In differs in colour from its conjugate base `(In^(-))` . The human eye is sensitive to colour differences only when the ratio `[In^(-)]//[H In]` is greater than 10 or smaller than `0.1`. What should to observe a complete colour change ? `(K_(a)= 1.0xx10^(-5))`

Explanation

To solve the problem, we need to determine the pH range in which the color change of the acid-base indicator \( HIn \) can be observed. The key points to consider are the given \( K_a \) value and the conditions for the ratio of the conjugate base \( [In^-] \) to the acid \( [HIn] \).

### Step-by-Step Solution:

1. **Identify the given values:**
- \( K_a = 1.0 \times 10^{-5} \)
- The human eye can detect color changes when the ratio \( \frac{[In^-]}{[HIn]} \) is greater than 10 or less than 0.1.

2. **Calculate \( pK_a \):**
\[ pK_a = -\log(K_a) = -\log(1.0 \times 10^{-5}) = 5 \]

3. **Set up the equations for the two conditions:**
- **Condition 1:** When \( \frac{[In^-]}{[HIn]} \geq 10 \)
\[ pH = pK_a + \log(10) = 5 + 1 = 6 \]

- **Condition 2:** When \( \frac{[In^-]}{[HIn]} \leq 0.1 \)
\[ pH = pK_a + \log(0.1) = 5 + (-1) = 4 \]

4. **Determine the pH range for color change:**
- The pH must be between 4 and 6 for the color change to be observed.

5. **Calculate the change in pH:**
\[ \text{Change in pH} = 6 - 4 = 2 \]

### Conclusion:
To observe a complete color change of the indicator \( HIn \), the pH should change from 4 to 6, which corresponds to a change of 2 units in pH.