The acid from of an acid base indicator is yellow in acid and red in basic from. What is the change in `pH` in order to change the indicator form `80%` yellow to `80%` red.

Explanation

To solve the problem of determining the change in pH required to change the indicator from 80% yellow to 80% red, we can follow these steps:

### Step 1: Understand the Indicator System
The indicator in question exists in two forms:
- Acid form (HIn) which is yellow in acidic conditions.
- Base form (In-) which is red in basic conditions.

### Step 2: Define the Initial and Final Conditions
- **Initial Condition (Case 1)**: 80% yellow (acid form) and 20% red (base form).
- **Final Condition (Case 2)**: 80% red (base form) and 20% yellow (acid form).

### Step 3: Set Up the Equations for Each Case
Using the Henderson-Hasselbalch equation:
1. For Case 1 (80% yellow, 20% red):
\[ \text{pH}_1 = \text{pKa} + \log\left(\frac{[\text{In}^-]}{[\text{HIn}]}\right) = \text{pKa} + \log\left(\frac{20}{80}\right) = \text{pKa} + \log\left(\frac{1}{4}\right) \]
This simplifies to:
\[ \text{pH}_1 = \text{pKa} - \log(4) \]

2. For Case 2 (20% yellow, 80% red):
\[ \text{pH}_2 = \text{pKa} + \log\left(\frac{[\text{In}^-]}{[\text{HIn}]}\right) = \text{pKa} + \log\left(\frac{80}{20}\right) = \text{pKa} + \log(4) \]

### Step 4: Calculate the Change in pH
To find the change in pH, we calculate:
\[ \Delta \text{pH} = \text{pH}_2 - \text{pH}_1 \]
Substituting the expressions we derived:
\[ \Delta \text{pH} = \left(\text{pKa} + \log(4)\right) - \left(\text{pKa} - \log(4)\right) \]
This simplifies to:
\[ \Delta \text{pH} = \log(4) + \log(4) = 2 \log(4) \]
Using the property of logarithms, \(2 \log(4) = \log(16)\).

### Step 5: Numerical Calculation
Since \(\log(4) \approx 0.602\):
\[ \Delta \text{pH} \approx 2 \times 0.602 = 1.204 \approx 1.2 \]
Thus, the change in pH required to change the indicator from 80% yellow to 80% red is approximately **1.2**.

### Summary
The change in pH required to change the indicator from 80% yellow to 80% red is **1.2**.