In an acidic indicator HIn has an ionization constant is `10^(-8).` The acid form of indicator is yellow and alkaline form is red. Which is correct statement?
(Given : log2= 0.3, log3 = 0.48)

Options

A. The pH range of indicator is 7 to 9
B. Change in pH is 0.96 when 75% yellow colour change to 75% red colour
C. This indicator is suitable for the titration of strong acid vs. strong base
D. pH of indicator is 8.3 when ration of acid from to alkaline from is 2.

Answer

To solve the problem regarding the acidic indicator HIn with an ionization constant of \( K_a = 10^{-8} \), we will analyze the statements provided and determine which are correct based on the information given.

### Step 1: Calculate pKa
The ionization constant \( K_a \) is given as \( 10^{-8} \). We can calculate \( pK_a \) using the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(10^{-8}) = 8 \]

### Step 2: Analyze the Indicator Color Forms
The indicator HIn has two forms:
- Acidic form (HIn) is yellow.
- Alkaline form (In^-) is red.

### Step 3: Determine pH for 75% Yellow (Acidic Form)
In the case where 75% of the indicator is in the acidic form (yellow):
- Concentration of HIn = 75%
- Concentration of In^- = 25%

Using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[In^-]}{[HIn]}\right) \]
Substituting the values:
\[ pH_1 = 8 + \log\left(\frac{25}{75}\right) = 8 + \log\left(\frac{1}{3}\right) \]
Using the logarithmic property:
\[ pH_1 = 8 - \log(3) \]
Given \( \log(3) \approx 0.48 \):
\[ pH_1 = 8 - 0.48 = 7.52 \]

### Step 4: Determine pH for 75% Red (Alkaline Form)
In the case where 75% of the indicator is in the alkaline form (red):
- Concentration of HIn = 25%
- Concentration of In^- = 75%

Using the Henderson-Hasselbalch equation again:
\[ pH_2 = pK_a + \log\left(\frac{[In^-]}{[HIn]}\right) \]
Substituting the values:
\[ pH_2 = 8 + \log\left(\frac{75}{25}\right) = 8 + \log(3) \]
Using the logarithmic property:
\[ pH_2 = 8 + 0.48 = 8.48 \]

### Step 5: Calculate Change in pH
Now, we can find the change in pH:
\[ \Delta pH = pH_2 - pH_1 = 8.48 - 7.52 = 0.96 \]

### Step 6: Determine pH Range
From the calculations:
- \( pH_1 = 7.52 \)
- \( pH_2 = 8.48 \)

Thus, the pH range of the indicator is from 7.52 to 8.48, which is indeed between 7 and 9.

### Step 7: Evaluate Statements
1. **pH range of indicator is 7 to 9**: **Correct** (as calculated).
2. **Indicator is suitable for titration of strong acid vs. strong base**: **Correct** (since the pH range covers neutral pH).
3. **pH of indicator is 8.3 when the ratio of acid to alkaline form is 2**: **Incorrect** (as shown in the calculations, it results in 7.7).

### Conclusion
The correct statements are:
- The pH range of the indicator is 7 to 9.
- The indicator is suitable for titration of strong acid vs. strong base.