An acid-base indicator has `K_(a) = 3.0 xx 10^(-5)`. The acid form of the indicator is red and the basic form is blue. Then:

Options

A. `pH` is `4.04` when indicator is `75%` red.
B. `pH` is `5.00` when indicator is `75%` blue.
C. `pH` is `5.00` when indicator is `75%` red.
D. `pH` is `5.05` when indicator is `75%` blue.

Explanation

To solve the problem regarding the acid-base indicator with a given \( K_a = 3.0 \times 10^{-5} \), follow these steps:

### Step 1: Understand the Indicator
The indicator is a weak acid (HIn) that can dissociate in water:
\[ \text{HIn} \rightleftharpoons \text{In}^- + \text{H}^+ \]
The acid form (HIn) is red, and the basic form (In^-) is blue.

### Step 2: Calculate \( pK_a \)
The \( pK_a \) is calculated using the formula:
\[ pK_a = -\log(K_a) \]
Substituting the given value:
\[ pK_a = -\log(3.0 \times 10^{-5}) \]

### Step 3: Calculate \( pH \) at 75% Red (HIn)
When the indicator is 75% red, it means:
- 75% is in the form of HIn
- 25% is in the form of In^-

Using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[\text{In}^-]}{[\text{HIn}]}\right) \]
Substituting the concentrations:
\[ pH = pK_a + \log\left(\frac{25}{75}\right) \]
This simplifies to:
\[ pH = pK_a + \log\left(\frac{1}{3}\right) \]

### Step 4: Substitute \( pK_a \) into the Equation
From Step 2, we substitute \( pK_a \):
\[ pH = -\log(3.0 \times 10^{-5}) + \log\left(\frac{1}{3}\right) \]

### Step 5: Simplify the Logarithmic Expression
Using properties of logarithms:
\[ pH = -\log(3.0) + 5 + \log(1) - \log(3) \]
Since \( \log(1) = 0 \):
\[ pH = 5 - \log(3) \]

### Step 6: Calculate the Final Value
Using \( \log(3) \approx 0.477 \):
\[ pH \approx 5 - 0.477 \approx 4.523 \]
This means the pH at 75% red is approximately 4.52.

### Step 7: Calculate \( pH \) at 75% Blue (In^-)
When the indicator is 75% blue:
- 75% is in the form of In^-
- 25% is in the form of HIn

Using the Henderson-Hasselbalch equation again:
\[ pH = pK_a + \log\left(\frac{[\text{In}^-]}{[\text{HIn}]}\right) \]
Substituting the concentrations:
\[ pH = pK_a + \log\left(\frac{75}{25}\right) \]
This simplifies to:
\[ pH = pK_a + \log(3) \]

### Step 8: Substitute \( pK_a \) into the Equation
From Step 2, we substitute \( pK_a \):
\[ pH = -\log(3.0 \times 10^{-5}) + \log(3) \]

### Step 9: Simplify the Logarithmic Expression
Using properties of logarithms:
\[ pH = -\log(3.0) + 5 + \log(3) \]
The \( -\log(3) \) and \( +\log(3) \) cancel out:
\[ pH = 5 \]

### Final Answers
- The pH at 75% red is approximately **4.52**.
- The pH at 75% blue is **5**.