An acid-base indicator has a `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 `4.05` when indicator is `75%` red.
B. `pH` is `5.00` when indicator is `75%` blue.
C. Both (a) and (b) are correct.
D. None of these

Explanation

To solve the problem, we need to determine the pH of an acid-base indicator at two different concentrations: when 75% of the indicator is in its acidic form (red) and when 75% is in its basic form (blue). The dissociation constant \( K_a \) is given as \( 3.0 \times 10^{-5} \).

### Step 1: Calculate \( pK_a \)
The first step is to find the \( pK_a \) from the given \( K_a \).

\[ pK_a = -\log(K_a) = -\log(3.0 \times 10^{-5}) \]

Using the logarithmic properties:

\[ pK_a = -\log(3.0) - \log(10^{-5}) = -\log(3.0) + 5 \]

Calculating \( \log(3.0) \) (approximately 0.477):

\[ pK_a \approx -0.477 + 5 = 4.523 \]

### Step 2: Calculate pH when 75% is in acidic form (red)
When 75% of the indicator is in the acidic form (HIn), we have:

- \( [HIn] = 0.75 \) (75% of total)
- \( [In^-] = 0.25 \) (25% of total)

Using the Henderson-Hasselbalch equation:

\[ pH = pK_a + \log\left(\frac{[In^-]}{[HIn]}\right) \]

Substituting the values:

\[ pH = 4.523 + \log\left(\frac{0.25}{0.75}\right) \]

Calculating the ratio:

\[ \frac{0.25}{0.75} = \frac{1}{3} \]

Now, substituting this into the equation:

\[ pH = 4.523 + \log\left(\frac{1}{3}\right) \]

Using the property \( \log\left(\frac{1}{3}\right) = -\log(3) \):

\[ pH = 4.523 - \log(3) \approx 4.523 - 0.477 = 4.046 \]

### Step 3: Calculate pH when 75% is in basic form (blue)
When 75% of the indicator is in the basic form (In^-), we have:

- \( [In^-] = 0.75 \) (75% of total)
- \( [HIn] = 0.25 \) (25% of total)

Using the same equation:

\[ pH = pK_a + \log\left(\frac{[In^-]}{[HIn]}\right) \]

Substituting the values:

\[ pH = 4.523 + \log\left(\frac{0.75}{0.25}\right) \]

Calculating the ratio:

\[ \frac{0.75}{0.25} = 3 \]

Now, substituting this into the equation:

\[ pH = 4.523 + \log(3) \]

Using the value of \( \log(3) \):

\[ pH = 4.523 + 0.477 = 5.000 \]

### Final Results
- The pH when 75% of the indicator is in its acidic form (red) is approximately **4.05**.
- The pH when 75% of the indicator is in its basic form (blue) is **5.00**.

### Summary
- pH at 75% red (acidic form): **4.05**
- pH at 75% blue (basic form): **5.00**