The `K_a` of an indicator HIn is `9` x `10^(-4)`, The percentage of the basic form of indicator is 10x in a solution of pH = 4. What is x ?

Answer

To solve the problem, we need to find the value of \( x \) given that the percentage of the basic form of the indicator \( \text{IN}^- \) is \( 10x \) in a solution with a pH of 4, and the \( K_a \) of the indicator \( \text{HIn} \) is \( 9 \times 10^{-4} \).

### Step-by-Step Solution:

1. **Calculate \( pK_a \)**:
\[ pK_a = -\log(K_a) = -\log(9 \times 10^{-4}) = -(\log(9) + \log(10^{-4})) = -\log(9) + 4 \]
Using \( \log(10^{-4}) = -4 \).

2. **Use the Henderson-Hasselbalch Equation**:
The Henderson-Hasselbalch equation for the dissociation of the indicator is given by:
\[ pH = pK_a + \log\left(\frac{[\text{IN}^-]}{[\text{HIn}]}\right) \]
Substituting the known values:
\[ 4 = pK_a + \log\left(\frac{[\text{IN}^-]}{[\text{HIn}]}\right) \]

3. **Rearranging the Equation**:
Rearranging gives:
\[ \log\left(\frac{[\text{IN}^-]}{[\text{HIn}]}\right) = 4 - pK_a \]

4. **Substituting \( pK_a \)**:
Substitute \( pK_a \) from step 1:
\[ \log\left(\frac{[\text{IN}^-]}{[\text{HIn}]}\right) = 4 - (4 - \log(9)) = \log(9) \]

5. **Exponentiating Both Sides**:
Exponentiating gives:
\[ \frac{[\text{IN}^-]}{[\text{HIn}]} = 9 \]
This implies:
\[ [\text{IN}^-] = 9 \times [\text{HIn}] \]

6. **Calculating the Percentage**:
The total concentration of the indicator is:
\[ [\text{Total}] = [\text{HIn}] + [\text{IN}^-] = [\text{HIn}] + 9[\text{HIn}] = 10[\text{HIn}] \]
The percentage of the basic form \( \text{IN}^- \) is given by:
\[ \text{Percentage of } \text{IN}^- = \frac{[\text{IN}^-]}{[\text{Total}]} \times 100 = \frac{9[\text{HIn}]}{10[\text{HIn}]} \times 100 = 90\% \]

7. **Setting Up the Equation**:
We know from the problem that this percentage is equal to \( 10x \):
\[ 10x = 90 \]

8. **Solving for \( x \)**:
\[ x = \frac{90}{10} = 9 \]

### Final Answer:
\[ x = 9 \]