Calculate pH at which an acid indicator Hin with concentration `0.1M` changes its colour (`K_(a)` for `"Hin" = 1 xx 10^(-5)`)

Answer

To calculate the pH at which the acid indicator \( \text{H}_{\text{in}} \) changes color, we can follow these steps:

### Step 1: Understand the dissociation of the indicator
The acid indicator \( \text{H}_{\text{in}} \) dissociates in water as follows:
\[ \text{H}_{\text{in}} \rightleftharpoons \text{H}^+ + \text{In}^- \]
This means that for every mole of \( \text{H}_{\text{in}} \) that dissociates, one mole of \( \text{H}^+ \) and one mole of \( \text{In}^- \) is produced.

### Step 2: Write the expression for \( K_a \)
The acid dissociation constant \( K_a \) is given by the expression:
\[ K_a = \frac{[\text{H}^+][\text{In}^-]}{[\text{H}_{\text{in}}]} \]
Given that \( K_a = 1 \times 10^{-5} \).

### Step 3: Calculate \( pK_a \)
To find \( pK_a \), we use the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(1 \times 10^{-5}) = 5 \]

### Step 4: Use the Henderson-Hasselbalch equation
The pH at which the indicator changes color can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + \log\left(\frac{[\text{In}^-]}{[\text{H}^+]}\right) \]
Since the concentration of the indicator \( [\text{H}_{\text{in}}] \) is \( 0.1 \, M \) and at the point of color change, the concentrations of \( [\text{In}^-] \) and \( [\text{H}^+] \) will be equal, we can say:
\[ \frac{[\text{In}^-]}{[\text{H}^+]} = 1 \]
Thus, the logarithm term becomes:
\[ \log(1) = 0 \]

### Step 5: Calculate the pH
Now substituting the values into the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + 0 = 5 + 0 = 5 \]

### Final Answer
The pH at which the acid indicator \( \text{H}_{\text{in}} \) changes color is:
\[ \text{pH} = 5 \]