Calculate the pH at which an acid indicator with `K_(a), = 1.0 xx 10^(-4)` changes colour when the indicator concentration is `2.0 xx 10^(-5)` M

Answer

To calculate the pH at which an acid indicator changes color, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- The dissociation constant \( K_a = 1.0 \times 10^{-4} \)
- The concentration of the indicator \( [Hx] = 2.0 \times 10^{-5} \, M \)

2. **Write the dissociation equation for the indicator:**
\[ Hx \rightleftharpoons H^+ + X^- \]
Here, \( Hx \) is the undissociated form of the indicator, \( H^+ \) is the hydronium ion, and \( X^- \) is the conjugate base of the indicator.

3. **Set up the expression for the dissociation constant \( K_a \):**
\[ K_a = \frac{[H^+][X^-]}{[Hx]} \]

4. **At the point of color change, the concentrations of \( Hx \) and \( X^- \) are equal:**
\[ [X^-] = [Hx] \]
Therefore, we can denote both as \( x \).

5. **Substituting into the \( K_a \) expression:**
Since \( [H^+] = x \) and \( [Hx] = [initial] - x \approx [initial] \) (as \( x \) is small compared to the initial concentration):
\[ K_a = \frac{x \cdot x}{[initial]} = \frac{x^2}{2.0 \times 10^{-5}} \]

6. **Rearranging to solve for \( x \):**
\[ x^2 = K_a \times [initial] \]
\[ x^2 = (1.0 \times 10^{-4}) \times (2.0 \times 10^{-5}) \]
\[ x^2 = 2.0 \times 10^{-9} \]
\[ x = \sqrt{2.0 \times 10^{-9}} = 1.414 \times 10^{-5} \, M \]

7. **Calculate the pH:**
The concentration of \( H^+ \) ions is equal to \( x \):
\[ [H^+] = 1.414 \times 10^{-5} \, M \]
Now, we can find the pH:
\[ pH = -\log[H^+] = -\log(1.414 \times 10^{-5}) \]
\[ pH \approx 4.85 \]

### Final Answer:
The pH at which the acid indicator changes color is approximately **4.85**.