At what pH will a 10 −4 M solution of an indicator with K b ​ =1×10 −11 change colour?

Answer

To solve the problem of determining at what pH a 10⁻⁴ M solution of an indicator with Kb = 1 × 10⁻¹¹ will change color, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the nature of the indicator**:
Since Kb is given, we can conclude that the indicator is a basic indicator. It can be represented as \( \text{InOH} \), which dissociates into \( \text{In}^+ \) and \( \text{OH}^- \).

2. **Write the dissociation equation**:
The dissociation of the indicator can be represented as:
\[ \text{InOH} \rightleftharpoons \text{In}^+ + \text{OH}^- \]

3. **Set up the expression for Kb**:
The expression for the base dissociation constant (Kb) is given by:
\[ K_b = \frac{[\text{In}^+][\text{OH}^-]}{[\text{InOH}]} \]
Since we are interested in the point where the indicator changes color, we can assume that at this point, the concentrations of \( \text{In}^+ \) and \( \text{InOH} \) are equal.

4. **Assume concentrations**:
Let \( [\text{In}^+] = [\text{OH}^-] = x \) and the initial concentration of \( \text{InOH} \) is \( 10^{-4} \, M \). Therefore, at equilibrium:
\[ [\text{InOH}] = 10^{-4} - x \]

5. **Substituting into the Kb expression**:
Since \( x \) is small compared to \( 10^{-4} \), we can approximate:
\[ K_b = \frac{x^2}{10^{-4}} \quad \text{(since \( 10^{-4} - x \approx 10^{-4} \))} \]
Substituting \( K_b = 1 \times 10^{-11} \):
\[ 1 \times 10^{-11} = \frac{x^2}{10^{-4}} \]

6. **Solving for x**:
Rearranging gives:
\[ x^2 = 1 \times 10^{-11} \times 10^{-4} = 1 \times 10^{-15} \]
Taking the square root:
\[ x = \sqrt{1 \times 10^{-15}} = 1 \times 10^{-7.5} \, M \]

7. **Finding the concentration of H⁺**:
The concentration of \( \text{OH}^- \) is \( x = 1 \times 10^{-7.5} \, M \). We can find the concentration of \( \text{H}^+ \) using the ion product of water:
\[ [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1 \times 10^{-14}}{1 \times 10^{-7.5}} = 1 \times 10^{-6.5} \, M \]

8. **Calculating the pH**:
Finally, we can calculate the pH:
\[ \text{pH} = -\log[\text{H}^+] = -\log(1 \times 10^{-6.5}) = 6.5 \]

### Conclusion:
The indicator will change color at a pH of approximately **6.5**.