A certain acid-base indicator is red in acid solution and blue in basic medium. At pH=5 75% of the indictor is present in the solution in its blue form. Calculate the pH at which indicator shows 90% red form ?

Explanation

To solve the problem, we need to determine the pH at which the indicator shows 90% in its red form. We will use the Henderson-Hasselbalch equation, which relates pH, pKa, and the ratio of the concentrations of the acid and its conjugate base.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- At pH = 5, 75% of the indicator is in its blue form (basic form).
- Therefore, 25% of the indicator is in its red form (acidic form).

2. **Calculate the Ratio of Concentrations at pH = 5:**
- Let [I^-] be the concentration of the blue form (75%) and [HIn] be the concentration of the red form (25%).
- The ratio of the concentrations can be expressed as:
\[ \frac{[I^-]}{[HIn]} = \frac{75}{25} = 3 \]

3. **Use the Henderson-Hasselbalch Equation:**
- The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[I^-]}{[HIn]} \right) \]
- Substituting the known values:
\[ 5 = \text{pKa} + \log(3) \]

4. **Calculate pKa:**
- Rearranging the equation to solve for pKa:
\[ \text{pKa} = 5 - \log(3) \]
- Using the value of \(\log(3) \approx 0.477\):
\[ \text{pKa} \approx 5 - 0.477 = 4.523 \]

5. **Determine the pH for 90% Red Form:**
- If the indicator is 90% red form, then 10% is in the blue form.
- The ratio of concentrations now becomes:
\[ \frac{[I^-]}{[HIn]} = \frac{10}{90} = \frac{1}{9} \]

6. **Apply the Henderson-Hasselbalch Equation Again:**
- Substitute the new ratio into the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[I^-]}{[HIn]} \right) \]
- This gives:
\[ \text{pH} = 4.523 + \log \left( \frac{1}{9} \right) \]
- Since \(\log \left( \frac{1}{9} \right) = -\log(9)\):
\[ \text{pH} = 4.523 - \log(9) \]
- Using \(\log(9) = 2 \log(3) \approx 2 \times 0.477 = 0.954\):
\[ \text{pH} = 4.523 - 0.954 = 3.569 \]

7. **Final Answer:**
- The pH at which the indicator shows 90% red form is approximately:
\[ \text{pH} \approx 3.56 \]