Let the colour of the indicator (HIn colourless) will be visible only when its ionised form (pink) is 25% or more in a solution. Suppose Hln `(pK_(In) = 8.0)` is added to a solution of pH = 8.3. Predict what will happen. (Take log 2 = 0.3)

Options

A. Pink colour will be visible
B. Pink colour will not be visible
C. % of ionised form will be less than 25%
D. % of ionised form will be more than 25%

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the dissociation of the indicator
The indicator HIn dissociates in solution as follows:
\[ \text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^- \]
Here, HIn is the non-ionized form (colorless), and In⁻ is the ionized form (pink).

### Step 2: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation relates the pH of the solution to the pKa of the indicator and the ratio of the concentrations of the ionized and non-ionized forms:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{In}^-]}{[\text{HIn}]}\right) \]

### Step 3: Substitute the known values
Given:
- pK_in = 8.0
- pH = 8.3

Substituting these values into the equation:
\[ 8.3 = 8.0 + \log\left(\frac{[\text{In}^-]}{[\text{HIn}]}\right) \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ \log\left(\frac{[\text{In}^-]}{[\text{HIn}]}\right) = 8.3 - 8.0 = 0.3 \]

### Step 5: Convert the logarithmic equation to exponential form
From the logarithmic form, we can convert it to the exponential form:
\[ \frac{[\text{In}^-]}{[\text{HIn}]} = 10^{0.3} \]
Given that \( \log 2 \approx 0.3 \), we can say:
\[ \frac{[\text{In}^-]}{[\text{HIn}]} = 2 \]

### Step 6: Express the concentrations in terms of HIn
Let the concentration of HIn be \( [\text{HIn}] = x \). Then:
\[ [\text{In}^-] = 2x \]

### Step 7: Calculate the total concentration
The total concentration of the indicator is:
\[ [\text{Total}] = [\text{HIn}] + [\text{In}^-] = x + 2x = 3x \]

### Step 8: Calculate the percentage of the ionized form
The percentage of the ionized form is given by:
\[ \text{Percentage of In}^- = \left(\frac{[\text{In}^-]}{[\text{Total}]}\right) \times 100 \]
Substituting the values:
\[ \text{Percentage of In}^- = \left(\frac{2x}{3x}\right) \times 100 = \frac{2}{3} \times 100 \approx 66.67\% \]

### Step 9: Conclusion
Since the percentage of the ionized form (66.67%) is greater than 25%, the color of the indicator will be visible, and the solution will appear pink.

### Final Answer
The color will be visible (pink) because the percentage of the ionized form is more than 25%.

---