An acid-base indicator which is a weak acid has a `pK_(In)` value =5.45. At what concentration ratio of sodium acetate to acctic acid would the indicator show a colour half-way between those of its acid and conjugate base forms ?
`[pK_(a)` of acetic acid =4.75, log 2=0.3]

Options

A. `4:1`
B. `6:1`
C. `5:1`
D. `3:1`

Answer

To solve the problem, we need to find the concentration ratio of sodium acetate (the conjugate base) to acetic acid (the weak acid) at which the indicator shows a color halfway between its acid and conjugate base forms.

### Step-by-Step Solution:

1. **Understand the Concept of pH and pKa**:
The pH of a solution can be expressed using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
Here, the base is sodium acetate and the acid is acetic acid.

2. **Set the Condition for Halfway Color Change**:
When the indicator shows a color halfway between its acid and conjugate base forms, the pH of the solution will equal the pKa of the indicator. Thus, we set:
\[ \text{pH} = \text{pK}_{\text{In}} = 5.45 \]

3. **Substitute Known Values**:
We know the pKa of acetic acid is given as 4.75. We can substitute this into the Henderson-Hasselbalch equation:
\[ 5.45 = 4.75 + \log\left(\frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]}\right) \]

4. **Rearranging the Equation**:
Rearranging the equation gives:
\[ 5.45 - 4.75 = \log\left(\frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]}\right) \]
\[ 0.7 = \log\left(\frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]}\right) \]

5. **Convert Logarithmic Form to Exponential Form**:
To find the concentration ratio, we take the antilog of both sides:
\[ \frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]} = 10^{0.7} \]

6. **Calculate the Antilog**:
Using the approximation \(10^{0.7} \approx 5\) (since \(10^{0.3} \approx 2\)):
\[ \frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]} \approx 5 \]

7. **Final Concentration Ratio**:
Therefore, the concentration ratio of sodium acetate to acetic acid is:
\[ \text{Ratio} = 5:1 \]

### Conclusion:
The concentration ratio of sodium acetate to acetic acid at which the indicator shows a color halfway between those of its acid and conjugate base forms is **5:1**.