At `25^(@)C` the enthalpy change, for the ionization of trichloroacetic acid is `+"6.3 kJ mol"^(-1)` and the entropy change, is `+"0.0084 kJ mol"^(-1)K^(-1)`. Then pKa of trichloro acetic acid is

Options

A. 1.74
B. 2.52
C. 0.66
D. 4.72

Answer

To find the pKa of trichloroacetic acid given the enthalpy change (ΔH) and the entropy change (ΔS), we can follow these steps:

### Step 1: Understand the relationship between ΔG, ΔH, and ΔS
We know that the Gibbs free energy change (ΔG) is related to enthalpy (ΔH) and entropy (ΔS) by the equation:
\[ \Delta G = \Delta H - T \Delta S \]

### Step 2: Convert units
Given:
- ΔH = +6.3 kJ/mol
- ΔS = +0.0084 kJ/mol·K

We need to convert these values into joules:
- ΔH = 6.3 kJ/mol = 6300 J/mol
- ΔS = 0.0084 kJ/mol·K = 8.4 J/mol·K

### Step 3: Calculate the temperature in Kelvin
The temperature is given as 25°C. To convert this to Kelvin:
\[ T = 25 + 273 = 298 \text{ K} \]

### Step 4: Calculate ΔG
Now we can substitute the values into the ΔG equation:
\[ \Delta G = 6300 \text{ J/mol} - (298 \text{ K} \times 8.4 \text{ J/mol·K}) \]
Calculating the second term:
\[ 298 \times 8.4 = 2503.2 \text{ J/mol} \]
Now substituting back:
\[ \Delta G = 6300 - 2503.2 = 3796.8 \text{ J/mol} \]

### Step 5: Relate ΔG to Ka
We know that:
\[ \Delta G = -RT \ln K_a \]
Where R = 8.314 J/(mol·K). Rearranging gives:
\[ \ln K_a = -\frac{\Delta G}{RT} \]
Substituting the values:
\[ \ln K_a = -\frac{3796.8}{8.314 \times 298} \]
Calculating the denominator:
\[ 8.314 \times 298 = 2477.572 \text{ J/mol} \]
Now substituting:
\[ \ln K_a = -\frac{3796.8}{2477.572} \approx -1.53 \]
Now we can find Ka:
\[ K_a = e^{-1.53} \approx 0.216 \]

### Step 6: Calculate pKa
Finally, we can find pKa:
\[ pK_a = -\log K_a \]
Calculating:
\[ pK_a = -\log(0.216) \approx 0.66 \]

### Conclusion
Thus, the pKa of trichloroacetic acid is approximately **0.66**.

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