The standard enthalpy of neutralization of strong acid and strong base is `-57.3 kJ "equiv"^(-1)`. If the enthalpy of neutralization of the first proton of aqueous `H_(2)S` is `-33.7 kJ mol^(-1)` then the `(pK_a)_(1)` of `H_(2)S` is

Options

A. `((23.6 xx 10^(3) - T Delta s^@)/(2.303 RT))`
B. `((23.6 xx 10^(3) - T Delta s^@)/(2.303 RT))`
C. `((TDelta S^@ - 23.6)/(RT))`
D. `2.303((T DeltaS^(@) - 23.6)/(RT))`

Answer

To find the \( pK_a \) of the first proton of \( H_2S \), we will follow these steps:

### Step 1: Understand the Given Information
We know the standard enthalpy of neutralization for a strong acid and strong base is \( -57.3 \, \text{kJ} \, \text{equiv}^{-1} \). The enthalpy of neutralization for the first proton of \( H_2S \) is given as \( -33.7 \, \text{kJ} \, \text{mol}^{-1} \).

### Step 2: Convert Units
Since the enthalpy of neutralization for the strong acid and base is given in \( \text{kJ} \, \text{equiv}^{-1} \), we can convert it to \( \text{kJ} \, \text{mol}^{-1} \) for consistency. Since 1 equivalent of a strong acid or base corresponds to 1 mole of \( H^+ \) or \( OH^- \), we can directly use the value:
\[ \Delta H_{\text{neutralization}} = -57.3 \, \text{kJ} \, \text{mol}^{-1} \]

### Step 3: Calculate the Change in Enthalpy
Now, we can find the change in enthalpy (\( \Delta H \)) for the first proton of \( H_2S \):
\[ \Delta H = \Delta H_{\text{neutralization}} - \Delta H_{\text{H}_2S} \]
Substituting the values:
\[ \Delta H = (-33.7) - (-57.3) = -33.7 + 57.3 = 23.6 \, \text{kJ} \, \text{mol}^{-1} \]

### Step 4: Convert to Joules
To convert \( \Delta H \) to Joules:
\[ \Delta H = 23.6 \, \text{kJ} \, \text{mol}^{-1} = 23.6 \times 10^3 \, \text{J} \, \text{mol}^{-1} \]

### Step 5: Use the Gibbs Free Energy Equation
We will use the Gibbs free energy equation:
\[ \Delta G^0 = \Delta H^0 - T \Delta S^0 \]
And we know:
\[ \Delta G^0 = -2.303RT \log K_a \]
Setting these equal gives:
\[ 23.6 \times 10^3 = -2.303RT \log K_a \]

### Step 6: Solve for \( K_a \)
Rearranging gives:
\[ \log K_a = -\frac{23.6 \times 10^3}{2.303RT} \]
Where \( R = 8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1} \) and assuming \( T = 298 \, \text{K} \):
\[ \log K_a = -\frac{23.6 \times 10^3}{2.303 \times 8.314 \times 298} \]

### Step 7: Calculate \( pK_a \)
Finally, since \( pK_a = -\log K_a \):
\[ pK_a = -\left(-\frac{23.6 \times 10^3}{2.303 \times 8.314 \times 298}\right) \]

### Final Calculation
Perform the calculation to find \( pK_a \).