Determine the degree of hydrolysis and hydrolysis constant of aniline hydrochloride in `M//32` solution of salt at `298K` from the following cell data at `298K`.
`Pt|H_(2)(1atm)|H^(+) (1M)||M//32 C_(6)H_(5)NH_(3)CI |H_(2)(1atm)|Pt, E_(cell) = -0.188V`.

Explanation

To determine the degree of hydrolysis and hydrolysis constant of aniline hydrochloride in a \( \frac{1}{32} \) M solution at 298 K, we can follow these steps:

### Step 1: Analyze the Cell Data
The cell representation is given as:
\[ \text{Pt} | \text{H}_2(1 \text{ atm}) | \text{H}^+ (1 \text{ M}) || \frac{1}{32} \text{C}_6\text{H}_5\text{NH}_3\text{Cl} | \text{H}_2(1 \text{ atm}) | \text{Pt} \]
From this, we can identify the half-reactions occurring at the anode and cathode.

**Anode Reaction:**
\[ \text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \]

**Cathode Reaction:**
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]

### Step 2: Write the Net Cell Reaction
The overall cell reaction can be written as:
\[ \text{H}_2 + 2\text{H}^+ \rightarrow \text{H}_2 + 2\text{H}^+ \]
This shows that the cell is at equilibrium.

### Step 3: Apply the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{H}^+]^2}{P_{\text{H}_2}} \]

Given that \( E_{\text{cell}} = -0.188 \, \text{V} \) and \( E^\circ_{\text{cell}} = 0 \), and \( n = 2 \) (number of electrons transferred), we can substitute the values:
\[ -0.188 = -\frac{0.0591}{2} \log \frac{[\text{H}^+]^2}{1} \]

### Step 4: Solve for \([\text{H}^+]\)
Rearranging gives:
\[ -0.188 = -0.02955 \log [\text{H}^+]^2 \]
\[ \log [\text{H}^+]^2 = \frac{0.188}{0.02955} \]
\[ \log [\text{H}^+]^2 = 6.35 \]
Taking the antilog:
\[ [\text{H}^+]^2 = 10^{6.35} \]
Thus,
\[ [\text{H}^+] = 10^{3.175} \approx 1498.5 \, \text{M} \]

### Step 5: Calculate pH
Using the concentration of \([\text{H}^+]\):
\[ \text{pH} = -\log [\text{H}^+] \approx 3.18 \]

### Step 6: Determine \(pK_b\) of Aniline
We know that for a salt of a weak base (aniline) and a strong acid (HCl):
\[ \text{pH} = 7 - \frac{1}{2} pK_b - \frac{1}{2} \log C \]
Where \(C = \frac{1}{32} \, \text{M}\).

Substituting the known values:
\[ 3.18 = 7 - \frac{1}{2} pK_b - \frac{1}{2} \log \left(\frac{1}{32}\right) \]

### Step 7: Solve for \(pK_b\)
Calculating \(\log \left(\frac{1}{32}\right) = -\log(32) \approx -1.505\):
\[ 3.18 = 7 - \frac{1}{2} pK_b + 0.7525 \]
\[ \frac{1}{2} pK_b = 7.7525 - 3.18 \]
\[ pK_b = 2(4.5725) \approx 9.145 \]

### Step 8: Calculate \(K_b\)
Using \(pK_b\):
\[ K_b = 10^{-pK_b} = 10^{-9.145} \approx 7.15 \times 10^{-10} \]

### Step 9: Calculate Hydrolysis Constant \(K_h\)
Using the relation:
\[ K_h = \frac{K_w}{K_b} \]
Where \(K_w = 10^{-14}\):
\[ K_h = \frac{10^{-14}}{7.15 \times 10^{-10}} \approx 1.39 \times 10^{-5} \]

### Step 10: Determine Degree of Hydrolysis
The degree of hydrolysis \(h\) can be calculated as:
\[ h = \sqrt{\frac{K_h}{C}} \]
Where \(C = \frac{1}{32} \, \text{M} = 0.03125 \, \text{M}\):
\[ h = \sqrt{\frac{1.39 \times 10^{-5}}{0.03125}} \approx 2.1 \times 10^{-2} \]

### Final Results
- Hydrolysis constant \(K_h \approx 1.39 \times 10^{-5}\)
- Degree of hydrolysis \(h \approx 2.1 \times 10^{-2}\)

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