Calculate the e.m.f. of the cell
`Pt|H_(2)(1.0atm)|CH_(3)COOH (0.1M)||NH_(3)(aq,0.01M)|H_(2)(1.0atm)|Pt`
`K_(a) (CH_(3)COOH) =1.8 xx 10^(-5), K_(b),(NH_(3)) = 1.8 xx 10^(-5)`

Explanation

To calculate the e.m.f. of the given electrochemical cell, we will follow these steps:

### Step 1: Calculate the hydrogen ion concentration for acetic acid (CH₃COOH)

The dissociation of acetic acid can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

Using the formula for the concentration of hydrogen ions:
\[ [\text{H}^+] = \sqrt{K_a \cdot C} \]
where:
- \( K_a = 1.8 \times 10^{-5} \) (given)
- \( C = 0.1 \, \text{M} \) (concentration of acetic acid)

Calculating:
\[ [\text{H}^+] = \sqrt{1.8 \times 10^{-5} \cdot 0.1} = \sqrt{1.8 \times 10^{-6}} \]
\[ [\text{H}^+] \approx 1.34 \times 10^{-3} \, \text{M} \]

### Step 2: Calculate the hydroxide ion concentration for ammonia (NH₃)

The dissociation of ammonia can be represented as:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

Using the formula for the concentration of hydroxide ions:
\[ [\text{OH}^-] = \sqrt{K_b \cdot C} \]
where:
- \( K_b = 1.8 \times 10^{-5} \) (given)
- \( C = 0.01 \, \text{M} \) (concentration of ammonia)

Calculating:
\[ [\text{OH}^-] = \sqrt{1.8 \times 10^{-5} \cdot 0.01} = \sqrt{1.8 \times 10^{-7}} \]
\[ [\text{OH}^-] \approx 4.24 \times 10^{-4} \, \text{M} \]

### Step 3: Convert hydroxide ion concentration to hydrogen ion concentration

Using the relationship:
\[ [\text{H}^+] \cdot [\text{OH}^-] = 10^{-14} \]

We can find the concentration of hydrogen ions from the hydroxide concentration:
\[ [\text{H}^+] = \frac{10^{-14}}{[\text{OH}^-]} \]
Calculating:
\[ [\text{H}^+] = \frac{10^{-14}}{4.24 \times 10^{-4}} \]
\[ [\text{H}^+] \approx 2.36 \times 10^{-11} \, \text{M} \]

### Step 4: Calculate the e.m.f. of the cell using the Nernst equation

The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[\text{H}^+]_{acid}}{[\text{H}^+]_{base}} \right) \]

For our cell:
- \( E^\circ_{cell} = 0 \) (as both half-cells involve hydrogen)
- \( n = 1 \) (number of electrons transferred)

Substituting the values:
\[ E_{cell} = 0 - \frac{0.0591}{1} \log \left( \frac{1.34 \times 10^{-3}}{2.36 \times 10^{-11}} \right) \]

Calculating the logarithm:
\[ \log \left( \frac{1.34 \times 10^{-3}}{2.36 \times 10^{-11}} \right) = \log(5.67 \times 10^{7}) \approx 7.754 \]

Now substituting back:
\[ E_{cell} = -0.0591 \times 7.754 \]
\[ E_{cell} \approx -0.458 \, \text{V} \]

### Final Answer:
The e.m.f. of the cell is approximately \( -0.46 \, \text{V} \).

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