Voltage of the cell `Pt, H_(2) (1atm)|HOCN(1.3 xx 10^(-3)M)||Ag^(+) (0.8M) |Ag(s)` is `0.982V`. Calculate the `K_(a)` for `HOCN`, neglect `[H^(+)]` because of oxidation of `H_(2)(g)`
`Ag^(+) +e^(-) rarr Ag(s) = 0.8V`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Electrode Potential for Silver Electrode
The standard electrode potential for the silver electrode reaction is given as:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
with \( E^\circ = 0.80 \, \text{V} \).

Using the Nernst equation:
\[ E = E^\circ - \frac{0.0591}{n} \log \frac{[ \text{products} ]}{[ \text{reactants} ]} \]
For the silver electrode:
- \( n = 1 \) (one electron is transferred)
- Concentration of \( \text{Ag}^+ = 0.8 \, \text{M} \)

Substituting the values:
\[ E_{\text{Ag}} = 0.80 - \frac{0.0591}{1} \log \frac{1}{0.8} \]
Calculating the logarithm:
\[ \log \frac{1}{0.8} = -\log(0.8) \approx 0.09691 \]
Now substituting back:
\[ E_{\text{Ag}} = 0.80 - 0.0591 \times 0.09691 \approx 0.80 - 0.00573 \approx 0.794 \, \text{V} \]

### Step 2: Calculate the Anode Potential for Hydrogen Electrode
The overall cell voltage is given as:
\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]
Where:
- \( E_{\text{cell}} = 0.982 \, \text{V} \)
- \( E_{\text{cathode}} = E_{\text{Ag}} = 0.794 \, \text{V} \)

Rearranging gives:
\[ E_{\text{anode}} = E_{\text{cathode}} - E_{\text{cell}} = 0.794 - 0.982 = -0.188 \, \text{V} \]

### Step 3: Calculate the Concentration of \( [H^+] \)
Using the Nernst equation for the hydrogen electrode:
\[ E_{\text{H}} = E^\circ_{\text{H}} - \frac{0.0591}{n} \log \frac{[H^+]^2}{P_{H_2}^{0.5}} \]
Where:
- \( E^\circ_{\text{H}} = 0 \, \text{V} \)
- \( P_{H_2} = 1 \, \text{atm} \)

Substituting the values:
\[ -0.188 = 0 - \frac{0.0591}{1} \log \frac{[H^+]^2}{1^{0.5}} \]
This simplifies to:
\[ -0.188 = -0.0591 \log [H^+]^2 \]
Rearranging gives:
\[ \log [H^+]^2 = \frac{0.188}{0.0591} \approx 3.18 \]
Taking the antilogarithm:
\[ [H^+]^2 = 10^{3.18} \implies [H^+] \approx 6.6 \times 10^{-4} \, \text{M} \]

### Step 4: Calculate the Ionization Constant \( K_a \) for HOCN
Using the relationship:
\[ K_a = \frac{[H^+][CNO^-]}{[HOCN]} \]
Assuming \( C \) is the initial concentration of HOCN which is \( 1.3 \times 10^{-3} \, \text{M} \) and \( \alpha \) is the degree of ionization, we have:
\[ [H^+] = C \alpha \quad \text{and} \quad [CNO^-] = C \alpha \]
Thus:
\[ K_a = \frac{(C \alpha)(C \alpha)}{C(1 - \alpha)} = \frac{C \alpha^2}{1 - \alpha} \]
Substituting \( C = 1.3 \times 10^{-3} \, \text{M} \) and \( [H^+] = 6.6 \times 10^{-4} \):
\[ \alpha = \frac{[H^+]}{C} = \frac{6.6 \times 10^{-4}}{1.3 \times 10^{-3}} \approx 0.5 \]
Now substituting back into the \( K_a \) expression:
\[ K_a = \frac{(1.3 \times 10^{-3})(0.5)(0.5)}{1 - 0.5} = \frac{1.3 \times 10^{-3} \times 0.25}{0.5} = 6.5 \times 10^{-4} \]

### Final Answer
Thus, the \( K_a \) for HOCN is:
\[ K_a \approx 6.5 \times 10^{-4} \]

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