One ecologically important equilibrium is that between carbonate and hydrogen carbonate ions in natural water. This standard Gibb's energies of formation of `CO_(3)^(-2)(aq)` and `HCO_(3)^(Theta)(aq)` are `-527.8 kJ mol^(-1)` and `-586.8 kJ mol^(-1)` respectively.
For water,
`2H_(2)O(l) +2e^(Theta) rarr H_(2)(g) +2OH^(Theta)(aq) , E_(RP)^(Theta) =- 0.83 V`
`2H_(2)O rarr O_(2) +4H^(+) +4e^(Theta), E_(OX)^(Theta) = - 1.23V`
The value of `pK_(a)` for `HCO_(3)^(Theta)` (aq) is (approx):

Options

A. `9.3`
B. `10.3`
C. `8.3`
D. `11.3`

Answer

To find the value of \( pK_a \) for \( HCO_3^- \) (aq), we can follow these steps:

### Step 1: Write the equilibrium reaction
The equilibrium between bicarbonate and carbonate ions can be represented as:
\[ HCO_3^- (aq) \rightleftharpoons H^+ (aq) + CO_3^{2-} (aq) \]

### Step 2: Calculate the change in Gibbs free energy (\( \Delta G \))
Using the standard Gibbs energies of formation:
- For \( CO_3^{2-} (aq) \): \( \Delta G_f = -527.8 \, \text{kJ/mol} \)
- For \( HCO_3^- (aq) \): \( \Delta G_f = -586.8 \, \text{kJ/mol} \)

The change in Gibbs free energy for the reaction can be calculated as:
\[ \Delta G = \Delta G_f (\text{products}) - \Delta G_f (\text{reactants}) \]
Substituting the values:
\[ \Delta G = [-586.8 \, \text{kJ/mol}] - [-527.8 \, \text{kJ/mol}] = -586.8 + 527.8 = -59 \, \text{kJ/mol} \]

### Step 3: Convert \( \Delta G \) to Joules
\[ \Delta G = -59 \, \text{kJ/mol} = -59000 \, \text{J/mol} \]

### Step 4: Relate \( \Delta G \) to the cell potential (\( E \))
Using the formula:
\[ \Delta G = -nFE \]
Where:
- \( n \) = number of moles of electrons transferred (for this reaction, \( n = 1 \))
- \( F \) = Faraday's constant \( \approx 96500 \, \text{C/mol} \)

Rearranging gives:
\[ E = -\frac{\Delta G}{nF} \]
Substituting the values:
\[ E = -\frac{-59000 \, \text{J/mol}}{1 \times 96500 \, \text{C/mol}} = \frac{59000}{96500} \approx 0.61 \, \text{V} \]

### Step 5: Use the Nernst equation to find \( K_a \)
Using the Nernst equation:
\[ E = \frac{0.059}{n} \log K_a \]
Substituting \( E \) and \( n \):
\[ 0.61 = \frac{0.059}{1} \log K_a \]
Rearranging gives:
\[ \log K_a = \frac{0.61}{0.059} \approx 10.34 \]

### Step 6: Calculate \( pK_a \)
Since \( pK_a = -\log K_a \):
\[ pK_a = -10.34 \approx 10.3 \]

### Final Answer
The value of \( pK_a \) for \( HCO_3^- (aq) \) is approximately **10.3**.

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