Equinormal solution of two weak acids, `HA(pK_(a) =3)` and `HB(pK_(a) =5)` are each placed in contact with standard hydrogen electrode at `25^(@)C`. When a cell is constructed by interconnecting them thorugh a salt bridge find the e.m.f. of the cell.

Explanation

To find the e.m.f. of the cell constructed from equinormal solutions of two weak acids, HA and HB, we can follow these steps:

### Step 1: Understand the Cell Representation
The cell consists of two half-cells:
- The anode half-cell with acid HA (pKa = 3)
- The cathode half-cell with acid HB (pKa = 5)

The overall cell can be represented as:
\[ \text{Pt} | \text{H}_2(g) | \text{HA}^+ || \text{HB}^+ | \text{H}_2(g) | \text{Pt} \]

### Step 2: Write the Half-Reactions
At the anode (oxidation):
\[ \text{H}_2(g) \rightarrow 2\text{H}^+ + 2e^- \]

At the cathode (reduction):
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g) \]

### Step 3: Write the Net Cell Reaction
The net cell reaction can be expressed as:
\[ 2\text{H}^+ \text{(from HB)} \rightarrow 2\text{H}^+ \text{(from HA)} \]

### Step 4: Determine Standard Electrode Potential
The standard electrode potential \( E^0 \) for hydrogen is defined as 0 V. Therefore, the standard electrode potential for the net cell reaction is also 0 V.

### Step 5: Apply the Nernst Equation
The Nernst equation is given by:
\[ E = E^0 - \frac{0.0591}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]

Here, \( n = 2 \) (number of electrons transferred).

### Step 6: Calculate the Concentrations of H⁺ Ions
Using the dissociation constant \( K_a \):
- For acid HA:
\[ K_a = 10^{-3} \Rightarrow [\text{H}^+]^2 = K_a \cdot C \]
\[ [\text{H}^+] = \sqrt{K_a \cdot C} = \sqrt{10^{-3} \cdot C} \]

- For acid HB:
\[ K_a = 10^{-5} \Rightarrow [\text{H}^+]^2 = K_a \cdot C \]
\[ [\text{H}^+] = \sqrt{K_a \cdot C} = \sqrt{10^{-5} \cdot C} \]

### Step 7: Substitute into the Nernst Equation
Substituting the concentrations into the Nernst equation:
\[ E = 0 - \frac{0.0591}{2} \log \left( \frac{\sqrt{10^{-3} \cdot C}}{\sqrt{10^{-5} \cdot C}} \right)^2 \]

This simplifies to:
\[ E = -\frac{0.0591}{2} \log \left( \frac{10^{-3}}{10^{-5}} \right) \]
\[ E = -\frac{0.0591}{2} \log (10^{2}) \]
\[ E = -\frac{0.0591}{2} \cdot 2 \]
\[ E = -0.0591 \text{ V} \]

### Step 8: Interpret the Result
Since the e.m.f. is negative, it indicates that the cell operates in the reverse direction. Thus, the e.m.f. of the cell constructed is:
\[ E = +0.0591 \text{ V} \]

### Final Answer
The e.m.f. of the cell is \( 0.0591 \text{ V} \).

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