Equinormal solutions of two weak acids, HA (pKa = 3) and HB (pKa = 5) are each placed in contact with equal pressure of hydrogen electrode at 25 °C. When a cell is constructed by interconnecting them through a salt bridge, find the emf of the cell.

To determine the electromotive force (emf) of a cell constructed from two weak acids, HA and HB, we first need to calculate the concentrations of the hydrogen ions in each solution using their respective pKa values.

Calculating Hydrogen Ion Concentrations

The concentration of hydrogen ions \([H^+]\) can be found using the formula:

• \([H^+] = 10^{-\text{pKa}}\)

For Acid HA

Given that the pKa of HA is 3:

• \([H^+]_{HA} = 10^{-3} = 0.001 \, \text{M}\)

For Acid HB

For HB, with a pKa of 5:

• \([H^+]_{HB} = 10^{-5} = 0.00001 \, \text{M}\)

Calculating the Cell Potential

The cell potential can be calculated using the Nernst equation:

• EMF = \(E^\circ - \frac{RT}{nF} \ln Q\)

For our case, we can simplify this since we are dealing with hydrogen ions:

• EMF = \(0.0591 \, \text{V} \cdot \log \left( \frac{[H^+]_{HA}}{[H^+]_{HB}} \right)\)

Substituting Values

Now, substituting the concentrations:

• EMF = \(0.0591 \, \text{V} \cdot \log \left( \frac{0.001}{0.00001} \right)\)
• EMF = \(0.0591 \, \text{V} \cdot \log(100)\)
• EMF = \(0.0591 \, \text{V} \cdot 2\)

Final Calculation

Thus, the emf of the cell is:

• EMF = \(0.1182 \, \text{V}\)

In summary, the emf of the cell constructed from the two weak acids is approximately 0.118 V.