The Ka of hydrofluoric acid (HF) at 25.0 °C is 6.8 × 10-4. What is the pH of a 0.15 M aqueous solution of HF?

To calculate the pH of a 0.15 M aqueous solution of hydrofluoric acid (HF), we can use the formula for the pH of a weak acid:

\[ pH = -\log_{10}[A^-] + \log_{10}K_a \]

where \( [A^-] \) is the concentration of the conjugate base (F^- in this case) and \( K_a \) is the acid dissociation constant.

Given:
- \( [HF] = 0.15 \) M (concentration of HF)
- \( K_a \) of HF at 25.0°C = \( 6.8 \times 10^{-4} \)

The concentration of F^- ions will be equal to the concentration of HF because HF is a strong enough acid to fully ionize in water:

\[ [F^-] = [HF] = 0.15 \text{ M} \]

Now, plugging these values into the pH equation:

\[ pH = -\log_{10}(0.15) + \log_{10}(6.8 \times 10^{-4}) \]

Let's calculate it.

pH = -log(0.15) + log(6.8e-4)

Using a calculator or Python's `math` module, we find the pH.

import math

pH = -math.log10(0.15) + math.log10(6.8e-4)
pH

After calculation, you'll get the pH value for the given solution. Note that the actual value needs to be computed numerically.