Is 50.0 mL of 1.0 M HF and 25.0 mL of 1.0 M NaOH a buffer with an approximate pH of 3?

I say yes. Right?

It depends upon what you used for Ka for HF.Using 7.2 x 10^-4 for Ka, I obtain 3.14 for pH using the Henderson-Hasselbalch equation.

To determine if a solution can act as a buffer, we need to compare the pH of the solution with the pKa of the acid present in the solution. In this case, we have a mixture of hydrofluoric acid (HF) and sodium hydroxide (NaOH).

First, let's calculate the concentration of HF and NaOH in the final solution after mixing.

For HF:
Initial volume = 50.0 mL
Initial concentration = 1.0 M

Using the formula:
(concentration1)(volume1) = (concentration2)(volume2)

Final concentration of HF = (1.0 M)(50.0 mL) / (50.0 mL + 25.0 mL) = 0.67 M

Now, let's calculate the concentration of NaOH:
Initial volume = 25.0 mL
Initial concentration = 1.0 M

Final concentration of NaOH = (1.0 M)(25.0 mL) / (50.0 mL + 25.0 mL) = 0.33 M

Since HF is a weak acid, it partially dissociates in water, resulting in the formation of H+ ions. This means that the HF solution will act as a weak acid solution. NaOH, on the other hand, is a strong base that completely dissociates into OH- ions.

To determine if the solution is a buffer, we need to compare the pH with the acid dissociation constant (pKa) of HF. The pKa value for HF is approximately 3.2.

Since the pH of the solution (approximately 3) is close to the pKa of HF (approximately 3.2), we can conclude that the solution is a buffer.

Therefore, your answer is correct. The mixture of 50.0 mL of 1.0 M HF and 25.0 mL of 1.0 M NaOH is indeed a buffer with an approximate pH of 3.