A 1.50 L buffer solution is 0.250 M HF and 0.250 M NaF. What is the pH of the solution after the addition of 0.0500 moles of solid NaOH? Assume no volume change. Ka for HF is 3.5x10^-4

You can anticipate that the strong base and the weak acid will neutralize one another when the sodium hydroxide solution is added, assuming that the buffer's overall volume remains unchanged.

Furthermore, more conjugate bases will be produced as a result of this reaction.

Thus, the reaction between hydrofluoric acid and sodium hydroxide can be represented by the following balanced chemical equation.

To calculate the number of moles in solution, use the buffer's volume and the hydrofluoric acid's molarity.

Similarly, deal with the conjugate base.

The quantity of hydrofluoric acid in moles will be adjusted to

The same amount will be added to the moles of fluoride anions.

To determine the new molarities of the weak acid and its conjugate base, use the buffer's volume.

Finally, determine the buffer's pH using the Henderson-Hasselbalch equation.

Consequently, the solution's pH will be

Does this outcome make sense in the end?

Observing that the weak acid and conjugate base concentrations in your starting buffer were equal, the H-H equation can be reduced to

Following the addition of the strong base, the pH of the buffer rises because the buffer transforms the strong base into a weak base, resulting in a conjugate base concentration that is greater than the weak acid concentration.

To find the pH of the buffer solution after adding NaOH, we need to consider the reaction that occurs when NaOH is added to the buffer solution:

NaOH + HF → NaF + H2O

First, calculate the initial moles of HF and NaF in the buffer solution:

Initial moles of HF = 1.50 L x 0.250 mol/L = 0.375 moles Initial moles of NaF = 1.50 L x 0.250 mol/L = 0.375 moles

Since NaOH is a strong base, it will react completely with HF to form NaF and water. Therefore, the moles of HF will decrease by 0.0500 moles, and the moles of NaF will increase by 0.0500 moles.

0.375 moles HF - 0.0500 moles = 0.325 moles HF remaining 0.375 moles NaF + 0.0500 moles = 0.425 moles NaF

Now, calculate the concentrations of HF and NaF after the reaction:

[HF] = 0.325 moles / 1.50 L = 0.2167 M [NaF] = 0.425 moles / 1.50 L = 0.2833 M

Next, calculate the concentration of H+ ions from the remaining HF:

Ka for HF = 3.5 x 10^-4 [H+] = sqrt(Ka x [HF]) = sqrt(3.5 x 10^-4 x 0.2167) = 0.00602 M

Finally, calculate the pH using the formula pH = -log[H+]:

pH = -log(0.00602) ≈ 2.22

So, the pH of the solution after the addition of 0.0500 moles of solid NaOH is approximately 2.22.