What is the pH of a buffer solution that is made by adding 21.0 grams of sodium fluoride to 1.00 Liter of 0.15M hydrofluoric acid? (see the Table above for the Ka of hydrofluoric acid; calculate the pH to 2 decimal places) hint: you will have to find the molarity of NaF in this solution.

Acid Dissociation Constant:
HF 6.6 x 10-4

Use the Henderson-Hasselbalch equation.

How? Im seeing the equation but what and where do i plug in?

pH = pKa + log (base)/(acid)

pH = ?
pKa you have (or can calculate it from Ka).
base = NaF and molarity NaF = mols/L. You have grams, convert to mols.
acid you have in M now.
Plug and chug.

thanks

To find the pH of the buffer solution, we first need to determine the concentration of sodium fluoride (NaF) in the solution. Here are the steps to calculate it:

1. Calculate the number of moles of hydrofluoric acid (HF) using the given concentration and volume:
moles of HF = concentration of HF * volume of solution
moles of HF = 0.15 M * 1.00 L

2. Convert the mass of sodium fluoride (NaF) to moles using its molar mass. The molar mass of NaF is 22.99 g/mol (sodium) + 18.99 g/mol (fluorine):
moles of NaF = mass of NaF / molar mass of NaF
moles of NaF = 21.0 g / (22.99 g/mol + 18.99 g/mol)

3. Calculate the molarity of NaF in the solution using the moles of NaF and the volume of the solution:
molarity of NaF = moles of NaF / volume of solution

Now that we have the concentration of NaF, we can calculate the pH of the buffer solution using the Henderson-Hasselbalch equation:

pH = pKa + log([A-] / [HA])

In this case:
- pKa is the negative logarithm (base 10) of the acid dissociation constant (6.6 x 10^-4), which is given in the question.
- [A-] is the concentration of the conjugate base (NaF in this case).
- [HA] is the concentration of the acid (HF in this case).

Plug in the values and calculate the pH of the buffer solution.

Note: The [A-] is the concentration of NaF, and the [HA] is the concentration of HF. To find the concentration of HF in the buffer solution, subtract the concentration of NaF from the original concentration of HF.