What is the pH of a 1.00 L solution of the following buffer

0.500 M propanoic acid (HC3H5O2 Ka = 1.30 × 10^−5) / 0.800 M sodium propanoate

after the addition of 0.700 mol solid NaOH.

I tried using the Henderson-Hasselbach equation

pH = pKa + log[A-/HA]

I used it in a previous problem that used .260 mol NaOH and got a PH of 5.54 but I'm doing something wrong because I can't figure out the pH for this question no matter what I try

To determine the pH of the solution after the addition of NaOH, you need to consider the equilibrium reaction between propanoic acid (HC3H5O2) and its conjugate base (C3H5O2-), as well as the reaction with NaOH.

First, let's write the balanced chemical equation for the reaction between HC3H5O2 (propanoic acid) and NaOH:

HC3H5O2 + NaOH -> NaC3H5O2 + H2O

Now, let's calculate the concentration of propanoate ion ([A-]) and propanoic acid ([HA]) after the addition of NaOH.

Before the addition of NaOH:
[A-] = 0.800 M (concentration of sodium propanoate)
[HA] = 0.500 M (concentration of propanoic acid)

After the addition of NaOH (0.700 mol):
Since NaOH is a strong base, it will react completely with propanoic acid to form sodium propanoate and water. This means that the concentration of propanoic acid will decrease while the concentration of propanoate will increase.

To calculate the new concentrations, start by calculating the moles of propanoic acid reacted with NaOH:
moles of HC3H5O2 = moles of NaOH

moles of NaOH = 0.700 mol

Since the reaction is 1:1 between propanoic acid and NaOH, the moles of propanoic acid reacted will also be 0.700 mol.

Now, subtract the moles of propanoic acid reacted from the initial moles of propanoic acid to get the remaining moles:
remaining moles of HC3H5O2 = initial moles - moles reacted
remaining moles of HC3H5O2 = 0.500 mol - 0.700 mol

Since the result is negative, it means that all the propanoic acid has been consumed, and only propanoate ions are present. Therefore, the new concentration of propanoate ion ([A-]) is 0.800 M + 0.700 M = 1.500 M.

Now, we can use the Henderson-Hasselbalch equation to calculate the pH:

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

Given that pKa = -log(Ka) = -log(1.30 × 10^−5), you can substitute the values:

pH = (-log(1.30 × 10^−5)) + log(1.500 M / 0.500 M)

Now, you can solve this equation to find the pH of the solution.