A hypothetical weak acid HA, was combined with NaOH in the following proportions: 0.20 mol HA, 0.08 mol NaOH. The mixture was then diluted to a total volume of 1L, and the pH measured.

(a) If pH=4.80, what is the pKa of the acid

(b) how many additional moles of NaOH should be added to the solution to increase the pH to 5.00

Duplicate post. answered the original.

Sorry bout that. I didn't notice part b.
Assuming not all of the HA is neutralized and that you still have a buffered solution,
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)].
OR just use the simple Ka expression you used for part a, plug in (H^+) equivalent to pH = 5.00.

To determine the pKa of the weak acid, we can use the Henderson-Hasselbalch equation:

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

In this case, the base is NaOH and the acid is HA. We are given the pH as 4.80.

(a) To find the pKa of the acid:
Rearrange the equation to solve for pKa:
pKa = pH - log [(base) / (acid)]

Since we are given the pH as 4.80, we can substitute the values:
pKa = 4.80 - log [(0.08 mol) / (0.20 mol)]

Calculate the logarithm and simplify:
pKa = 4.80 - log (0.40)
pKa = 4.80 - (-0.3979)
pKa = 5.1979

Therefore, the pKa of the acid HA is approximately 5.20.

(b) To determine how many additional moles of NaOH should be added to increase the pH to 5.00:
We want to find the difference in base concentration needed to change the pH from 4.80 to 5.00.

Using the Henderson-Hasselbalch equation again:
pH = pKa + log [(base) / (acid)]

In this case, we can rearrange the equation to solve for the base concentration:
(base) = (acid) * 10^(pH - pKa)

Substituting the given values:
(base) = (0.20 mol) * 10^(5.00 - 5.20)

Calculate the exponential part and simplify:
(base) = (0.20 mol) * 10^(-0.20)
(base) = (0.20 mol) * 0.631

(base) = 0.1262 mol

Since we already have 0.08 mol of NaOH in the mixture, we can subtract it from the additional amount needed:
Additional moles of NaOH = 0.1262 mol - 0.08 mol
Additional moles of NaOH = 0.0462 mol

Therefore, approximately 0.0462 moles of NaOH should be added to the solution to increase the pH to 5.00.