500.0 mL of 0.120 M NaOH is added to 595 mL of 0.250 M weak acid (Ka = 2.95 × 10-5). What is the pH of the resulting buffer?

The answer that i got using HH equation is 4.14 but apparently it's wrong. Am I approaching this problem from a different angle ?

I think it can be done that way. Show your work and I'll try to find the error. I think your answer is close.

.5*.12=.06 mols NaOH

.595*.25=.14875 mols Acid
(.14875)/(1.095)=.1358M Acid
(.06/1.095)=.05479M NaOH

pH=4.53+log(base/acid)=4.14

The acid and base react. Why didn't you let them do that.

...........NaOH + HA ==> NaA + H2O
I.........0.055..0.135....0.....0
C........-0.055..-0.055..+0.055
E...........0.....?........?
Then plug those numbers into except I would plug in the "real" numbers and not my estimates. I get 4.35 for the pH.

The main problem with the way you worked it is that the (HA) is decreased by the amount of NaOH added and you didn't count that.

OHH Great !!! It makes much more sense now.Thanks !

To calculate the pH of the resulting buffer, you are correct in using the Henderson-Hasselbalch equation. However, it seems like there might be a mistake in your calculation or the data you used.

To approach this problem correctly, follow these steps:

1. Calculate the number of moles of NaOH:
moles of NaOH = volume of NaOH (in liters) × molarity of NaOH
= 500.0 mL × (1 L/1000 mL) × 0.120 M
= 0.060 mol

2. Calculate the number of moles of weak acid:
moles of weak acid = volume of weak acid (in liters) × molarity of weak acid
= 595 mL × (1 L/1000 mL) × 0.250 M
= 0.14875 mol

3. Calculate the moles ratio of NaOH to weak acid:
moles ratio = moles of NaOH / moles of weak acid
= 0.060 mol / 0.14875 mol
≈ 0.403

4. Calculate the pKa (negative logarithm of the acid dissociation constant Ka) using the given Ka value:
pKa = -log10(Ka)
= -log10(2.95 × 10^-5)
≈ 4.53

5. Now, apply the Henderson-Hasselbalch equation to calculate the pH of the resulting buffer:
pH = pKa + log10(moles ratio)
= 4.53 + log10(0.403)
= 4.53 + (-0.394)
≈ 4.14

So, the pH of the resulting buffer, based on the calculations using the Henderson-Hasselbalch equation, should be approximately 4.14. If this value does not match the provided answer, please double-check your calculations and make sure you used the correct values for the given concentrations and Ka.