[spartanR]

Hey guys,

I've been working on this exam practice question but can't seem to grasp the correct answer or really I'm just going around not knowing lol

"A chemist wishes to prepare 250mL of a buffer that is pH = 4.30. Beginning with 100mL of 0.12mol L-1 acetic acid and a supply of 0.10mol L-1 NaOH.
a) explain how this could be done.
b) How much 0.20mol L-1 NaOH must be added to this buffer to raise the pH to 5?
c) If the same amount of 0.20mol-1 NaOH were added to 250mL of dionized water, what would the new pH be?"

ok so far,
given: final volume of 250mL and final pH = 4.30
_        initial volume of 100mL of 0.10mol L-1 acetic acid
_        supply of 0.10mol L-1 NaOH.

so presuming we have to express it:
  CH3COOH   +   NaOH  --------->   CH3COONa  +  H20

Total mmol CH3COONa needed:   
      (.250L)(0.012mol L-1) = .003 mol

Total grams of CH3COONa needed:   
      (22.98 g/mol)(.003 mol) = 0.069 gram

I think, first we have to use the Henderson – Hasselbalch equation since it is very helpful in calculating an expected pH for a buffer. [pH = pKa  +  log (A-) / (HA)]
(confusing here, is that the question doesn't specifically tell us the Ka value... but i'm guessing its from the question before where we had to calculate the Ka... so Ka = 2.29x10^-8)

pKa = -log(2.29x10^-8) = 7.64
so,
4.30 = (7.64) + log (A-) / (HA)
log (A-) / (HA) = 4.30 - 7.64 = -3.34
(A-) / (HA) = 10^ -3.34 = 0.00045  bcuz this number would then be the ratio

so I pretty much blank out there... help
*insert headache*

[Borek]

Your Ka is several orders of magnitude off.

You are right that you need Henderson-Hasselbalch equation, but you are doing things that don't make sense to me. For example I have no idea where did you get 0.003 mole of CH₃COONa from.