posted by John .
What is the pH of the solution created by combining 2.40 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HC2H3O2(aq)?
Can somebody explain how to do this? I am stuck. I've tried solving for H+ concentration. I got .00024M and subtracted it from [OH-]. Then I solved for pOH and used that to get pH but still got the wrong answer. Somone plz explain this to me.
I think it's a simple problem IF you recognize what you have. To simplify, let's call HC2H3O3, acetic acid, HAC. Then
NaOH + HAc ==> NaAc + H2O
moles NaOH initially = L x M = 0.00240 x 0.1 = 0.000240
moles HAc = L x M = 0.008 x 0.1 M = 0.0008.
Note the NaOH is the smaller amount; therefore, HAc will neutralize all of the NaOH and some HAc will be left over.
NaOH = 0
HAc = 0.0008-0.00024 = 0.00056
NaAc formed = 0.00056
H2O formed but not worry with that.
So what do you have when all of this is over? We have a weak acid (HAc) and its salt (NaAc) which makes a buffer. Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log[(base)/(acid)]
base = concn of Ac^- which is moles/L.
acid = concn of HAc which is moles/L.
Plug and chug. I'll post this, then repost with an approximate answer to help you check your answer.
You should get close to 4.3 or so for pH.
How do you find the concentration of the conjugate base?
base = mols acetate/L
moles acetate = 0.00056
volume = 10.4 mL = 0.0104 L.
concn = 0.00056/0.0104 = ??
How are we finding pKa for the Henderson-Hasselbalch equation?
Never-mind decided just to use a table of pKa values. By the way the NaAc formed wouldn't be 0.00056 it would be 0.00024 because you can only make as much product as your limiting reagent allows (in this case OH- ions) I seriously tried using these numbers a ton of times and it was the NaAc value that screwed everything up.