AL Chem: Dissociation constant

Ryan, this is NOT an experiment about titration. it's about determination of pKa value of a weak acid.

1.
no, NaOH is not in excess. when end-point is reached, indicator changes its color, like in normal titrations. phenolphthalein, the indicator, is used to ensure all acid present reacts with NaOH.

amount of NaOH is added just enough to react with ALL ethanoic acid to give ethanoate, i.e. complete neutralization.

2.
the aim of reaction is NOT to determine the pH of titration between NaOH and ethanoic acid.
as you've mentioned, [CH3COOH] = [CH3COO(-)] after extra portion of acid is added.

yes, pH may change a little bit as new equilibrium is attained, but this is SMALL. the concentration of ethanoate is already high while ethanoic acid is a weak acid. thus not much acid will ionize ------ of course, this would be one of your possible errors in Q.3.

3.
NaOH is NOT in excess throughout whole reaction.
in step 3, NaOH is titrated against acid, so equal no. of mole of both are present and react.
in step 4, excess acid is added.

another error is that, quite a long period of time is required to attain the equilibrium. remember to stir and wait for longer time.

4.
as you've mentioned, Ka = [H+][CH3COO-] / [CH3COOH]
log(Ka) = log[H+] + log( [CH3COO-] / [CH3COOH] )
pKa = pH - log( [CH3COO-] / [CH3COOH] )

in step 3, y mole of acid is turned to the salt, CH3COOH;
in step 4, same no. of mole of acid is added.
therefore, the final concentration of both acid and anion is the same.
[CH3COO-] / [CH3COOH] = 1
log( [CH3COO-] / [CH3COOH] ) = 0

then, the calculation becomes pKa = pH
by measuring the resultant pH of mixture in step 4 with pH meter, pH is found ---- which is as same as the pKa value.

A: 冇錯, NaOH in excess

A: 加 acid 因為想減低pH值, 唔會, 準架!

A: 冇錯!