i've spent entirely too long doing this.
oh, but it's been fun.
1. roches being a bastard about pH
your method won't work. that's for an aqueous solution of NaOH. well, that whole pH thing is just ridiculously high-school. if methods of measuring acidity were ravers, pondus hydrogenii would wear plastic bracelets dyed with phenolphthalein and methyl orange up to its elbows.
however, how much NaOH DOES make the difference between pH 8 and pH 14? assuming that NaOH dissociates fully, the difference would be 1 mol/L - 1x10^-6 mol/L = 0.9999 mol/L. pH is wack, remember. i can't give the answer in mL without guessing a concentration and the volume of your reaction mixture. and--this is the really cool part--you can't determine the pH or concentration of an NaOH solution by weighing it, because it absorbs significant amounts of water from the air. you have to standardize it with (usually) a known concentration of potassium hydrogen phthalate.
so, can i have a bracelet? =)
3. roches on the henderson-hasselbach equation
note that none of this could possibly have any application to anything *bad*. it's just a random, off-topic discussion of acid-base chemistry.
first, the equilibrium constant of an acid-base reaction is:
(1) pKa = - log ([H+][A-]/[HA]) where A represents any anion
this can be related to the pH of the reaction mixture thus:
(2) pH = pKa + log ([A-]/[HA]) the Henderson-Hasselbach equation
here, [A-] can also be called [base] or even [product] (if it's a unimolecular reaction with all coefficients = 1) and [HA] can be called [acid] or [reactant].
in this *cough* reaction, you can use this to estimate both the pH of the solution containing a given ratio of reactants and products and the ratio at a given pH, if you assume that the reaction itself goes to completion (that is, that all the anion would be converted to the product). also, you need to know the pKa for the reactant here.
you can get a lot more out of that equation than just the answer to your question, but these are the two you asked. i'm assuming the pKa is 4.5, but i'm only basing this on an MSDS sheet saying the pH of the 99% solution is 4.5, so i could be totally wrong.
the pH of the reaction mixture after addition of 0.9 molar equivalents of NaOH
pH = pKa + log ([A-]/[HA])
pH = 4.5 + log (9/1)
pH = 4.5 + 0.95
pH = 5.45
the composition of the reaction mixture at pH 8
pH = pKa + log ([A-]/[HA])
8 = 4.5 + log ([A-]/[HA])
3.5 = log ([A-]/[HA])
([A-]/[HA]) = 3162 as far as you're concerned this means [HA] = 0
interpret these however you want. this will not be totally accurate, for several reasons (e.g. the equilibrium constant of the overall reaction, and the possible presence of ethanol or some other protic solvent).
the end.
