1. Find pH.
30cm^3 of 0.1mol/dm^3 CH3COOH mix with
10cm^3 of 0.1mol/dm^3 NaOH.
[ Ka of CH3COOH = 1.7 * 10^-5 ]
2. Calculate
25cm^3 of 0.050mol/dm^3 ethanoic acid titrated with
0.100mol/dm^3 NaOH
(i) initial pH
(ii) final pH
(iii) pH at equivalence point
(iv) after 10.00cm^3 NaOH was added
(v) after 15cm^3 NaOH was added.
This is a classic acid-base equilibria question,
and (this very same question, though with varying molarities
and volumes) is in fact one of my favourite classic questions
that I frequently force... I mean kindly offer, my students to
practice on.
Now, as Moderators of the Homework Forum, we're not here to do
your homework, but are here to moderate, assist and support your
understanding, learning and personal academic evolutionary process.
As such, for the expressed purpose of helping you understand (as
opposed to doing your homework for you and writing out a complete
worked solution), I will be guiding you along with hints and tips
on what you need to do to solve this (and all other) problems.
Q1.
Write out the ICF (Initial Change Final) table, in no. of
moles. Note the final volume (ie. volume of acid + volume of
alkali).
You will have a buffer system made of up the weak acid (ethanoic
acid) and the conjugate base (ethanoate ions).
Using acid dissociation constant Ka = ([protons][conjugate
base])/[acid], make molarity of protons the subject, and hence
determine pH.
Q2.
(i)
Using acid dissociation constant, Ka, find molarity of protons,
hence pH.
(ii)
Final pH would be when excess NaOH is added till there is no
change in pH. This would eventually approximate (assuming large
quantities of excess) the pH of the NaOH(aq) used. To determine the
pH, given the molarity of OH- (which is the given molarity of
NaOH), calculate pOH, hence obtain pH.
(iii)
Work out ICF table (in no. of moles for Initial Change
Final) for equivalence point. Note the final volume (ie.
volume of acid + volume of alkali). Use this final volume, as
well as the no. of moles of the salt at Final, calculate
molarity of salt. The salt is the product of a strong alkali (Na+
has no tendency to covalent bond OH- ions) with and
a weak acid (CH3COO- has tendency to covalent bond with H+), hence
sodium ethanoate is a basic salt (since OH- will exceed H+ in
solution). Therefore, the next step is to calculate the pH of the
solution resulting from the hydrolysis of the conjugate base, the
ethanoate ion. Using (Ka)(Kb)=Kw, and assuming this is at room
temperature (don't forget that Kw = 1x10^-14 *only* at room
temperature! The dissociation of water molecules into protons and
hydroxide ions is endothermic, so can you tell if the pH of neutral
boiling water is greater or less than 7? How about pH of neutral
melting ice?), calculate the value of Kb. Using this value of Kb,
and the formula Kb = ([hydroxide ions][conjugate acid])/[base],
work out the molarity of hydroxide ions, and hence pOH, and
therefore pH.
(iv)
Write out the ICF (Initial Change Final) table, in no. of
moles. Note the final volume (ie. volume of acid + volume of
alkali).
You will have a buffer system made of up the weak acid (ethanoic
acid) and the conjugate base (ethanoate ions).
Using acid dissociation constant Ka = ([protons][conjugate
base])/[acid], make molarity of protons the subject, and hence
determine pH.
(v)
Write out the ICF (Initial Change Final) table, in no. of
moles. Note the final volume (ie. volume of acid + volume of
alkali).
You will have two species that are responsible for making
the solution basic or alkaline : excess hydroxide ions
and the basic salt sodium ethanoate. However, because OH- is
a much stronger base than the CH3COO- ion (because
the negative formal charge is delocalized by resonance over 2
electronegative oxygen atoms in ethanoate anion, compared to only 1
oxgyen atom in OH- ion; hence OH- ion is significantly more
unstable and thus a significantly stronger base),we can safely
neglect the hydrolysis of the ethanoate ions and calculate the pH
of the solution using only the molarity of the OH- ions. From the
molarity of hydroxide ions in the final solution (again, note
the final volume correctly!), work out pOH, and hence pH.
Everyone
Loves Chemistry!