ModuleAnalytical MeasurementTutorCodeICAExcel si.docx

Module
Analytical Measurement
Tutor
Code
ICA
Excel simulation of a titration involving strong acid and a weak
base (or v.v.)
2nd Marker
Weight
50%
Type
Online submission
Produce an Excel spreadsheet based on simulation to illustrate
the variation of the pH of the solution that occurs when strong
acid is added to a weak base (or strong alkali and weak acid).
The spreadsheet should demonstrate the change in the pH of the
mixture as the titrant is added for a range of concentration of
both components. Particular attention should be paid to the
behaviour close to the end point (or equivalence point). Graphs
should be generated on a range of scales showing in detail the
likely behaviour around the end point.
The exact calculations for all situations can be quite involved so
it may be useful to divide the titration into various segments in
which various equations and approximations are valid and then
stitch these together.
Write a report explaining the design of your spreadsheet and

describe the graphs, which you should import.
Write a short description of the theory of buffer solutions and
give examples of their applications.
The report should be a word document and an Excel
spreadsheet. Describe the background theory and the
methodology behind the construction of the simulation, and
include the realisation of the system equations in Excel
formulae. Graphs of pH against added solution over various
ranges and around particular key points – such as the
equivalence point or “end point”.
[60 marks]
The report should also include a discussion of the concept of
buffer solutions and give examples of their use either in
industrial processes or naturally occurring phenomena. Any
sources should be appropriately referenced (Harvard).
[40 marks]
assignhints(1).pptx
Hints for the Assignment Weak Monoprotic
AcidsNameFormulaValue of KaMonochloracetic acid
Hydrofluoric acid
Nitrous acid
Formic acid
Lactic acid
Benzoic acid
Acetic acid
Hydrated aluminum(III) ion
Propanoic acid
Hypochlorous acid

Hypobromous acid
Hvdrocyanic aid
Boric acid
Ammonium ion
Phenol HC2H3ClO2
HF
HNO2
HCO2H
HC3H5O3
HC7H502
HC2H3O2
[Al (H20)6]3+
HC3H5O2
HOCl
HOBr
HCN
H3BO3
NH4+
HOC6H5
HOI 1.35 x 10 -3
7.2 x 10-4
4.0 x 10-4
1.8 x 10-4
1.38 x 10-4
6.4 x 10-5
1.8 x 10-5
1.4 x 10-5
1.3 x 10-5
3.5 x 10-8
2 x 10-9
6.2 x 10-10
5.8 x 10-10
5.6 x 10-10
1.6 x 10-10
2 x 10-11

Weak Alkalis or BasesFormulaNameKbNH3ammonia
1.8 x 10 -5CH3NH2methylamine
4.4 x 10 -4C5H5Npyridine
1.5 x 10 -9C6H5NH2aniline4.3 x 10 -10
It is possible to set up a simulated titration between a weak acid
and a strong base.
It is easier to work backwards from pH to the added volume of
titrant using the equations below as [H+] = 10(-pH)
It is possible to generate the curve with just one formula going
backwards.
However we know by working forwards from added volume to
pH that there are certain key points of interest, namely where φ
= 0, 0.5, 1 and >1.
Solver in excel
Looking at the phi we have slightly missed the key point
working backwards.
Using solver excel can vary the pH until the equations give a
target value.
The blue cell is the set objective.
Excel will change the value in cell A58 until the objective cell
reads 0.5. To speed things up we constrain the trial values to lie

between 4.75 and 4.8.
Hence we can tweak the pH values to give the key phi values as
these may be of academic interest.
The formulae for titrating a weak alkali with a strong base are
similar to the above.
Some sources use more difficult notation.
The treatment is for monoprotic acid (those that release at most
one hydrogen ion per formula unit).
Polyprotic acids have different dissociation constants for the
first, second etc. hydrogen ions and hence produce a titration
curve with many steps.
There are relatively few common bases and no issues
corresponding to the polyprotic complexity.
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buffer1.pptx
Buffer solutions:
Either a weak acid like Ethanoic Acid HE mixed with a salt
having a COMMON ion e.g. Sodium Ethanoate NaE.
Or a similar set up with a weak base.
Recall in aqueous solution HE = H+ + E-
Normally [H+] =[E-] but when the salt NaE is added the
concentration of [E-] is dominated by the concentration of the
added salt.
COMMON ION EFFECT
Consider an initial concentration of HE of 0.1M. If NaE is
added such that it would by itself have a concentration of 0.2M
in the same volume, we may assume that as the dissociation of
weak acids is small that the concentration of [E-] is almost
0.2M. Putting [H+] =x we have the residual concentration of
[HE] = 0.1-x

pH = -logx = 5.06
An alternative calculation is called the Henderson-Hasselbalch
equation.
This reveals the buffer action. If “acid” meaning extra H+ ions
are added (e.g. if a strong acid is added) most of these H+ ions
will combine with E- to form more HE so [HE] rises and [E-]
falls. As these are in the log term the change in these only has
a weak effect on pH.
Conversely adding OH- breaks down the HE into H2O and E-
Calculations.
As only the ratios of the concentrations matter and the reactants
(products) are in the same volume, we can work with the ratios
of the moles.
Consider a volume V containing 1mmol 0f HE and 2mmol of
NaE.
pH = pKa + log(2/1) = 5.1
Adding 0.3mol of H+ ions will convert 0.3 mmol of E- to HE
giving new quantities of 2-0.7 = 1.7mmol of E- and 1.3mmol of
HE
New pH = pKa +log(1.7/1.3) = 4.759 +0.117 =4.9
Conversely adding 0.3mmol of OH to the original mixture
would have the reverse effect.
pH = pKa + log(2.3/0.7) =5.3

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moretitration.pptx
Titration of weak acid with strong base.
The weak acid is only slightly dissociated so the hydrogen ion
concentration is much lower than for a strong acid and hence
the pH is higher.
As the base is added the acid is neutralised and the salt is
produced. The salt of the corresponding acid is what is required
to make a buffer solution. Hence there is a region in which the
pH changes very slowly as further base is added.
Near the equivalence point the hydrogen ion concentration due
to the acid is so low that the background hydrogen ion
concentration from the dissociation of water becomes important.
Beyond the equivalence point the treatment can be the same as
for the strong/strong titration.
Titration of weak acid with Strong Base.
Step 1 Work out the volume of base required to neutralise the
acid.
Remember any reaction between a strong base and a weak acid
goes to completion.
Consider 50ml of 0.02M weak acid HA.
The total number of moles is o.o2 x 50 = 1 mmol (1 millimole)
If this is titrated with 0.1 M NaOH
Given the reaction HA + NaOH = NaA + H2O
It is apparent that 1 mol of NaOH reacts with 1 mol of HA
Hence we require 1mmol of NaOH
Moles of NaOH = concn x volume, volume = 1mmol/0.1 = 10ml
Step 2 Find the initial pH of the weak acid before any base is

added.
If pKa = 6 and F=0.02M, x = 1.409 x 10 -4 pH=3.85
Step 3 add smaller amounts of Na OH less than the volume
required for equivalence and calculate the pH
Recall HA + NaOH = NaA + H2O
Initially we had 1mmol of HA
This is neutralised by 1oml of NaOH
Hence adding 3ml of NaOH will react with 3/10 = 0.3mmol of
HA
The new Formula concentration of HA is 0.7 mmol
However the presence of 0.3mmol of NaA in solution will
disturb the equilibrium.
pH= 6 + log (0.3/0.7) = 5.15
Note as the various species are in the same total volume and we
only need the ratio of the 2 concentrations we can just use the

moles in the argument of the logarithm.
Step 4 The equivalence point.
At the equivalence point all the acid has been converted to the
salt.
HA + NaOH → NaA + H2O
HA + OH- → A- +H2O
However, we know that water dissociates very slightly
H2O = H+ + OH- recall the equilibrium constant is Kw
Clearly the presence of any H+ ions will drive the reverse
reaction
A- + H+ = HA for which the reaction constant is 1/Ka
Hence the combined reaction A- + H2O = HA + OH- has a
reaction constant Kw/Ka
Step4 continued
Using the reaction A- +H2O = HA + OH- with K = Kw/Ka
Note the solution at the equivalence point is actually slightly

alkaline. Nearly all of the Hydrogen ions have been combined
with added Ohs, but the A- ions that are still in the solution
make the mixture slightly basic.
Step 5
Beyond the equivalence point we simply calculate the excess
OH
If the added volume of NaOH is 10+y ml the excess OH is due
to the yml
The concentration of excess OH is y x 0.1/(50+10+y)
For example if y = 0.5ml
[OH-] = 0.5 x 0.1/(60.5) = 8.26 x 10 -4
pH = 14 + log[OH-] = 10.92
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Solution
s, concentration, acids, pH (1).pptx

ModuleAnalytical MeasurementTutorCodeICAExcel si.docx

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