SOLVING PROBLEMS BY SUCCESSIVE APPROXIMATIONS AND CIRCULAR REFERENCE:
Calculation of the pH of a solution of a weak acid:
For the dissociation of a weak acid (HA) with initial molar concentration of CA we can write:
….
Initial Concentration CHA
nil
+ H2O  H3O+ + A-
HA
At Equilibrium
10-7
CHA – [X]
[X]
+ [X]
If the Ka value is known, the equilibrium molar concentrations can be solved from the equilibrium constant
expression by:
1. solving for the positive root of the resulting quadratic equation (conc. cannot be negative)
2. the method of successive approximations
1. Ka = (X)(X)/(CHA-X)
Then

(CHA-X)Ka = X2
X = [-b + (b2 - 4ac)] / (2a)

X2 + KaX - KaCHA = 0
where a = 1, b = Ka, c = - KaCHA
2. Assume [X]  CHA, then (CHA – X)  CHA 
Ka = (X)(X)/(CHA)

(CHA)Ka = X2

X1 = (KaCHA)
This 1st approximation is good to within 5% if CHA/Ka  102
When CHA/Ka  102, perform successive approximations:
X2 = [Ka(CHA-X1)]
X3 = [Ka(CHA-X2 )]
X4 = [Ka(CHA-X3 )]
Etc.
This process is repeated until successive approximations agree.
Once [X] is determined, [H3O+], = [A-], = [X] and [HA] = CHA – [X].
pOH can similarly be calculated for dissociation of a weak base and then pH since: pH + pOH = 14.
SUCCESSIVE APPROXIMATIONS IN EXCEL:
Excel is well suited to successive calculations as formulas can be easily copied by dragging to adjacent
cells.
Consider a 0.1M concentration of the weak acid NaHSO4 with Ka = 1.20E-02.
Roughly speaking, weak acids with Ka values greater than 10-4 will require the quadratic or successive
approximations approach.
The [H+] is calculated on the following spreadsheet.
CIRCULAR REFERENCE, DIFFERENTIATION and GRAN’S METHOD
1
Cell C1 contains the Ka of the acid (1.20E-02).
Cell A4 contains the initial concentration of HA (0.1 M).
Cell C4 contains the formula ‘=SQRT($C$1*A4)’ or
‘‘=SQRT(0.012*A4)’ and displays the 1st approximation of
[H+]. Note the absolute cell reference ($C$1) for the Ka, which
is constant in all calculations. When this formula is dragged
down column C, the relative cell reference, A4, changes to A5,
A6, A7, and so on.
Cell A5 contains the formula ‘0.1-C4’ and displays the 1st
approximation of [HA] at equilibrium. Note the absolute cell
reference ($A$4) for the initial HA concentration, which is used
in all calculations. When this formula is dragged down column
A, the relative cell reference C4 becomes C5, C6, C7, and so
on.
Cell D4 contains the formula ‘=-LOG10(C4)’ and displays the
log10 of C4. When this formula is dragged down column D the
cell reference
changes
to C5,and
C6,completely
C7 and so on.
Note that the equilibrium [H+] concentrationrelative
is essentially
constantC4
after
3 iterations
constant after 6 iterations.
The pH at equilibrium in an aqueous solution of NaHSO4 is seen to be 1.54.
Exercise:

Reproduce these calculations on an Excel spreadsheet.

Use successive approximations on an Excel spreadsheet to calculate the equilibrium pH of a 0.20 M
aqueous solution of HIO3 (Ka = 1.70E-1). Answer: [H+] = 0.118, pH = 0.928 (after 20 iterations).
SOLVING PROBLEMS BY INTENTIONAL CIRCULAR REFERENCE:
When a formula refers to itself, either directly or indirectly, it creates a circular reference. If a circular
reference occurs, Excel issues a ‘Cannot resolve circular references’ message; and displays a zero value in
the cell when you click OK in the dialog box.
Usually circular references occur unintentionally, because the user
incorrectly entered a cell reference in an equation. But
occasionally a problem can be solved by intentionally creating a
circular reference.
As an example, the [H+] calculated by successive approximations
in the previous problems can also be solved by circular reference.
The accompanying spreadsheet fragment illustrates the calculation
via circular reference.
Cell A17 contains the initial [HA], i.e., 0.1 M.
Cell B17 contains the formula ‘$A$17-C17’. It is not required but
shows the residual [HA] at equilibrium.
Cell C17 contains the formula ‘=SQRT($C$14*($A$17-C17))’.
When Enter is hit, the ‘Cannot resolve circular references’ message
may appear and a zero is displayed when you click OK. Choose
Options from the Tools menu and on the Calculation tab select
(check) the Iteration box and enter 1E-15 in the Maximum Change
box.
You will rarely have to increase the Maximum Iterations parameter. The current iteration is displayed in the
formula bar. To terminate calculations, press ESC. (Be sure the iteration box is checked.)
CIRCULAR REFERENCE, DIFFERENTIATION and GRAN’S METHOD
2
DIFFERENTIATION:
The process of finding the derivative or slope of a function is the fundament of differential calculus.

The simplest method of obtaining the first derivative of a function, represented in a table of x,y data
points, is to calculate the slope (y/x).

The second derivative, d2y/dx2 is determined by calculating (y/x)/x.
Calculating the first or second
derivative of a data set
emphasizes rapid changes in
measured data. Points on a
curve of x,y values for which
the first derivative is either a
maximum, a minimum or zero
are often of particular
importance and are called
‘critical points’.
Excel is well equipped to
simplify the tedious task of
plotting and interpreting
potentiometric titration data.
A sample containing both
iodide and chloride was titrated
with standardized AgNO3
titrant. The data and calculated
values are shown on the
accompanying spreadsheet and
plots.
Formulas used:
Column A: titrant volume (mL)
Column B: Emf. (volts)
Cell C6: =A6-A5 (x)
Cell D6: =B6-B5 (y)
Cell E6: =AVERAGE(A6,A5)
Cell F6: =D6/C6 (y/x)
Cell G6: =E6-E5 (x)
Cell H6: =F6-F5 (y/x)
Cell I6: =AVERAGE(E6,E5)
Cell J6: =H6/G6 (y/x)/x
The maxima in the 1st derivative plot indicate the inflection points (end points) of the titration.
The 2nd derivative can be used to locate the inflection points more precisely. The second derivative passes
through zero at the end points. Subsections of data can be plotted on expanded scales to read the end point
more accurately. In addition, linear interpolation can be used to determine the end point.
On the 1st derivative chart, the slope between two points from the titration curve is plotted against the mL
of titrant (average of the two points used). On the 2nd derivative chart, the slope between consecutive
points on the 1st derivative graph is plotted against the averaged volumes from the 1st derivative chart.
CIRCULAR REFERENCE, DIFFERENTIATION and GRAN’S METHOD
3
Potentiometric Iodide and Chloride
Potentiometric Iodide and Chloride
2
0
1
-0.5
-1
Emf (volts)
Emf (volts)
0
-1
-2
-3
-1.5
-2
-2.5
-3
-3.5
-4
-4
-5
-4.5
-5
-6
0
10
20
30
40
50
27
60
27.5
14
12
12
10
10
8
8
E/v
E/ v
14
6
6
4
4
2
2
0
0
20
30
40
50
24
60
25
26
27
28
29
vol. (mL)
vol. (mL)
2nd Derivative Plot
2nd Derivative Plot
20
20.00000
0
0.00000
-20
-20.00000
-40
-40.00000
(E/v)
(E/v)
29
1st Derivative of Tit'n
1st Derivative of Tit'n
10
28.5
0.0491 M AgNO3 Titre (mL)
0.0491 M AgNO3 Titre (mL)
0
28
-60
-80
-60.00000
-80.00000
-100
-100.00000
-120
0
10
20
30
40
50
60
-120.00000
23
24
vol (mL)
CIRCULAR REFERENCE, DIFFERENTIATION and GRAN’S METHOD
25
26
27
28
29
vol (mL)
4
ANALYSIS OF TITRATION DATA:(Gran’s Method)
The location of the end point of a titration using either the 1st or 2nd derivative has just been discussed.
These methods use only the data around the end point.
Another approach, Gran’s method, makes use of data just prior to the end point. It is useful when inflection
at the end point is not clearly defined or when the student has missed or overshot the end point.
This method is applicable to acid-base as well as potentiometric titrations. Note that in emf is relatively
unaffected by dilution during potentiometric titrations so volume corrections are unnecessary here.
The following data relates to titration of Cl- with standard AgNO3. The potential of a silver electrode in
combination with a saturated calomel reference was used to follow the course of the titration. As the Cl- ion
concentration decreases, the electrode potential increases.
Potentiometric Cl- Titration
400
350
Emf (mV)
300
250
200
150
100
50
0
24
29
34
39
Titre (mL)
Within the solution, free Cl- ion is precipitated as the titration proceeds (Ag+ + Cl-  AgCl).
Within the calomel electrode the reverse reaction occurs; solid HgCl is converted to Cl- (HgCl  Hg+ + Cl-).
With respect to the reaction in the calomel electrode, from the Nernst equation, we know that
E - E0 = -59.2log[Cl-] for the Cl- half cell, where E is the measured emf, E0 is the standard emf and
-59.2 is the theoretical coefficient (electrode slope or Nernst slope) at 25°C. The logarithmic relationship of
[Cl-] (or titre) and Emf is apparent.
We can combine E and E0, rearrange the resulting equation, and taking antilogs, which yields:
[Cl-] = 10(E/-59.2) when E is mV (as in this data set).
In Gran’s method a linear plot is obtained by plotting the antilog of (emf/ electrode slope) vs. titre. In the
event that such a calculation produces numbers that are so large as to exceed the allowable limits of Excel
(as well as all hand held calculators), one can divide E by any large negative value (other than –59.2) to
achieve the same effect, i.e., generate an inverse linear plot approaching zero Cl- concentration as titre
increases.
CIRCULAR REFERENCE, DIFFERENTIATION and GRAN’S METHOD
5
-
Potentiometric Cl
Antilog(Emf/-59.2)
0.02
0.015
0.01
0.005
0
27
28
29
30
31
32
33
34
Titre (mL)
Rather than trying to read the end point from the graph, use LINEST or Data Analysis, Regression on the
linear data points to obtain the equation of the straight portion of the curve. Then use the equation of the
line to calculate the value of x (titre) when the straight line would intersect the x-axis, i.e., when y = 0.
The results from LINEST are shown below:
x
vol (mL)
27
29.7
31.1
31.73
32.13
32.4
32.54
32.67
32.81
32.94
33.08
33.21
33.35
y
Emf
104
120
135
147
159
171
180
191
230
313
337
348
355
antilog y
10^(Emf/-59.2)
0.01750827
0.009396648
0.005243178
0.003287698
0.002061528
0.001292666
0.000910876
0.000593812
0.000130276
5.16224E-06
2.0297E-06
1.32319E-06
1.00781E-06
parameters
std. dev.
R2, SE(y)
F, df
ss(reg), ss(resid)
m
-0.002994183
8.95843E-06
0.999937342
111710.2375
0.000258665
b
0.098336007
0.000281235
4.81197E-05
7
1.62085E-08
Note that only the bordered data was used in LINEST since these data points occurred before the end point
and were linear. Do not include the data after the end point, i.e., data in the nonlinear portion of the curve.
Using the formula y = -0.002994183titre + 0.098336007 and solving this for y = 0 gives x (titre) = 32.84 mL
as the end point in the titration.
In theory, one could accurately determine the end point without even finishing the titration, or if one
overshot the end point, or if one could not see the color change (color blind or dirty sample).
CIRCULAR REFERENCE, DIFFERENTIATION and GRAN’S METHOD
6
GRAN’S METHOD FOR ACID-BASE TITRATIONS:
Consider an acid-base titration. At any point before the end point the unreacted H+ is given by 10-pH. To
estimate the volume required to reach the end point, plot 10-pH versus titre (mL) and extrapolate to 10-pH = 0.
This assumes that one is using a pH meter in place of or in addition to an acid/base indicator.
If dilution by the titrant is important, then the value of [H+] should be multiplied by (V0+V)/V0, where V0 is
the initial volume. This correction factor increases the [H+] slightly to the value it would have been if
dilution were not occurring during the titration. V = mL titrant, i.e., titre.
The following data arises from the titration of 100 mL of 0.100M CH3COOH (a weak acid) with 0.100 M
NaOH (a strong base). We know the correct titre at the end point will be 100.0 mL so we can compare our
answer to the correct value. We also know that the pH at the end point will be 8.72. Note that the end point
was missed. Gran’s method will allow us to closely estimate the end point using the data obtained just
before the end point.
HAc tit'n w. NaOH
14
12
pH
10
8
6
4
2
0
0
50
100
150
mL .100 M NaOH
Column A contains the volume (mL) of 0.100 M NaOH. Column B contains pH readings.
Formulas:
Column C: =POWER(10,-B2) giving 10-pH, i.e., [H+].
Column D: is shown in the spreadsheet formula bar. It corrects for dilution.
Plotting pH vs. mL NaOH yields the usual S-shaped titration curve (above).
Plotting corrected [H+] vs. mL NaOH linearizes the plot near the end point.
[H+]
HAc vs. NaOH-volume corrected
0.0014
0.0012
0.001
0.0008
0.0006
0.0004
0.0002
0
0
50
100
mL NaOH
150
After studying the plot, it is apparent that the 1st
data point is not linear with subsequent data so
it should not be included in the data used to
estimate the end point.
To remove this data from the plot (without
eliminating it from the spreadsheet), simply
make a copy of this graph (CTRL+ drag), then
change the data range via Chart, Source Data,
Series tab and use the ‘go to spreadsheet’
buttons to reenter the x and y values (excluding
the first data point). Use Format, Selected axis,
Scale tab to change the minimum X axis value.
CIRCULAR REFERENCE, DIFFERENTIATION and GRAN’S METHOD
7
HAc vs. NaOH-volume corrected
HAc vs. NaOH-volume corrected
0.00003
0.0001
0.000025
0.00002
0.00006
[H+]
[H+]
0.00008
0.00004
0.000015
0.00001
0.00002
0.000005
0
0
20
70
120
170
50
70
mL NaOH
HAc vs. NaOH-volume corrected
0.00001
[H+]
0.000008
0.000006
0.000004
0.000002
0
85
95
105
110
130
The process of studying the plots and removing non
linear data is repeated until a straight section of plot is
obtained. As in the example on the left, data prior to
75 mL has been removed from the plot and what
remains ‘appears’ linear before the end point. At this
point, use LINEST or Data Analysis to determine the
slope and intercept of the remaining data prior to the
end point. If the R2 is very good, use these parameters
to write the equation of a straight line and use that
equation to solve for the value of x (mL NaOH) when
y = 0, i.e,. when [H+] ‘appears to’ reach 0.
0.000012
75
90
mL NaOH
115
125
mL NaOH
m
LINEST gave the following results utilizing the
data from 75 to 99 mL NaOH. An end point of
99.43 mL is predicted (not bad but not as good as
we would like). Since R2 was only 0.9983, it was
decided to eliminate the 75 mL point and
recalculate using only the data at 90, 95 and 99 mL.
parameters
std. dev.
R^2, SE(y)
F, dF
ss(reg), ss(resid)
0
For 90 through 99 mL data only, 99.93 mL end point.
m
Parameters
std. dev.
R^2, SE(y)
F, dF
ss(reg), ss(resid)
0
b
-3.81182E-07
3.04495E-09
0.999936193
15671.2988
5.90885E-12
3.80903E-05
2.88473E-07
1.94178E-08
1
3.77049E-16
99.92693433
CIRCULAR REFERENCE, DIFFERENTIATION and GRAN’S METHOD
b
-4.27699E-07
1.23389E-08
0.998338191
1201.507615
6.0503E-11
4.25261E-05
1.11308E-06
2.24401E-07
2
1.00712E-13
99.42978052
The final result is very close
to the true value (-0.075%
error). R2 = 0.99994.
There is no substitute for
good data, however, Gran’s
method can give excellent
results when poor technique
occurs or weak end point
inflection is unavoidable.
Caution: If you eliminate all
but 2 data points, an R2 =1
will always be obtained
even if the data is poor.
8