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Title Potentiometry: Titration of a Halide Ion Mixture

Name Manraj Gill (Lab partner: Tanner Adams)

Abstract

Potentiometric titrations of a sample using a system of a electrolytic cell can be used to

analyze the concentration and identify species of halides present in any sample. We

identified I- and Cl- halides in Unknown #21 and accurately determined their

concentrations in this sample. And we additionally determined that around 0.5M Cl- is

present in seawater and it is the most prominent halide present in seawater.

Purpose

In this series of experiments, we use potentiometric titrations (gradual addition of titrant

to a solution while measuring the potential of a electric cell) to measure the

concentrations of halides in solutions. This is achieved by titrating the sample containing

the halides with a standardized silver ion solution (in this case, silver nitrate, AgNO3).

This titration leads to the halides precipitating out of the solution (described in detail in

Theory and Methods) and thereby allows us to potentiometrically determine the

concentrations of the halides that were initially present. We use this approach to

determine the identity of an unknown solution in terms of its specific constituents. And

we then extend this approach, in Part II, to the analysis of seawater to determine the most

prominent halide in the seawater sample and the total salinity of this sample!

Theory and Methods

The approach used towards understanding the composition of the samples analyzed relies

on measuring the potential by observing a voltmeter. The half-cells created for this

purpose consist of an indicator electrode and a reference electrode.

The indicator electrode is this case is the unknown (or seawater in Part II). We create the

reference electrode is immersed in a solution of 1M KNO3 and we use this reference

electrode (or reference half-cell) because of it’s ability to maintain a reproducibly high

and constant potential (1). Additionally, the used of a salt-bridge allows us to separate the

two electrodes (reference and test/indicator) from each other and allows for the reference

electrode potential to be constant!

This approach, described above, allows us to correlate any chemical alterations reflected

in the measured potential of the entire cell to be reflective of the test electrode!

Additionally, we use the reduction-oxidation (redox) couple of silver metal (submerged

in the test electrode) and silver ion (the titrant or silver nitrate (AgNO3 mentioned in the

purpose section) as an “internal standard” (2). This allows us to derive the concentrations

of the halides present in the indicator electrode because of the different potentials of the

redox reactions between silver and the halides!

The titrant added dissociates into silver ions and these silver ions interact with the halides

in solution. This interaction leads to the concentrations of the halide ions decreasing as

the Ag-X (X denoting a halide species) compound precipitates out of the solution. The

potential of the cell starts off at a negative value as the flow of electrons is from the silver

wire to the reference electrode. This potential remains constant (for an experimentally

relative time) and becomes progressively positive as more titrant is added. It is due to this

reversal in the flow of electrons that we observe plating on the silver wire used in the test

electrode because once the potential is positive, the silver wire is serving as the cathode!

The important measurement is of the volume of the titrant added at the equivalence points

because it from these endpoints that we can calculate the amount of each halide present.

The equivalence points can be determined using the first-derivative method of analysis as

the equivalence point is the small amount of volume that leads to the drastic increase in

measurement (pH or in this case, voltage). Therefore, the volume at the maximum of the

first-derivative corresponds to the volume at the equivalence point.

Results and Application of Theory

In Part I of this experiment, we determine the concentration of each halide ion in the

unknown sample. Unknown sample number: #21

The halides present (of varying concentrations) in each unknown are Chlorine (Cl-),

Bromine (Br-) and Iodine (I-). The following page shows a preliminary titration

performed to gain a rough idea of where the equivalence points are found because this

then allows us to be more precise in the three consequent “accurate” measurements of the

titrations that can be used for statistical analysis.

As evident by the performed first derivative analysis of this preliminary titration, the

equivalence points lie around silver nitrate volumes of ~14.75ml and ~42.75ml (the two

maxima of the 1st derivative plot).

Based on this preliminary observation, three similar titrations were performed to obtain

accurate values (with small standard deviations) for the equivalence points. The three

repeat measurements give us the following values for the equivalence points (the graphs

are over-laid to represent consistency in measurements and to avoid redundancy in the

represented data)

1st Equivalence Point (ml) 2nd Equivalence Point (ml)

1st Measurement 15.15 42.82

2nd Measurement 15.03 42.65

3rd Measurement 15.35 42.05

Mean Value 15.17 42.51

Standard Deviation 0.16 0.40

95% Confidence Interval +/- 0.27 +/- 0.68

Therefore,

1st equivalence point at 15.17 +/- 0.27ml and

2nd equivalence point at 42.51 +/- 0.68ml

Analysis of the potentiometric data and determination of the halide concentrations is

continued after the Voltage to Volume titration plots and 1st derivative graphs below.

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Titration Plot of Actual Voltage Measured:

Preliminary Titration of Unknown

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1st Derivitive Plot:

Preliminary Titration of Unknown

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1st Derivite Plot:

Accurate Titrations of Unknown

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These potentiometric titration curves plot the voltage of the entire cell as milliliters silver

nitrate is added. Upon addition of this silver nitrate titrant, the reaction of silver ions with

the halide ions results in the formation of the solid precipitate that we observe (yellowish

in color, see Observations column in the excel data sheet for more details).

Therefore, in measuring the potential of the test electrode, we need to keep in mind [Ag+]

(i.e. the concentration of the Ag+ ions). We focus on the Ag+ ions and not only on the

concentrations of the halide ions because the indicator electrode responds to [Ag+]!

The potential for the test electrode is given by the Nernst Equation (3) as follows:

E(test) = E°(Ag/AgX) – (RT/F) ln[X-]

Which, if we consider the variation in the different solubilities of the AgX compounds

and take into account the solubility products (Ksp), turns to:

E(test) = E°(Ag+/Ag) + (RT/F) lnKsp(AgX) – (RT/F) ln[X-]

It is because of the different Ksp values for silver with the halides (specifically Cl, Br and

I) that we are able to differ between them chemically! That is, the least soluble silver

halide is expected to precipitate first!

Here are the known Ksp values (4) for the three Ag-X species we are investigating

Silver Halide (Ag-X) Species Solubility Product Ksp

Silver-Chloride 1.77 x 10-10

Silver-Bromide 5.35 x 10-13

Silver-Iodide 8.52 x 10-17

Additionally, we know the value of the E°(Ag+/Ag) term in the equation above (citation

needed) to be 0.799V

Now, before we can correlate the measured voltage (i.e. the voltage of the cell) to the text

electrode’s potential (i.e. the E(test) in the modified Nernst equation above), we need to

factor in the reference electrode and the standardization in the measurements.

To this end, we define the measured cell potential as follows:

E(cell) = E(test) – [E(ref) + constant]

This equation can be modified to the following:

E(cell) = E(test) – constant

This modification can be done because the reference electrode is separated from the test

electrode and therefore the E(ref) term is also a constant term in our experiment.

Then, to calculate the value of this constant and directly relate the E(cell) measured

voltages to the E(test) and by extension, the [X-], we need to analyze the experimental

data after the final end point. i.e. the plateau of the titration curve after the second

equivalence point!

This constant can be determined based on this region of the plot after the equivalence

point because that is when the halide ions have been consumed and the potential observed

(the measured voltage of the cell) is simply dependent upon the concentration of the

silver ions. This [Ag+] can then be used to determine the value of E(test) and thereby

subtracted from E(cell) to give us the value of the constant for calibration of the data

we’re interested in!

In the above analysis, we calculated the second equivalence point to occur at 42.51 +/-

0.68 milliliters of AgNO3. Using one of the points post this equivalence point, we can

determine the expected value of E(test) because, in this case, E(test) can be described as

follows:

E(test) = E°(Ag+/Ag) + (RT/F) ln[Ag+]

E(test) = 0.799V + [(8.314J/molK) (298.15K) / (96485.332C/mol)] ln[Ag+]

Therefore, E(test)…

When 44.33ml AgNO3 (from 1st Accurate Titration in attached Raw Data excel sheet) is

added,

moles AgNO3 = (44.33 – 42.51ml) (0.10028moles/L AgNO3) (1L/1000ml)

moles AgNO3 = 1.825 x 10-4 moles

Therefore, [Ag+] = 1.825 x 10-4moles / (25.00mlunknown + 0.5mlnitric acid + 44.33mlsilver nitrate)

[Ag+] = 2.613 x 10-3 moles/L

Which gives us

E(test) = 0.799V – 0.1527V = 0.646V

Now, to derive the value of the constant, we look to the actual measured voltage of the

cell, E(cell), at 44.33ml. This value, is 0.407V.

So,

E(cell) = E(test) – Constant

Constant = E(test) – E(cell)

Constant = 0.646V – 0. 407V = 0.239V

Now, we can go back to analyzing the obtained E(cell) potential measurements and

derive the E(test) values which can be used to analyze the identity of the analytes using

the following equation (mentioned once previously):


The important question now is of choosing the areas of the titration curve where this

E(test) value should be measured from for obtaining concentrations of the halides, [X-].

We know that our unknown solution contains only 2 halides. That is also explained by

the titration curve we observe because of the two measured equivalence points. The

potential of the indicator/test electrode is based on the halides being precipitated. Since

the Ksp values are so different, one component of the mixture of halides dominate the

potential until it is completely precipitated (i.e. until the equivalence point).

Therefore, the first halide (the least soluble one) governs the potential until the first

equivalence point. Then the second (and last) halide governs the potential after the first

equivalence and before the second equivalence point. And no halides are present in

solution after the second equivalence point (this was the fundamental idea used to

determine the constant, see analysis above).

The initial step in determining the concentration of a halide is looking at how much

titrant was required to reach equivalence point. Because it is at the equivalence points

that all of one species of halides is titrated. Therefore, based on the data analyzed

above…

The concentration of the initial halide (the one that is least soluble) in our unknown

(unknown #21) is determined by:

Volume of AgNO3 at first equivalence point = 15.17 +/- 0.27ml

Moles of AgNO3 at first equivalence point =

(15.17 +/- 0.27ml AgNO3) (0.10028moles/L AgNO3) (1L/1000ml)

= 1.52 x 10-3 (+/- 2.7 x 10-5) moles

Therefore, moles of Ag+ = 1.52 x 10-3 (+/- 2.7 x 10-5) moles = moles of first X-

Concentration of first X- in the unknown solution =

1.52 x 10-3 (+/- 2.7 x 10-5) moles / 25.00mlunknown

= 6.08 x 10-5 (+/- 1.08 x 10-6) moles/ml = 0.0608 +/- 0.0011M

Similarly, performing the same calculation for the second X-…

Volume of additional AgNO3 at second second point =

(42.51 +/- 0.68ml) – (15.17 +/- 0.27ml) = 27.34 +/- 0.41ml

Which gives us


= 2.74 x 10-3 (+/- 4.11 x 10-5) moles AgNO3 added since the first equivalence point

And the concentration of the second X- in the unknown solution =

2.74 x 10-3 (+/- 4.11 x 10-5) moles / 25.00mlunknown

= 1.096 x 10-4 (+/- 1.64 x 10-6) moles/ml = 0.1096 +/- 0.0016M

Now, since we have confirmed the concentrations of the halides in the unknown solution,

we can use the following equation (mentioned above):


In combination with


To get

E(cell) = E°(Ag/AgX) – (RT/F) ln[X-] – Constant

Solving for E°(Ag/AgX), we get

E°(Ag/AgX) = E(cell) + (RT/F) ln[X-] + Constant

We now obtain values for E(cell) and for [X-] from the horizontal regions of the titration

curves before each of the equivalence points. This would allow us to solve for

E°(Ag/AgX) and compare the obtained value with the known potentials for the silver

halide half cells (5):

Silver Halide Half-cells (Ag/AgX) Standard Potentials (V)

(Ag/AgCl) + 0.222

(Ag/AgBr) + 0.071

(Ag/AgI) - 0.151

For the first halide, plateau of the titration curve is at the addition of 5ml of AgNO3

titrant. At this point, E(cell) = -0.278V (see excel sheets, 3rd accurate titration)

To obtain the value of [X-], we need to know the moles of the halide ions left in solution

(i.e. moles that have not precipitated). We know, from the analysis, above that the total

number of moles of the first halide is 1.52 x 10-3 (+/- 2.7 x 10-5) moles X-first.

In 5.00ml of 0.10028M AgNO3, there are 5.01 x 10-4 moles.

Assuming full reactivity of the Ag+ with the X-first, the number of moles of X-

first

remaining in solution =

1.52 x 10-3 (+/- 2.7 x 10-5) moles - 5.01 x 10-4 moles = 1.02 x 10-3 (+/- 2.7 x 10-5) moles

Therefore, the [X-] =

1.02 x 10-3 (+/- 2.7 x 10-5) moles / (25.00mlunknown + 0.5mlnitric acid + 5mlsilver nitrate)

= 0.0334 +/- 0.0009M

Which gives us E°(Ag/AgX) =

E(cell) + (RT/F) ln[X-] + Constant

= (-0.278V) + [(8.314J/molK) (298.15K) / (96485.332C/mol)] ln(0.0334) + 0.239V

= (-0.278V) + (-0.0873) + (0.239V)

= -0.126V

Since our experimental conditions are not entirely standardized (i.e. not ideal), there is an

observed off set from the known values but regardless, the E°(Ag/AgX) value obtained is

very close to the E°(Ag/AgI) value of -0.151V. Therefore, we can conclude that in the

unknown solution (unknown #21), there is 0.0608 +/- 0.0011M Iodide

Now, analyzing the second halide and following a similar procedure of analyzing the

plateau before the second halide’s equivalence point…

At 27.00ml AgNO3, E(cell) = 0.094V (see excel sheets, 3rd accurate titration)

At this volume of titrant added, number of moles of AgNO3 =

[27.00ml – (15.17 +/- 0.27ml)] (0.10028M AgNO3) (1L/1000ml)

= (11.83 +/- 0.27ml) (0.10028M AgNO3) (1L/1000ml)

= 1.186 x 10-3 +/- 2.707 x 10-5 moles

This is also the number of moles of X-second, the second halide that have precipitated out

of solution at this titrant volume.

Therefore, the number of moles of the second halide at this volume is given by:

2.74 x 10-3 (+/- 4.11 x 10-5) moles - 1.186 x 10-3 +/- 2.707 x 10-5 moles

= 1.55 x 10-3 +/- 1.40 x 10-5 moles

Which gives us [X-second] =

[1.55 x 10-3 +/- 1.40 x 10-5 moles / (25.00mlunknown + 0.5mlnitric acid + 27.00mlsilver nitrate)]

= 0.0295 +/- 0.0003M

Now, E°(Ag/AgX) =


= (0.094V) + [(8.314J/molK) (298.15K) / (96485.332C/mol)] ln(0.0295) + 0.239V

= (0.094V) + (-0.091) + 0.239V

= 0.242V

Which is a value close to (considering experimental error), the E°(Ag/AgCl) of 0.222V!

Therefore, we can additionally conclude that there is 0.1096 +/- 0.0016M Chloride in

unknown #21!

Results and Application of Theory in Part II of the experiment

This is an analysis of a commercial seawater sample and we are attempting to analyze the

presence of halides in this solution. Using the technique similar to part I in terms of

analysis, we identify one halide ion present in this sample because of the one observed

equivalence point (see the preliminary and accurate titration plots below).

Deviation from the experimental conditions in part I is the concentration of the

standardized silver nitrate solution used: 0.10189M AgNO3

We first determine the volume needed to reach the equivalence point. As evident by the

performed first derivative analysis of the preliminary titration, the equivalence point lies

around silver nitrate volume of ~28.75ml (the maxima of the 1st derivative plot).

Based on this preliminary observation, three similar titrations were performed to obtain

accurate values (with small standard deviations) for the equivalence point:

1st Measurement 28.65ml

2nd Measurement 28.05ml

3rd Measurement 28.25ml

Mean Value 28.32ml

Standard Deviation 0.31ml

95% Confidence Interval +/- 0.52ml

Therefore the equivalence point can be described accurately to occur at titrant volume of

28.32 +/- 0.52ml AgNO3

Now, we determine the value of the Constant used for correlating the measured E(cell)

values to E(test).

E(test) = E°(Ag+/Ag) + (RT/F) ln[Ag+]

E(test) = 0.799V + [(8.314J/molK) (298.15K) / (96485.332C/mol)] ln[Ag+]

Therefore, E(test)…

When 44.00ml AgNO3 (from Part II: Preliminary Titration in attached Raw Data excel

sheet) is added,

moles AgNO3 = (44.00 – 28.32) (0.10189moles/L AgNO3) (1L/1000ml)

moles AgNO3 = 1.598 x 10-3 moles

Therefore, [Ag+] = 1.598 x 10-3moles / (25.00mlunknown + 0.5mlnitric acid + 44.00mlsilver nitrate)

[Ag+] = 0.02299 moles/L

Which gives us

E(test) = 0.799V – 0.0969V = 0.702V

Now, to derive the value of the constant, we look to the actual measured voltage of the

cell, E(cell), at 44.00ml. This value, is 0.477V.

So,


Constant = E(test) – E(cell)

Constant = 0.702V – 0.477V = 0.225V

Now, analyzing the concentration of the halide present,

Volume of AgNO3 at equivalence point = 28.32 +/- 0.52ml

Moles of AgNO3 at first equivalence point =


= 2.89 x 10-3 (+/- 5.3 x 10-5) moles

Therefore, moles of Ag+ = 2.89 x 10-3 (+/- 5.3 x 10-5) moles = moles of X-

Concentration of X- in the seawater solution =

2.89 x 10-3 (+/- 5.3 x 10-5) moles / 5.00mlseawater

= 0.578 +/- 0.011M

Now, to determine the identity of this halide, we analyze the region before this

equivalence point.

The plateau of the titration curve is at the addition of 15ml of AgNO3 titrant. At this

point, E(cell) = 0.075V (see excel sheets, Part II: 1st accurate titration)

To obtain the value of [X-], we need to know the moles of the halide ions left in solution

(i.e. moles that have not precipitated). We know, from the analysis, above that the total

number of moles of the halide is 2.89 x 10-3 (+/- 5.3 x 10-5) moles X-.

In 15.00ml of 0.10189M AgNO3, there are 1.528 x 10-3 moles.

Assuming full reactivity of the Ag+ with the X-first, the number of moles of X-

first

remaining in solution =

2.89 x 10-3 (+/- 5.3 x 10-5) moles - 1.528 x 10-3 moles= 1.362 x 10-3 (+/- 5.3 x 10-5) moles

Therefore, the [X-] =

1.362 x 10-3 (+/- 5.3 x 10-5) moles / (5.00mlsea-water + 15.00mlsilver nitrate)

= 0.0681 +/- 0.0027M

Analysis continued after the following plots

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Preliminary Titration of Sea Water Sample

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Which gives us E°(Ag/AgX) =


= (0.075V) + [(8.314J/molK) (298.15K) / (96485.332C/mol)] ln(0.0681) + 0.225V

= (0.075) + (-0.069) + (0.239V)

= 0.244V

Therefore, since this value is close to (considering experimental error), the E°(Ag/AgCl)

of 0.222V, we can conclude that there is 0.578 +/- 0.011M Chloride in the sea-water

sample!

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Accurate Titrations of Sea Water Sample

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Accurate Titrations of Sea Water Sample

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Discussion

As elaborately explained in the data/application of theory section above, the first of the

two experiments allowed us to not only determine, to some degree of accuracy, not only

the concentration of 2 halides in an unknown solution but also their identities. This was

all done by using a set-up of a reference electrode and a test electrode and using

techniques of potentiometric titration established in previous lab experiments. It allows us

to obtain an accurate measurement of the equivalence points of the titrations and just that

information is sufficient, once all the groundwork is established in terms of calibrating

the obtained measurements, to determine that in our Unknown Solution #21, there was

0.0608 +/- 0.0011M Iodide and 0.1096 +/- 0.0016M Chloride halides!

There are a lot factors that bring about uncertainties in our measurements from the

graduated cyclinder used for the volumetric transfer of 25.00ml of unknown not being the

most accurate, to the precipitate most likely interfering with the chemistry in the test

electrode and the unlikelihood of all of the AgNO3 completely dissociating. Perhaps it is

a misrepresentation to simply measure the number of moles expecting not only complete

dissociation but also complete precipitation of all added Ag+ with the halide.

We extended this experimental approach, i.e. using the same techniques and analysis, to

analyze the identity of halides in a sample of sea water. In this case, we observed only

one equivalence point which, based on the experimental techniques established, tells us

that there is only one halide present in the sample. We measured that to be 0.578 +/-

0.011M Chloride. But our experimental procedure probably has a limitation in regards to

the minimum concentration of a halide species to be detected. Because other halides

could be present but at a concentration much lower than what can “truly” and completely

precipitate out in a manner to affect the voltage of the test electrode. Therefore, while we

cannot be sure that the chloride halide is the only halide present, we can definitely be

certain that it is the most prominent. This makes sense since we are detecting common

salt’s (NaCl) halide!

Conclusions

Based on this experiment, we can conclude that potentiometric titration using a system of

a electrolytic cell can be used to analyze the concentration and identify species of halides

present in any sample. As explained in detail in the discussion section above, we

identified I- and Cl- halides in Unknown #21 and additionally determined that around

0.5M Cl- is present in seawater!

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