Title Potentiometry: Titration of a Halide Ion Mixture. Name Manraj Gill (Lab partner: Tanner Adams)

Transcription

1 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.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

2 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 1 st 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) 1 st Equivalence Point (ml) 2 nd Equivalence Point (ml) 1 st Measurement nd Measurement rd Measurement Mean Value Standard Deviation % Confidence Interval +/-.27 +/-.68 Therefore, 1 st equivalence point at /-.27ml and 2 nd equivalence point at /-.68ml

3 1st Derivative (d-voltage / d-volume) Voltage Measured Analysis of the potentiometric data and determination of the halide concentrations is continued after the Voltage to Volume titration plots and 1 st derivative graphs below..5.4 Titration Plot of Actual Voltage Measured: Preliminary Titration of Unknown Silver Nitrate (ml) 1st Derivitive Plot: Preliminary Titration of Unknown Silver Nitrate (milliliters)

4 1st Derivative (d-voltage / d-volume) Volatge Measured.5 Titration Plot of Actual Voltage Measured: Accurate Titrations of Unknown Silver Nitrate (ml) 1st 2nd 3rd.7 1st Derivite Plot: Accurate Titrations of Unknown st 2nd 3rd Silver Nitrate (ml)

5 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 1-1 Silver-Bromide 5.35 x 1-13 Silver-Iodide 8.52 x 1-17 Additionally, we know the value of the E (Ag + /Ag) term in the equation above (citation needed) to be.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.

6 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 /-.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) =.799V + [(8.314J/molK) (298.15K) / ( C/mol)] ln[ag + ] Therefore, E(test) When 44.33ml AgNO3 (from 1 st Accurate Titration in attached Raw Data excel sheet) is added, moles AgNO3 = ( ml) (.128moles/L AgNO3) (1L/1ml) moles AgNO3 = x 1-4 moles Therefore, [Ag + ] = x 1-4 moles / (25.mlunknown +.5mlnitric acid mlsilver nitrate) [Ag + ] = x 1-3 moles/l Which gives us E(test) =.799V.1527V =.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.47v. So, E(cell) = E(test) Constant Constant = E(test) E(cell) Constant =.646V. 47V =.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): E(test) = E (Ag/AgX) (RT/F) ln[x - ]

7 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 = /-.27ml Moles of AgNO3 at first equivalence point = ( /-.27ml AgNO3) (.128moles/L AgNO3) (1L/1ml) = 1.52 x 1-3 (+/- 2.7 x 1-5 ) moles Therefore, moles of Ag + = 1.52 x 1-3 (+/- 2.7 x 1-5 ) moles = moles of first X - Concentration of first X - in the unknown solution = 1.52 x 1-3 (+/- 2.7 x 1-5 ) moles / 25.mlunknown = 6.8 x 1-5 (+/- 1.8 x 1-6 ) moles/ml =.68 +/-.11M Similarly, performing the same calculation for the second X - Volume of additional AgNO3 at second second point = Which gives us ( /-.68ml) ( /-.27ml) = /-.41ml ( /-.41ml AgNO3) (.128moles/L AgNO3) (1L/1ml) = 2.74 x 1-3 (+/ x 1-5 ) moles AgNO3 added since the first equivalence point And the concentration of the second X - in the unknown solution =

8 2.74 x 1-3 (+/ x 1-5 ) moles / 25.mlunknown = 1.96 x 1-4 (+/ x 1-6 ) moles/ml =.196 +/-.16M Now, since we have confirmed the concentrations of the halides in the unknown solution, we can use the following equation (mentioned above): E(test) = E (Ag/AgX) (RT/F) ln[x - ] In combination with E(cell) = E(test) Constant 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) (Ag/AgBr) +.71 (Ag/AgI) For the first halide, plateau of the titration curve is at the addition of 5ml of AgNO3 titrant. At this point, E(cell) = -.278V (see excel sheets, 3 rd 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 1-3 (+/- 2.7 x 1-5 ) moles X - first. In 5.ml of.128m AgNO3, there are 5.1 x 1-4 moles. Assuming full reactivity of the Ag + with the X - first, the number of moles of X - first remaining in solution = 1.52 x 1-3 (+/- 2.7 x 1-5 ) moles x 1-4 moles = 1.2 x 1-3 (+/- 2.7 x 1-5 ) moles Therefore, the [X - ] = 1.2 x 1-3 (+/- 2.7 x 1-5 ) moles / (25.mlunknown +.5mlnitric acid + 5mlsilver nitrate)

9 =.334 +/-.9M Which gives us E (Ag/AgX) = E(cell) + (RT/F) ln[x - ] + Constant = (-.278V) + [(8.314J/molK) (298.15K) / ( C/mol)] ln(.334) +.239V = (-.278V) + (-.873) + (.239V) = -.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 -.151V. Therefore, we can conclude that in the unknown solution (unknown #21), there is.68 +/-.11M Iodide Now, analyzing the second halide and following a similar procedure of analyzing the plateau before the second halide s equivalence point At 27.ml AgNO3, E(cell) =.94V (see excel sheets, 3 rd accurate titration) At this volume of titrant added, number of moles of AgNO3 = [27.ml ( /-.27ml)] (.128M AgNO3) (1L/1ml) = ( /-.27ml) (.128M AgNO3) (1L/1ml) = x 1-3 +/ x 1-5 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 1-3 (+/ x 1-5 ) moles x 1-3 +/ x 1-5 moles = 1.55 x 1-3 +/- 1.4 x 1-5 moles Which gives us [X - second] = [1.55 x 1-3 +/- 1.4 x 1-5 moles / (25.mlunknown +.5mlnitric acid + 27.mlsilver nitrate)] =.295 +/-.3M Now, E (Ag/AgX) = E(cell) + (RT/F) ln[x - ] + Constant = (.94V) + [(8.314J/molK) (298.15K) / ( C/mol)] ln(.295) +.239V = (.94V) + (-.91) +.239V =.242V Which is a value close to (considering experimental error), the E (Ag/AgCl) of.222v!

10 Therefore, we can additionally conclude that there is.196 +/-.16M 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:.1189m 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 1 st derivative plot). Based on this preliminary observation, three similar titrations were performed to obtain accurate values (with small standard deviations) for the equivalence point: 1 st Measurement 28.65ml 2 nd Measurement 28.5ml 3 rd Measurement 28.25ml Mean Value 28.32ml Standard Deviation.31ml 95% Confidence Interval +/-.52ml Therefore the equivalence point can be described accurately to occur at titrant volume of /-.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) =.799V + [(8.314J/molK) (298.15K) / ( C/mol)] ln[ag + ] Therefore, E(test) When 44.ml AgNO3 (from Part II: Preliminary Titration in attached Raw Data excel sheet) is added, moles AgNO3 = ( ) (.1189moles/L AgNO3) (1L/1ml) moles AgNO3 = x 1-3 moles Therefore, [Ag + ] = x 1-3 moles / (25.mlunknown +.5mlnitric acid + 44.mlsilver nitrate) [Ag + ] =.2299 moles/l Which gives us E(test) =.799V.969V =.72V

11 Now, to derive the value of the constant, we look to the actual measured voltage of the cell, E(cell), at 44.ml. This value, is.477v. So, E(cell) = E(test) Constant Constant = E(test) E(cell) Constant =.72V.477V =.225V Now, analyzing the concentration of the halide present, Volume of AgNO3 at equivalence point = /-.52ml Moles of AgNO3 at first equivalence point = ( /-.52ml AgNO3) (.1189moles/L AgNO3) (1L/1ml) = 2.89 x 1-3 (+/- 5.3 x 1-5 ) moles Therefore, moles of Ag + = 2.89 x 1-3 (+/- 5.3 x 1-5 ) moles = moles of X - Concentration of X - in the seawater solution = 2.89 x 1-3 (+/- 5.3 x 1-5 ) moles / 5.mlseawater =.578 +/-.11M 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) =.75V (see excel sheets, Part II: 1 st 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 1-3 (+/- 5.3 x 1-5 ) moles X -. In 15.ml of.1189m AgNO3, there are x 1-3 moles. Assuming full reactivity of the Ag + with the X - first, the number of moles of X - first remaining in solution = 2.89 x 1-3 (+/- 5.3 x 1-5 ) moles x 1-3 moles= x 1-3 (+/- 5.3 x 1-5 ) moles Therefore, the [X - ] = x 1-3 (+/- 5.3 x 1-5 ) moles / (5.mlsea-water + 15.mlsilver nitrate) =.681 +/-.27M Analysis continued after the following plots

12 1st Derivative (d-voltage / d-volume) Voltage Measured.6 Titration Plot of Actual Voltage Measured: Preliminary Titration of Sea Water Sample Silver Nitrate (ml) Preliminary Titration of Sea Water Sample Silver Nitrate (ml)

13 1st Derivative (d-voltage / d-volume) Voltage Measured Titration Plot of Actual Voltage Measured: Accurate Titrations of Sea Water Sample Silver Nitrate (ml) 1st 2nd 3rd Accurate Titrations of Sea Water Sample 1st 2nd 3rd Silver Nitrate (ml) Which gives us E (Ag/AgX) = E(cell) + (RT/F) ln[x - ] + Constant = (.75V) + [(8.314J/molK) (298.15K) / ( C/mol)] ln(.681) +.225V = (.75) + (-.69) + (.239V) =.244V Therefore, since this value is close to (considering experimental error), the E (Ag/AgCl) of.222v, we can conclude that there is.578 +/-.11M Chloride in the sea-water sample!

14 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.68 +/-.11M Iodide and.196 +/-.16M 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.ml 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.578 +/-.11M 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.5m Cl - is present in seawater!