Experiment 13. QUANTITATIVE DETERMINATION OF TOTAL ION CONCENTRATION BY ION EXCHANGE E XCHANGE CHROMATOGRAPHY CHROMATOGRAPHY
Chem 26.1 WFQR1 Mandaing, Shejay G. 15 March 2013
Introduction Chromatographic techniques have been valuable in the separation and analysis of highly complex mixtures and revolutionized the capabilities of analytical chemistry. These techniques involves the physical method of separation in which the components to be separated are distributed between two phases, one of which is stationary (stationary phase), and the other which moves in a definite direction (mobile phase). Ion exchange chromatography is one of these techniques, which uses an ion exchange resin as stationary phase (Christian, 2004) Cation-exchange resins contain acidic groups, while anion-exchange resins contain basic groups (Skoog, 2004). In this experiment, the total ion concentration of Cu (II) from a provided sample was determined using cation ion exchange column chromatography. A wad of absorbent cotton was placed at the bottom of a clean ion-exchange column (in this case a burette was used as the column). A prepared Dowex 50 cation exchange resin was gently poured to the column while suspended in the concentrated acid solution until ¼ of the column was filled up. The stopper was opened to allow some liquid to flow out and the excess acid was washed out of the column with distilled water until the pH of the eluate was equal to the pH of the distilled water being used. This was checked using pH paper. To determine the total cation concentration of the unknown sample, a 16 mL aliquot of the sample was diluted to 50 mL. 10 mL of this working solution was poured into the column and an Erlenmeyer flask was placed under the column to collect the eluate. The stopcock was opened until the rate of the flow of the eluate is about 30 drops per minute. The column was washed with distilled water until the pH of a drop of the eluate is equal to the pH of the distilled water being used. The washings and the eluate were combined in the same Erlenmeyer flask, which was then titrated with the standardized sodium hydroxide (NaOH) solution, using phenolphthalein as indicator (ACAG, 2012).
Results and Discussion Table 1 below shows the data obtained from the standardization of NaOH. From these data obtained, the average concentration of NaOH was calculated from the stoichiometric reaction of NaOH with KHP as follows:
NaOH + KHP NaKP + H2O
(1)
Table 1. Standardization of Sodium Hydroxide
Trial
1
2
3
Primary standard weight, g
0.1 g
0.0999
0.1026
Final volume of NaOH, mL
3
6
9.2
Initial volume of NaOH, mL
0
3.1
6.1
Net volume of NaOH, mL
3
2.9
3.1
M NaOH
0.1624
0.1679
0.1613
Average M NaOH
0.1639
This average concentration was then used to calculate the equivalent moles of H+ from the volume of NaOH used in titration througt the stoichiometric reaction:
2OH- + 2H+
2H2O
(2)
Now, the equivalent moles of Cu2+ can be calculated from the number of moles of H+ following the equation 2rSO3 – H+ + Cu2+ (rSO3)2Cu + 2H+
(3)
which shows that for every 1 mole of Cu2+, 2 moles of H+ are displaced.
Table 2 gives us the computed confidence interval and RSD for the data results. The calculated experimental Cu(II) concentration of value of experiment.
ppm showed a deviation from the theoretical
ppm of Cu(II). This gives us a percentage error of
% for the
Table 2. Determination of the total cation concentration
Trial 1
Trial 2
Trial 3
Volume of sample, mL
50
50
50
Net volume of NaOH, mL
2.175
2.075
2.2
[H+], mol
3.56 x 10^-4,
3.401 x 10^-4,
3.6058 x 10^-4,
[Cu2+] , mol
1.78 x 10^-4
1.700 x 10^-4
1.8029 x 10^-4
Average [Cu2+] , mol
1.7953 x 10^-4,
RSD Confidence Interval
Answers to Questions 1. Discuss the basic principles of ion-exchange chromatography. All chromatographic techniques are based on an equilibrium between stationary and mobile phase. In ion exchange chromatography, a small volume of sample is placed at the top of the column which is filled with chromatographic particles (stationary phase) and solvent as it is moved along the column(Christian, 2004). Here, the stationary phase is a porous, essentially insoluble solid that is exchanged for ions in the sample solution that is brought into contact with the solid stationary phase, usually an ion exchange resin. This synthetic ion-exchange resin is a high-molecular-weight polymer that contains a large numbers of an ionic functional group per molecule. A Cation exchange resin contains acidic groups, while anion exchange resins have basic groups (Skoog, 20040). Anions such as —SO3- or cations such as —N(CH3)# 3 are covalently attached to the resin and solute ions of the opposite charge are attracted to the stationary phase (Harris, 2010). In cation exchange, the resin contains acidic functional groups added to its aromatic ring. The strong-acid cation excahngers have sulfonic acid groups-SO 3H, which are strong acids much like sulphuric acid. The protons on these groups can exchange with other cations (Christia, 2004). The exchange reaction that is followed by the cation exchange is given by: xRSO-3H+ + Mx+
(RSO-3)xMx+ + xH+
(4)
(Skoog, 2004) where Mx+ represents the cation, R represents that part of a resin molecule that contains one sulfonic acid group, and x is the number of moles of ion or charge equivalent (Skoog, 2004). As
the components of the sample solution are carried through the stationary phase by the flow of a mobile phase, the separations are based on differences in migration rates among the mobile phase components (Skoog, 2004). Elution is process by which the eluent (the solvent used to carry the components of the sample)enters the column and is washed through a stationary phase(fixed in place in the column) by the movement of a mobile phase(the one that moves over or through stationary phase, carrying with it the analyte). The mobile phase that exits the column is called the eluate. The column consists of a narrow
tubing packed with a finely
divided inert solid that holds the stationary phase on its surface. The mobile phase occupies the open spaces between the particles of the packing. Initially, a solution of the sample containing A and B in the mobile phase is introduced at the head of the column. At the start, the two components distribute themselves between the mobile and stationary phase. Elution then occurs by forcing the sample components through the column by continuously adding more of the fresh mobile phase. Further addition of solvent, can carry solute molecules down the column in a continuous series of transfers between the two phases. Because solute movement can occur only in the mobile phase, the average rate at which the solute migrates depends on the fraction of time it spends on that phase.
Solutes that are strongly retained by the
stationary phase have slow rate and fast when its retention in the mobile phase is more likely. Ideally, the resulting differences in rates causes the components in a mixture to separate in to bands or zones along the length of the column. Isolation of the separated species is then accomplished by passing a sufficient quantity of mobile phase through the column to cause the individual bands to pass out the end
(to be eluted from the column), where they can be
collected or detected (Skoog, 2004).
2. What are the factors that can affect ion-exchange? Length and width of the column. Band broadening is inevitable while addition of more mobile phase in the column takes place. However, this can be lessened if the column width is kept at a shorter radius. While, separation of bands or zones will be better if the column length is kept long. The relative rates at which the two species are eluted, which is determine by the ratio of the solute concentration in each of the two phases (stationary and mobile) also affects the ionexchange since the two species can be better separated if the differences between their rates is quite appreciable.
The retention time in both phases also affects ion exchange because this affects the elution rate. If one ion has more retention time in the stationary phase, then it is eluted more slowly and vice versa, while the other will be eluted more rapidly while travelling down the column.
The selectivity of the resin to the ions also affects ion exchange because this as well contributes to retention time in the stationary phase.
A proper balance of intermolecular forces among the analyte, mobile phase, and stationary phase I also required for a success in the separation process. These IMF are described qualitatively in terms of the relative polarity possessed by each of the 3 components. As a rule, most chromatographic separations are achieved by matching the polarity of the analyte to that of the stationary phase, a mobile phase of different polrity is then used. This procedure is generally more successful than in the one wherein the polarities of the analyte and the mobile phase are matched but are different from that of the stationary phase.
Because here, the
stationary phase, often cannot compete successfully for the sample components and retention times then become too short. But if the polarity of analyte and stationary phase are too much alike, retention time become extraordinarily long (Skoog, 2004)
The distribution equilibrium is described by the distribution constant
Kc= [Xs]/[Xm]
(5)
where Xs is the solute concentration in the stationary phase and Xm I the solute concentration in the mobile phase. Kc is governed by the temperature, type of compound, and the stationary and mobile phase. Solutes with large Kc will be retained more strongly by stationary phase than those with small value. The result is that the latter will move along the column (be eluted) more rapidly. Because true equilibrium between the two phases is not truly achieved, there will be some lag of the analyte molecules between the two phases which depends on the flow rate of the mobile phase and on the degree of interaction with the stationary phase and results in band broadening(Christian, 2004).
Lastly, the greater the cross linkage of the resin , the greater the difference in selectivities. Crosslinkage increases the rigidity of the resin, reduces swelling, reduces porocity, and reduces the solubility of the resin.(Christian, 2004)
3. Why was resin soaked in water for an hour before introducing to the ion-exchange column? Why can’t a dry resin be used?
The resin was soaked in water to lessen its concentration and to allow more solute from the concentrated acid to later enter its pores. We should note that because resins are made of polymers with high molecular weight, it would not dissolve in water. Dry resin cannot be used because formation of air pockets will cause an uneven flow and poor efficiency of ion-exchange. Also, in order for the ion-exchange reaction to occur, the ions must be subjected to an aqueous environment.
4. Why was the water level kept above the top of the resin? The liquid level of the ion-exchange was kept above the resin level to prevent air pockets to form and remain inside the column. The air pockets may react with the resins forming altered canals. These canals can lead to a low column capacity and adsorptive power. It may also hinder in the interaction of the solution with the resin since it may prevent its contact with the resin. The resin must be hydrated at all times because water serves as a very good medium for the reaction to take place and water also displaces entrapped gases.
5. Why was a strong acid added to the column during a) preparation of the resin and b) storage of the resin? The resin was soaked in concentrated acid in order to regenerate it with H+ ions needed to facilitate the ion-exchange. A strong acid added prolongs the activity of the cation exchanger. The resins are also perforated with very small pores for liquid pathways to increase surface area of adsorption. As the resins react with the acid, water forms and swells the beads making it a suitable medium for reaction. The addition of acid shows that the resin is activated and the H+ are now ready to be exchanged with another cation species.
6. Why was the rate flow maintained at 30 drops per minute?
To allow the completion of the replacement of H+ ions of the resin, the solution should be given enough time to establish an exchange equilibrium and to come in contact with all portions of the resin. Because of this, the rate of the expelling the solution which have already come in contact with the resin, should be kept at a certain slow flow rate. This constant flow rate is crucial for the ion exchange to take place. Water is the medium for reaction and a rapid flow will never assure the optimum completion of the reaction and the total liberation of H+. A modest rate of flow will ensure that at the end of the column, equilibrium is ― achieved‖ and that the flow of the eluate is constant and no pressure or disturbance is observed.
7. Why was the column washed until the eluate has the pH equal to the pH of distilled water during a) preparation of column and b) sample analysis? a) The solution was kept at a pH equal to that of the distilled water to be used later on to make sure that the H+ concentration is not affected with the constant addition of water. B) The same pH means that the cation exchanger has fully reacted with the cation analyte. It means that all H+ have been displaced and are now ready for titration with strong base.
8. Give the balanced exchange reaction. Replacing the general exchange reaction given in equation 1 by the particular species used in the experiment, the following reaction occurs 2rSO3 – H+ + Cu2+ (rSO3)2Cu + 2H+
(3)
Which tells us that 2 mol H+= 1 mol Cu2+
(6)
9. What are the possible sources of errors and their effect on the calculated parameters? Rationalize. If the flow rate of the elution (30 drops per minute) was not followed, the solution is not given enough time for the replacement of H+ ions to be completed. The completion of the reaction and the total liberation of H+ ions will not be optimized.
If the liquid level fell below the resin level, air pockets might form altered canals which in turn would interfere with the contact of the ions with the resins. This would cause a low efficiency of
adsorption. This will affect the time for the pH of the eluate to be equal to that of the distilled water being used.
Overtitration of the sample would cause an increase in the volume of NaOH used, and an increase in the calculated experimental value of the Cu(II) ion concentration. Also, this would cause an increase in the percentage error
an deviation from the theoretical Cu(II) ion
concentration.
If there was a cotton wad or any other absorbing impurities with the column, above the resin, same as the air pockets, it would interfere with the contact of the resin to the analyte and this may absorb either of the ions. This would lead to a lower concentration of the calculated experimental ion concentration.
If the pouring of water was not gentle that the resin was not tact or was disturbed, the purpose of the resin acting a sift would be depleted. This would lead to a lower efficiency of the column and disturbance in the exchange between the cations.
If the pH were not yet equal and titration was already performed, this would lead to a lower computed value for the Cu(II) concentration since nit all H+ ions have been liberated yet.
Wrong molarity of NaOH. If the molarity of NaOH is too high, less of it will be used for titration and vice versa. The computed value for Cu(II) ion concentration may increase or decrease if the molarity of NaOH is too high or too low.
REFERENCES: Analytical Chemistry Academic Group. (2012). Analytical chemistry laboratory manual . Quezon City: Institute of Chemistry, University of the Philippines, Diliman. Christian, G.D. (2004). Analytical chemistry (6th ed.). Somerset, NJ: John Wiley & Sons, Inc. Harris, D.C. (2010). Quantitative chemical analysis (8th ed.). New York, NY: W.H. Freeman and Company. Skoog, D.A., West, D.M., Holler, F.J., & Crouch, S.R. (2004). Fundamentals of analytical chemistry (8th ed.). Belmont, CA: Thomson-Brooks/Cole
