Chapter 15 Liquid Chromatography Problem 15.1: Albuterol (a drug used to fight asthma) has a lipid:water KOW (see “Profile—Other Applications of Partition Coefficients”) value of 0.019. How many grams of albuterol would remain in the aqueous phase if 0.001 grams of albuterol initially in a 100 mL aqueous solution were allowed to come into equilibrium with 100 mL of octanol? C (aq) ⇌ C (org) Initial (g) 0.001 g 0 Change -y +y Equil 0.001-y y 푦 100 푚퐿 퐾푂푊 = 0.019 = (0.001−푦) ⁄100푚퐿 1.9 x 10-5 – 0.019y = y 1.019y = 1.9 x 10-5 y = 1.864x10-5 g albuterol would be in the octanol layer Problem 15.2: (a) Repeat Problem 15.1, but instead of extracting the albuterol with 100 mL of octanol, extract the albuterol solution five successive times with only 20 mL of octanol per extraction. After each step, use the albuterol remaining in the aqueous phase from the previous step as the starting quantity for the next step. (Note: the total volume of octanol is the same in both extractions). (b) Compare the final concentration of aqueous albuterol to that obtained in Problem 15.1. Discuss how the extraction results changed by conducting five smaller extractions instead of one large one. What is the downside to the multiple smaller volume method in this exercise compared to the single larger volume method in Problem 15.1? 푦 (a) Each step will have this general form of calculation: 퐾 = 20 푚퐿 푂푊 (푔푖푛푖푡−푦) ⁄100푚퐿 (퐾 )(푔 ) so we can calculate the g transferred to octanol from 푦 = 푂푊 푖푛푖푡 5+퐾푂푊 We can use a spreadsheet to calculate the successive steps: First extraction: KOW= 0.019 Init (g) 0.001 g in Octanol: 3.78561E-06 g left in aq 0.000996214 Total albuteral transferred: 1.87853E-05 Second Extraction: Init (g) 0.000996214 g in Octanol: 3.77128E-06 g left in aq 0.000992443 Third Extraction: Init (g) 0.000992443 g in Octanol: 3.75701E-06 g left in aq 0.000988686 Fourth Extraction: Init (g) 0.000988686 g in Octanol: 3.74278E-06 g left in aq 0.000984943 Fifth Extraction: Init (g) 0.000984943 g in Octanol: 3.72862E-06 g left in aq 0.000981215 (b) In Prob 15.1, the final concentration of albuterol in the water layer would be (1-1.864 x 10-2)mg/0.1L = 9.8136 mg/L In part (a) above, the final concentration of albuterol in the aqueous layer is (1-1.878 x 10-2)mg/0.1L = 9.8122 mg/L In this particular case, the successive extractions only yielded slightly higher extraction of the albuterol, probably because the KOW is so low. In general, we would expect a higher net transfer, but the risk of doing multiple extractions is that errors associated with each step are propagated multiple times. Problem 15.3: Albuterol has a lipid:water KOW value of 0.019. If you were to conduct a chromatographic separation using a polar solvent (i.e., methanol) and a nonpolar stationary phase, would the albuterol’s distribution favor the mobile phase or the stationary phase? Explain your answer. [퐶]표푟푔 We know that 퐾퐷 = 퐾푂푊 = . A large KOW value would indicate a preferential affinity for [퐶]푎푞 an organic or nonpolar phase vs. an aqueous or polar phase. The very low value of KOW indicates that the albuterol would have a higher affinity for the polar solvent than the nonpolar stationary phase. Problem 15.4: Caffeine has a lipid:water KOW value of 0.79. If a mixture of albuterol and caffeine were to be separated using a polar solvent (i.e., methanol) and a nonpolar stationary phase, which component would elute first (having the shorter retention time)? Would the two components elute close to each other—or far apart? Explain. Caffeine (KOW=0.79) has a roughly 40 times greater affinity for the nonpolar stationary phase than does albuterol (KOW = 0.019). The albuterol will have a much shorter retention time than caffeine – they would not be expected to elute close to each other at all. Problem 15.5: Sometimes it is desirable to determine the retention (R) of an analyte as the difference between the observed retention time (tr) and the void time (tm). Using the chromatogram in Figure 15.4, determine the retention (R) of compounds A and B. Compound A: R = tr – tm = 6.6-1.6 = 5.0 min Compound B: R = tr – tm = 10.4 – 1.6 = 8.8 min Problem 15.6: Sometimes you will see chromatographic peaks identified by the volume of mobile phase that was required to elute the sample instead of the time it took for the sample to elute. Using the retention times seen in Figure 15.4, calculate the retention volumes of compounds A and B assuming a flow rate of 1 mL/min. Repeat this exercise assuming a flow rate of 0.7 mL/min. Flow rate of 1 mL/min: Compound A: Vr = tr x rate = (6.6 min)x(1 mL/min) = 6.6 mL Compound B: Vr = 10.4 min Flow rate of 0.7 mL/min: Compound A = (6.6 min)(0.7 mL/min) = 4.6 mL Compound B = 7.3 mL Problem 15.7: Determine the retention factors (kr values) for analytes A and B seen in Figure 15.4. tr−tm 6.6 min − 1.6 min Compound A: kr = = = 3.125 = 3.1 tm 1.6 min tr−tm 10.4 min − 1.6 min Compound B: kr = = = 5.5 tm 1.6 min Problem 15.8: Calculate the retention factor for each peak in the chromatograph seen in Example 15.3 and create a results table that might be used in a report. tr **Note: Met-Enk is labeled as having a tr=10.1, but it Substance (min) kr clearly comes before the 10-minute mark, estimated at t 2.1 --- m around 8.7 min. Gly-Tyr 3.3 0.57 Val-Tyr-Val 6.7 2.19 Met-Enk 10.1 3.81 Met-Enk (actual) 8.7 3.14 Problem 15.9: Use the chromatogram from Example 15.3 to determine the retention times and retention volumes of each of the peaks in this sample and create a results table that might be included in a report. tr Vr Substance (min) (mL) **Note: Met-Enk is labeled as having a tr=10.1, but it t 2.1 --- m clearly comes before the 10-minute mark, estimated at Gly-Tyr 3.3 3.30 around 8.7 min. Val-Tyr-Val 6.7 6.70 Met-Enk 10.1 10.10 Met-Enk (actual) 8.7 8.70 Problem 15.10: Use the chromatograms in panels (B) and (C) from Example 15.4 and estimate the resolution of the two-component system under each of these conditions. What challenges do you face for estimating σ for panel (B)? In terms of the value of the separation, discuss the significance of the resolution in panels (A), (B), and (C) with respect to the speed of the analysis (throughput), quantitative analysis, and qualitative analysis. *** Hint – scan or photo the image and then bring it up in Paint – you can use the pixel index (lower left) for each position and use interpolation to calculate the time of each point. Panel B: Wh,1 = 1.33 min Wh,1 = 2.355σ1 σ1 = 0.5656 Wb1 = 4σ = 2.262 Wh,2 = 1.35 min Wh,2 = 2.355σ1 σ2 = 0.5732 Wb2 = 4σ = 2.293 tr2−tr1 3.5−2 Rs = = = 0.66 (Wb2+Wb1)/2 (2.293+2.262)/2 The challenging thing about Panel B, and the thing that introduces error, is the need to mentally extend the peaks beyond the dip that lies between the two peaks Our measurement of width at half-height is based on an estimate of where the sides of the peaks are. Panel C: Wh,1 = 1.24 min Wh,1 = 2.355σ1 σ1 = 0.5265 Wb1 = 4σ = 2.106 Wh,2 = 1.25 min Wh,2 = 2.355σ1 σ2 = 0.5307 Wb2 = 4σ = 2.123 tr2−tr1 4.75−2.0 Rs = = = 1.3 (Wb2+Wb1)/2 (2.123+2.106)/2 The resolution increases from B to A to C. However, in order to improve resolution, we need to extend the run time, which will decrease throughput. Usually this will increase peak width. Depending upon your analytical needs, this might be an acceptable trade-off. Problem 15.11: In a chromatographic experiment, two components elute with retention times of 11.52 and 12.76 minutes, respectively. The half-height widths of the peaks were determined to be 0.37 and 0.42 minutes, respectively. Assuming Gaussian peak shapes, estimate the resolution between the two peaks. tr2−tr1 12.76−11.52 Rs = = = 3.1 (Wb2+Wb1)/2 (0.37+0.42)/2 Problem 15.12: Use Equation 15.15 to determine Nt for each peak in the chromatogram found in Example 15.4, panel (C). Using values found in Problem 15.10: 2 2 2 2 tr 2.0 tr 4.75 Nt = 16 ( ) = 16 ( ) = 14.4 Nt = 16 ( ) = 16 ( ) = 80.1 Wb 2.106 Wb 2.123 Problem 15.13: Use Equation 15.16 to determine Nt for each peak in the chromatogram found in Example 15.4, panel (A). 2 2 2 2 tr 2.0 tr 4.14 Nt = 5.54 ( ) = 5.54 ( ) = 13.1 Nt = 5.54 ( ) = 5.54 ( ) = 65.9 Wh 1.3 Wh 1.2 Problem 15.14: Relate the multistep extraction process described in Problem 15.2 to the concept of theoretical plates.