CHM 510 Topic 1 Chromatographic Separations PDF

CHM 510 TOPIC 1 CHROMATOGRAPHIC SEPARATIONS 1  Course Learning Outcomes Students should be able to: 1.Understand the general concepts and principles of separation in chromatographic technique. 2.Understand the efficiency of separation: a) Resolution b) Plate The...

CHM 510 TOPIC 1 CHROMATOGRAPHIC SEPARATIONS 1  Course Learning Outcomes Students should be able to: 1.Understand the general concepts and principles of separation in chromatographic technique. 2.Understand the efficiency of separation: a) Resolution b) Plate Theory (H & N) c) Rate Theory (van Deemter Theory) 3.Apply chromatographic equations in solving problems. 4.Plot & interpret chromatogram. 2  CHROMATOGRAPHIC SEPARATION Methodsused to separate and/or to analyze mixtures of compounds. The components to be separated are distributed between two phases: a stationary phase bed and a mobile phase which permeates through the stationary bed. Separation of compounds depends on the rates at which the components moves through the stationary phase. 3  Definition of chromatography ❑Chromatography is a separation method wherein the components in a mixture are separated on column in a flowing system (mobile phase). ❑The component are distributed between two phases, - a stationary phase - a mobile phase 4 ❑ Stationary phase – remains inside the column (hold stationary phase) or coated on a solid surface (flat sheet-Thin Layer Chromatography) ❑ Mobile phase – the solvent moving through stationary phase, carrying with it the component mixtures. ✔ GC - carrier gas (He, N2) ✔ HPLC - liquid (H2O, organic solvent-ACN, MeOH) ✔ SFC - supercritical fluid (CO2) 5  Pack column TLC Stationary phase Stationary phase fixed on a (solid particles) in a solid surface/plate (thin layer column (HPLC) chromatography, TLC) 6  Stationary phase Cross section Capillary column GC oven 7  General principle of chromatography ❑ Components of a mixture are carried through the stationary phase by the flow of a mobile phase. ❑ Separation occur because of differing affinities of components with stationary phase & mobile phase. 8 ❑ Those components that are strongly retained by the stationary phase move slowly with the flow of mobile phase. ❑ Components that are weakly held by the stationary phase travel rapidly. ❑ The differences in migration rate will cause the sample components to separate. 9  The separation process (Start) Flow of Mobile Phase Injector Detector T=0 T=10’ T=20’ 10 Most interaction with stationary phase Least Migration of Solutes sample mobile phase A+ B B A packed column B A B A B t0 t1 t2 t3 t4 detector A B signal time  Chromatogram A record of a separation produced by a recorder or integrator based on the signal obtained from the detector. Response Time Seven compounds were separated ❑ Peak positions, tR is used to identify components; peak areas to determine amounts of each component/analyte. Example of report obtained from GC: Peak RetTime Type Witdth Area Height Area (min) (min) (pA*s) (pA) (%) B t1 BB 0.05 327.098 754.641 6.5 A t2 BB 0.04 218.341 167.492 4.3 14 C t3 BV 0.09 4483.544 99.808 89.2  Some useful terms ❑ Process in which components are washed through a stationary phase by the movement of a mobile phase is called ELUTION ❑ Mobile phase is described as ELUENT ❑ Component leaving column is called ELUATE ❑ Solid that is dissolved in liquid is called SOLUTE ❑ The chemical/substance that is determined in an analytical procedure is named ANALYTE 15  Migration rates of solutes The relative rates at which the 2 analytes are eluted are determined by the magnitude of the equilibrium constant by which the solute distribute themselves between stationary & mobile phase. For solute A, Amp Asp The equilibrium constant, Kc for the distribution of species A between mobile phase & stationary phase is called the distribution constant/partition coefficient. 16  PRINCIPLE OF CHROMATOGRAPHIC SEPARATION Different distribution of analytes between the mobile and stationary phase results in different migration rates (velocities) [A] stationary K= [A] mobile K = Partition coefficient  Cs Kc = Cm Kc is defined as the molar concentration of analyte in the stationary phase divided by the molar concentration of the analyte in the mobile phase. Large Kc means analyte prefer stationary phase, while small Kc shows analyte prefer in mobile phase. The Kc for each analyte must differ to achieve separation. Kc can be manipulated by appropriate choice of mp, sp or both. 18  Retention time, tR Retention time (tR ): The time between sample injection and an analyte peak reaching a detector at the end of the column Each analyte will have a different tR. tR = retention time tM = void time tS = adjusted retention time 19  Dead Time/void time ❑ A compound which not retained by the stationary phase will elute out of the column at time tM (or to), is called the void time/dead time. Thus, tM means… ✔ the time of a non-retained compound spends in the mobile phase, or spends in the column, or ✔ the time taken for the mobile phase to pass through the column. ❑ tM provides a measure of the average rate of migration of the mobile phase. ❑ Important in identifying analyte peaks. ❑ All components spend time tM in the mobile phase 20 ❑ Significance of tR in chromatography : For qualitative analysis, under the same chromatographic conditions, same compounds should have the same tR. Adjusted retention time, tRʹ ❑ tʹR is the time a compound spends in the stationary phase. tʹR = tR - tM 21 Average Linear Velocity (µ)/flow rate of mobile phase ❑ Average ‘speed’ of the carrier gas in cm/sec (cm of column traveled per second/min by a carrier gas. ❑ Influence the retention time & efficiency. Flow rate: How much mobile phase passed / minute (mL/min). Linear velocity: Distance passed by mobile phase per 1 min in the column (cm/min). 22 AVERAGE LINEAR RATE - Average Linear Velocity of analyte migration through the column Average linear rate of migration of a solute, ν L = length of column (cm) Average linear rate of movement of mobile phase molecules, u (velocity) L = length of column (cm) tM = retention time of un-retained peak (sec)  Retention/Capacity factor, kʹ ❑ kʹ is more useful measure of partitioning because the value is related elution time. ❑ Compound with larger Kc, will have larger kʹ & elute later. ❑ For effective separations, a column must have the capacity to retain samples & the ability to separate sample component efficiently. ❑ kʹ of a column is a direct measure of the strength of the interaction of the solute with the stationary phase. ❑ Also used to compare the migration rates of solutes in the column. 24 ❑ kʹ for analyte A : or tR-tM (time spend in the stationary phase) 25 ✔ The higher the kʹ of the column, the greater is its ability to retain solutes (Higher kʹ - more practical for complicated samples). ✔ Using a column with higher kʹ can improve the resolution (Lower kʹ - desired for simpler samples to save time). ✔ If k’ is too low, elution occur so rapidly, therefore, accurate determination of the tR is difficult. ✔ kʹ > 20-30 helps a good separation, but increased elution time (longer tR). ✔ Good separation, 2< kʹ >10 (balance between elution time & resolution). 26  Sample problem 1 If an analyte has a small capacity factor (small k’), what does it mean? Small capacity factor indicates that the analyte was less retained by stationary phase, hence shorter tR. 27  Relative migration rate : Selectivity factor, α ❑ α is a measure of the difference in tR between 2 peaks on the column. ❑ Indication of how well the compounds will separate. ❑ α always greater than 1. ❑ If α = 1, the 2 compounds cannot be separated. ❑ The higher the α, the more separation between 2 compounds or peaks. 28  Selectivity factor, α (Formula) (Retention factor) (Retention time) B is most retained 29 A is least retained  Efficiency of separation ❑ Two factors affect how well two components are separated : ✔ difference in tR between peaks (farther apart, better separation) ✔ peak widths (an efficient separation will produce narrow peaks, wide peak; poor separation) ❑ Solutes in a column spread into a Gaussian profile. 30 Detector response Efficiency of separation Gaussian peak shape : σ σ w1/2=2.35σ h 1/2h w=4σ t0 tr 31 time  Resolution, Rs ❑ The degree (how well) of separation of 2 peaks. ❑ Rs provides a quantitative measure of the ability of the column to separate 2 anaytes. ❑ Although α, describes the separation of peak centres & retention, it does not take into account column efficiency or peak widths (Wb). ❑ Rs is preferred over α since both retention (tR) & column 32 efficiency (W ) are considered in defining peak separation.  tR2 Δt tR1 Detector response Time wb1 wb2 2Δt 2[tR2 - tR1] Rs = = wb1 + wb2 wb1 + wb2 Δt = Difference between tR of two peaks = tR2 - tR1  Overlap of 2 peaks with different degrees of Rs Rs=0.50 Rs=0.75 t 2 tim t 3 ti 0 σ e 0R σ me Rs=1.00 s =1.50 Rs≥1 Complete is good separation t tim 4 t 6 tim σe σ e 0 0 Higher Rs, better separation 34  Rs=1.50 t tim 0 e baseline separation / complete separation of two neighboring solutes ideal case. Rs = 1.0 is considered adequate separation (for quantitative analysis) Rs < 1, compounds are overlapped or not separated (can’t be used for quantitative analysis) 35  Sample problem 1 Ethanol & methanol are separated in a GC column with t of 370 s R & 385 s and a base widths of 16.0 s & 17.0 s, respectively. i. Calculate the Rs. ii. Sketch the chromatogram. iii. Is the separation acceptable for quantitative analysis? Explain. iv. Comment on the efficiency of separation. 2[tR2 - tR1] Rs = wb1 + wb2 Rs = 2(385-370) = 0.91 17.0+16.0 (Rs Less than 1.0, therefore not acceptable for quantitative analysis) 36  Sample problem 2 The separation of 4 compounds gave Rs values of 0.5, 1.8 & 10.5. Sketch the chromatogram & comment on the efficiency of separation. Peak 1 & 2, Rs 1.5, separated well/baseline separation. Peak 3 & 4, Rs too large, tR very long, may cause band broadening. 37  Column efficiency Column efficiency is related to : ❑ a solute’s peak width. An efficient column will produce narrow peaks ❑ various kinetic processes that are involved in solute retention & transport in the column. 2 theories to describe column efficiency: 1.Plate Theory - to measure column performance & efficiency 2.Rate Theory - measure the contribution to band 38 broadening The selectivity factor (α) is important to determining how compounds separate, but it is not enough by itself. Figure 1 shows portions of two different chromatograms where the α is the same (the peaks are the same distance apart), but in Fig 1a, the peaks are separated and in Fig. 1b the peaks are not separated. (a) (b) Figure 1: Illustration of the separation of 2 pairs of peaks with the same α value. a) peaks are fully separated because they are narrow. b) peaks are not separated because they are wide. 39  The important difference between Fig. 1a and Fig. 1b is the width of the peaks. If peaks are too wide, they won’t separate well. Therefore, we need a measure of peak width also known as efficiency. The term that is generally used to describe column efficiency is “number of theoretical plates” or N. Chromatography columns with high numbers of N produce very narrow peaks resulting in better separation. Efficiency (N) can be increased by: ⚫using longer columns, or ⚫using columns with small plate height (H). 40  A Quantitative Measures of Column efficiency 1. Theoretical plates, N Plate Theory 2. Plate height (H) or Height Equivalent to a Theoretical Plate (HETP) Column efficiency increases with N ↑ & H ↓ H & N are useful to compare the performance/efficiency of different columns for a given analyte. 41  Theoretical plates, N Column efficiency is expressed by the number of theoretical plates (N) Column divided into segment called theoretical plates 42 ❑ Equilibration of the solute between the stationary & mobile phase occur in these “plates”. ❑ The analyte moves down the column by transfering of equilibrated between mobile phase & stationary phase from 1 plate to the next. ❑ Plates do not really exist; they are a figment of the imagination that helps to understand the processes at work in the column. ❑ N ~ few hundred to several hundred thousand. ❑ Columns with high N produce narrow peak, resulting in better separation of 2 compounds (higher column efficiency) than a column with a lower N. 43 (Tangen method) (Peak half-height method) 44  Sample problem 3 A separation of three compounds on two types of column of similar dimensions (250 mm x 4.6 mm x 10 µm) produced the following data: Compound Column A Column B Retention Peak width Retention time Peak width time (min) (min) (min) (min) Unretained 1.2 - 1.2 - X 3.8 0.30 4.2 0.52 Y 4.5 0.55 4.7 0.65 a) Compare the number of theoretical plates of the two columns (based on compound X). Which column is more efficient? Explain. NA = 2568 NB = 1044 45 Higher N = more efficient b) Which column is more suitable in separating compound X and Y? Verify your answer based on quantitative approach (with calculations). RA = 1.65 RB = 0.86 RA = 1.65. Baseline resolution between two peaks requires an Rs>1.5. Good separation for column A. RB = 0.86. Resolution is less than 1.0, therefore overlapping of peaks for column B. Column A better than column B. 46 ❑ N depends upon the length of the column. L N = HETP How to increase N ? 1.Using longer column (L) 2.Using column with small H 47 ❑ N’s relation to Rs : t’R2 k’2 K2 α = relative retention = = = t’R1 k’1 K1 ❑ N required to obtain a certain Rs : N N N2>N 1 2 1 t0 tim t tim e 0 e ❑ Rs can be improved by lengthening the column, thus, increasing the N.  Plate height (H) or Height Equivalent to a Theoretical Plate (HETP) ❑ H or HETP is approximately the length of column that corresponds to one theoretical plate. ❑ This term refers to the length of column in which the analyte equilibrates between the two phases. ❑ The more efficient the column is, the smaller the height of the plate will be, & the more equilibrations will occur in the length of the column.gt 50 ❑ Compare efficiencies of columns with different lengths: H or HETP = L/N ❑ An increase in efficiency as a decrease in the width of each sample peak, showing that the bands of sample have not spread much as they passed through the column. ❑ Therefore, H is used to measure of band spreading per unit length of the column. 51 The smaller the H, the narrower the peak width As HETP ↓, Rs increases (N ↑) ❑ Greater separation occurs with : ✔ greater number of N ✔ H or HETP becomes smaller Chromatographic relationships  Sample problem 5 A separation of 3 compounds on a column (25 mm x 4.6 mm x 10 µm) produced the following data. Peak tR (min) Peak width (min) Unretained 1.2 - 1 3.4 0.11 2 3.8 0.13 3 4.7 0.17 a)The average number of plates from the above data. b)The average plate height for the column. 54 a) N = 16 (tR/W)2 N1 = 16 (3.4/0.11)2 =15285 N2 = 16 (3.8/0.13)2 = 13671 N3 = 16 (4.7/0.17)2 = 12229 Naverage = N1 + N2 + N3 / 3 = 13728 b) H = L/N = 25 mm/13728 = 1.821 x 10-3 mm 55  Zone/band broadening ❑ Band broadening reflects a loss of column efficiency. ❑ The slower the rate of elution occurring while a solute migrate through a column, the broader the band. 56 There are 2 basic theories applicable to chromatography, the Plate Theory & the Rate Theory. The plate theory describes the mechanism of retention & gives an equation that allows the calculation of the column efficiency. The rate theory describes the process of peak dispersion (band spreading (narrow/broad)) & provides an equation that allows the calculation of the variance per unit length of a column (the H) in terms of the mobile phase velocity & other physical chemical properties of the solute and distribution system. 57  The Rate Theory of Chromatography A more realistic description of the processes at work inside a column takes account of the time taken for the solute to equilibrate between the stationary & mobile phase (unlike the plate model, which assumes that equilibration is infinitely fast). The resulting band shape of a chromatographic peak is therefore affected by the rate of elution. It is also affected by the different paths available to solute molecules as they travel between particles of stationary phase. 58  van Deemter equation for plate height (H) : H = A + B/v + Cv ❑ van Deemter equation tell how the column & flow rate affect the H ❑ The smaller the H, the narrower the bandwidth. ❑ H represents the band broadening. 59  H = A + B/v + Cv Where, A = Multiple paths / Eddy diffusion term B = Longitudinal diffusion term C = Mass transfer between phases / equilibration time between phases term v = flow rate or mobile phase velocity 60  A term - Multiple Paths (or Eddy Diffusion) Is a band broadening due to molecule takes a different path through the packed column ❑ As a solute molecules passes through the column, it will take different paths around the stationary phase particles at random. ❑ Each of the path will be of a different length, so that as solute molecules follow different paths, they will exit the column at different times. ❑ This will cause broadening of the solute band.  A term - Multiple Paths tim e Some paths are longer than others, molecules entering the column at the same time (left) but eluted out at different times (right).  A term- Multiple paths A = 2λdp ❑ A is directly proportional to the packing particle diameter, dp. ❑ λ is a function of packing uniformity and column geometry (same shape). ❑ In packed columns, A ≠ 0, in capillary column, A =0. ❑ A is small when using smaller particles size & uniform size. ❑ Smaller dp will reduce the differences in the path lengths. 64  B term - Longitudinal diffusion Diffusion is a process in which analyte migrate from a more concentrated to a more dilute region Longitudinal diffusion in column chromatography is a band broadening process in which solute diffuse from the concentrated center of zone to more dilute zone in front & behind zone; inversely proportional to mobile phase flow rate.  B term - Longitudinal diffusion High Low High Low Concentration Conc. Conc. Conc. Low Concentration Component Molecule Concentration ❑ Solute diffuses out from the center to the edges. ❑ Solute [ ] is lower at the edges of a band. ❑ This causes band broadening. B term - Longitudinal diffusion S a r t t Time Time Time #1 #2 #3  B term - Longitudinal diffusion B/v = 2γDM/v ❑ The longitudinal diffusion effect on H is inversely proportional to v because the solute spends less time in the column at high/faster v & less diffusional broadening occurs.  B term - Longitudinal diffusion B/v = 2γDM/v ❑ The obstructive factor, γ shows that longitudinal diffusion is hindered by the packing. ❑ γ is lower for a packed column (ie γ = 0.6) than an unpacked (capillary) column (ie γ = 1). 69  B term - Longitudinal diffusion B/v = 2γDM/v ❑ B is directly proportional to the analyte diffusion coefficient in the mobile phase, DM. The faster the rate of diffusion of the analyte, the greater the extent of longitudinal diffusion. ❑ DM is smaller in liquids than in gases. B term is common source of band broadening in GC, where the rate at which molecules diffuse is 104 higher than in HPLC. So, B/v is less pronounced in LC than in GC. 70  Effect of mobile phase flow rate on H for GC and LC VGC is larger than VLC GC faster analysis time than HPLC H for HPLC smaller than H for GC HPLC more efficient (B term) Column length (L) GC longer column (more than 50m) than HPLC (~25cm). Larger N (more efficient) 71 C term-Mass transfer between phases The term Cv comes from the finite (or a certain amount) time required for solute to equilibrate between mobile & stationary phase. 72  Carrier Packed Column Gas Liquid Phase Capillary Column Carrier Gas Liquid Phase Column Wall The equilibrium between the mobile & stationary phase is established so slowly that a chromatographic column always operates under non-equilibrium conditions. Mobil e slow phase equil. Stationar y bandwidt bandwidt phase h h So, analyte molecules at the front of a band are swept ahead without equilibrating with the stationary phase & those at the trailing edge are left behind for a longer time in the stationary phase  C term-Mass transfer between phases ❑ If the velocity of the mobile phase is high & the molecule has a strong affinity for the stationary phase, then the molecule in the mobile phase will move ahead of the molecule in the stationary phase. ❑ The band of analyte is broadened. ❑ The higher the velocity of mobile phase, the worse the broadening becomes.  Mass transfer coefficient (Cs & CM) Hmass transfer = Cv = (Cs + CM)v ❑ CMv is the mass transfer in mobile phase. ❑ CSv is the mass transfer in stationary phase. 76  The mass-transfer effect on H is directly proportional to v because ✔ the solute residence time is longer at low v, the deviation from equilibrium is less & band broadening or H is smaller. 77  Mass transfer in stationary phase (Csv) 2 Csv = (fS(k’)df /DS)v df = stationary phase film thickness (most important factor) DS = diffusion coefficient of the solute in the film/stationary phase A complex function of fs(k’) of the retention factor k’ Csv is less if the df is smaller, or the solute DS is larger 78 ❑ Decreasing df, reduces H & increase efficiency because solute can diffuse faster from the farthest depth of the stationary phase into the mobile phase. ❑ E.g., ✔ With thick films, molecules must on the average travel farther to reach the surface; with smaller Ds, they travel more slowly. Therefore, require more time to reach equilibrium, increase C, increase band broadening. ✔ The consequence of both factors lower rate of mass transfer & an increase in H. 79  The mobile phase mass transfer (CMv) CMv = (fM(k’)dp2/DM)v dp = particle diameter of packing DM = diffusion coefficient of solute in mobile phase CMv is less if dp is smaller (hence greater surface area) or DM is larger Small particles (dp) reduce the distance solute must diffuse in the mobile phase ❑ For packed column, CM α dp2. 80 ❑ 2 For capillary column, CM α dc (column diameter). 81  Note: Both longitudinal & mass transfer broadening depend upon the rate of diffusion of analyte molecules but the direction of diffusion in the 2 cases is different. ❑Longitudinal broadening arises from the tendency of molecules to move in directions that tend to parallel the flow. ❑Mass transfer broadening occurs from diffusion that tends to be at right angles to the flow. 82  As a consequence, ✔The extend of longitudinal broadening is inversely related to flow rate. ✔For mass transfer, the faster the mobile phase moves, the less time there is for equilibrium to be approached. 83  van Deemter Plot ❑ Determine the optimum mobile phase flow rate to obtain the minimum H (maximum efficiency). H H = A + B/v + Cv B Cv The contribution to v the overall H of the Hmin column by each of these factors can A be plotted against flow velocity v vopt ❑ The minimum in the curve indicates the flow velocity which will give the maximum efficiency in that particular column. ❑ Using a carrier gas flow which is higher than the optimum will cause some loss of efficiency. ❑ However, this may be a satisfactory compromise, since it will also reduce the tR, & so may shorten the time required for an analysis. ❑ Using a flow slower than the optimum is, however, never a good idea, since it lengthens the analysis time & also causes a loss in efficiency. 85  Why optimization of flow rate/linear velocity of the mobile phase is one of the important steps in chromatographic analysis? ❑For B/v : Solute molecules diffuse to regions of lower solute concentration in front of & behind zone; inversely proportional to v. High flow rate is preferred so that solute residence time is shorter & the extent of diffusion is less. ❑But for Cv : The equilibrium between the mp & sp is established so slowly that a chromatographic column always operate under non-equilibrium conditions thus increasing Cv. Low flow rate is required to give the molecules enough time to reach equilibrium. ❑Thus, an optimum flow rate is high enough to reduce diffusion & low enough to give enough time for achieving equilibrium. 86 Decrease in flow rate will increase diffusion, H will increase Increase in flow rate – not enough time for equilibrium – inefficient mass transfer, H increases 87  Sample problem 6 Which term in van Deemter equation is most important at low flow rates? Why? B term-longitudinal diffusion. The lower the flow rates, the longer the analytes in the column, thus more time for the analytes to diffuse & cause band broadening. 88  Sample problem 7a Specify what would be the effect (increase, decrease or no change) of the change on the H. Explain. a) Change the particle size from 3µm to 5 µm (HPLC). Increasing the particle size of the packing : increase H. -A, multipath, bigger diameter will increase the differences between path, increases multipath, thus increase H. Increasing band broadening. -Cv, mass transfer coefficient, bigger diameter will reduce the surface area, equilibrium will be slower, increase C thus increase H. Increasing band broadening. 89 Effect of particles size 90  Sample problem 7b b) Increase thickness of liquid stationary phase (film thickness). Increase film thickness : Increase H Cv, mass transfer coefficient, increase film thickness will require more time for equilibrium thus increasing band broadening. 91  Effect of column ID ID (mm) 000 0.6 40, 0.53 n= 0,000 0.32 n=8 H (mm) 0.4 ,000 n = 125 0.25 0.2 20 40 60 Average Linear Velocity (ū cm/s)  Sample problem 4a Substances A and B were found to have tR of 6.4 &14.4 min, respectively, on a 22.6 cm column. An unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the: (a) column resolution 2((tR)B - (tR)A) 2(14.4 - 6.4) Rs = ----------------- = ---------------- = 10.5 WB + WA (0.45 + 1.07)  Sample problem 4b Substances A and B were found to have tR of 6.4 & 14.4 min, respectively, on a 22.6 cm column. An unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the: (b) the average no. of plates in the column N = 16 *(tR/W)2 For component A NA = 16 * (6.4/0.45)2 = 3.2 x 103 plates For component B NB = 16 * (14.4/1.07)2 = 2.9 x 103 plates Naverage = NA + NB / 2 = 3.05 x 103 plates  Sample problem 4c Substances A and B were found to have tR of 6.4 & 14.4 min, respectively, on a 22.6 cm column. An unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the: (c) the plate height for B H = L/N H = L/NB = (22.6 cm)/(2.9 x 103 plates) = 7.8 x 10-3 cm/plate THE END 97