Effects of Particle Size Distribution and Sampling on the
Determination of Chloride Content of a Binary Mixture
Introduction
Generally, any pure crystalline substance is composed of microscopic particles with a range of sizes. For example, small quantities of pure potassium chloride or sodium chloride may be composed of millions of micro-particles ranging in size from perhaps 50
mm to 500 mm (one mm = one "micron" = 10-4 cm). The purpose of this experiment is to determine the percentage chloride in a large chemical supply composed of a mixture of KCl and NaCl, where the two compounds have different ranges of particle size. The effect of sampling the mixture with preferential particle size selection for chemical analysis is demonstrated in this experiment by deliberately separating the original mixture (with sieves) into five different fractions differing in particle sizes as follows:Fraction Particle Size
A < 106
mmB 106 - 212
mmC 212 - 250
mmD 250 - 425
mmE > 425
mmConsequently, if the pure KCl and NaCl materials have different particle size distributions, each fraction obtained above contains a different percentage of KCl and NaCl, and each fraction will yield a different percentage chloride upon analysis.
Your unknown sample will come from one of the above particle size fractions. The objective of this experiment is to determine the percentage chloride and to compute the respective percentages of KCl and NaCl in your sample. Subsequently, the results of the entire class will be examined to determine the relationship between particle size and chloride percentage. The analytical technique to be used for chloride determination is direct potentiometry, measuring the potential of a Chloride Ion Selective Electrode (Cl-ISE) which is dependent on log[Cl-].
Special Equipment
Four sets of equipment are available. Each consists of a Combination type Chloride Ion Selective Electrode (Cl-ISE), (which includes a calomel reference electrode), and a millivolt (pH) meter.
Check out one each of the following: 100 ml beaker, powder funnel, 250.00 ml volumetric flask, 25.00 ml pipet, and 5.00 ml pipet.
Procedures
(1) Preparation of ISA Solution. Prepare 2 liters of 0.40M NaNO3 (no need for high accuracy) in a clean 2 liter bottle. (Filter this solution if undissolved foreign matter is observed.) Use this as the diluting solution for preparation of all standard and unknown chloride solutions in procedures (2) and (3) below. This solution will be referred to as the "ISA solution" (Ionic Strength Adjustment).
(2) Preparation of the chloride standards (NOTE: You will use the chloride salt you analyzed
previously by AgCl gravimetry as the standard reference material for this experiment.)
Dry about a gram of your standard chloride salt for at least two hours at 110oC. Cool in a desiccator. Prepare a Stock Chloride solution by weighing accurately 0.44 to 0.46 g. of your standard; transfer to a 250.00 mL volumetric flask through a powder funnel. Rinse the funnel and dilute to the mark with ISA solution. Transfer to a clean container which can be stoppered, e.g., 250 mL conical flask. Calculate and record the molarity. Pipet 50.00 mL of the Stock Chloride solution into a 100 mL volumetric flask; dilute to mark with ISA solution. (Refer to this as Solution I.)
(3) Preparation of Unknown Solutions. Obtain a sample from the instructor. Dry it at 110o C. for about two hours, then cool in a desiccator. To prepare sample A, weigh accurately 0.155 to 0.160 g. of sample, transfer to a 250.00 mL flask, dilute to the mark with ISA solution, mix, and transfer to a clean container which can be stoppered.
Prepare sample B similarly, except weigh 0.140 to 0.145 g. of sample.
Prepare sample C similarly, except weigh 0.131 to 0.135 g. of sample.
(4) Measurement Procedures
Calibration. The Combination Cl-ISE contains a choride indicator electrode and a calomel reference electrode. Connect this to the meter as indicated. Insert the electrode into solutions to be measured such that the solution covers the measurement probe tip of the electrode. NOTE: Do NOT stir solutions during measurements; the Cl-ISE is designed for unstirred solutions. Validate your Cl-ISE electrode response by making two measurements: Use a clean, dry 100 mL beaker for the test solutions. Pipet 25.0 mL of Solution 1 into the test beaker. Rinse the Cl-ISE with distilled water; blot dry and place in the test solution; measure the electrode potential. Then dilute the test solution by exactly 0.50 (by adding 25.0 mL of ISA solution) and measure again. The millivolt difference should be about 18 mV (0.018 V). Observe the I.D. No. for the Cl electrode, and record your observations in your lab notebook and the Chloride Electrodes Calibration Record Book. Contact the Lab Instructor if the meter responds sluggishly; if the reading does not stabilize; or if the mV difference is <12 mV or >21 mV.
Measuring Standard Solutions. Pipet 25.00 mL of Solution I into a clean, dry 100 mL beaker. Insert electrode and record millivolts until the readings stop drifting. (with the meter on "auto", record "locked in" millivolt readings until three consecutive readings are the same value.)
For solution II, pipet 5.00 mL of ISA solution into a beaker containing solution I: mix carefully by swirling: and read voltage. Do not remove electrodes when swirling, but be sure that the solution on the exposed electrode surfaces is well mixed with the diluted solution in the beaker.
For solution III, pipet an additional 5.00 mL of ISA solution into solution II, mix, and measure.
For solution IV, pipet an additional 5.00 mL of ISA solution into solution III, mix, and measure.
For solution V, pipet an additional 10.0 mL of ISA solution into solution IV, mix, and measure.
For solution VI, pipet an additional 15.0 mL of ISA solution into solution V, mix, and measure.
Graph the data, (see Calibration Curve on page 4), and calculate the equation of the line obtained, using least squares fitting. Write this equation on the data sheet.
Measuring Unknown Solutions. (When moving electrode to different solutions, rinse off with distilled water and gently blot off the water.)
Pipet 50.00 mL of solution A into a 100 mL beaker (refer to this as sample A.1; insert electrodes and record millivolts as above. For sample A.2, add 5.00 mL of ISA solution to A.1, mix, measure, and record mV. For sample A.3, add an additional 5.00 mL of ISA solution, measure and record millivolts.
Treat samples B and C similarly.
Calculate the percent chloride in each of your sample solutions (A.1, A.2, A.3, B.1, B.2,....C.3); report the average and the standard deviation for the nine sample solutions.
Calculate the percentages of KCl and NaCl in your sample, assuming it was a pure binary mixture. (See Theory section, pp 3 - 5).
Note 1: You should plan to read the voltages of all unknown and standard solutions in one uninterrupted session. If you have to complete this experiment on a different day, you must re-determine the voltages of all the standard and unknown solutions, because time and temperature affect the voltage readings. You may save for later use only solutions maintained in stoppered containers.
Note 2: If the voltage of any of the unknown solutions (A.1 to C.3) fall outside the range of the voltages of your calibration curve, there is a good chance you have made a solution preparation or dilution error for either calibration or unknown solutions.
Theory/Background
Particle Size Distribution and Sampling Errors
Solid particulate materials (even pure substances) have a range of particle sizes which typically follow a
gaussian type distribution as shown in Figure 1.

For mixtures of different chemical particles with differing particle size distributions, the measured composition will depend on how the bulk material is sampled! For example, Figure 2 shows hypothetical particle size distributions for two different sources of pure NaCl and pure KCl. If a 50/50 mixture of these two pure materials were prepared and thoroughly mixed, a subsequent chloride analysis of the mixture would depend on how the mixture was "sampled". For example, if the dry mixture were placed in a container which experienced periodic shaking or vibration, the finer particles would settle towards the bottom, with the larger particles predominating near the top. Thus, a sample taken from the top of the container would have a larger KCl composition, while particles taken from the bottom would have a larger NaCl composition. In both cases the sample taken would NOT give the correct chloride analysis for the overall mixture.
For this experiment, a real binary mixture of pure KCl and NaCl crystalline materials was prepared, where the pure KCl and NaCl materials have significantly different particle size distributions. This mixture was deliberately separated into five different fractions with differing particle sizes. Thus, it would be expected that each fraction has a different KCl/NaCl composition, and thus a different percentage chloride. The goal of this experiment is to illustrate the dependence of analytical results on physical properties of the sample. We will also use the overall class results to determine the different particle size distributions of the pure KCl and NaCl.
Sample Composition.
To calculate the sample composition when the percent chloride of the solid sample is known, use the following formula:
(G g.)(%Cl-) = (# moles Cl) = (fNa)(G g.) + (1 - fNa)(G g.) (1)
(A.Wt. Cl)(100%) (F.Wt. NaCl) (F.Wt. KCl)
where G = sample weight in grams; fNa = fraction of NaCl in mixture; (1 - fNa) = fraction of KCl in mixture.
The percent Cl is determined from the analysis; the sample weight is known; and the above equation can be solved for fNa.
Direct Potentiometry.
The potential of the Chloride ISE depends on the following electrode reaction:
AgCl(s) + e- == Ago + Cl- Eo = +0.222 V. vs. NHE (2)
The electrode potential depends on [Cl-], and can be predicted from the Nernst Equation:
EAg/AgCl = Eo - Slog[Cl-] (3)
where S
» 0.059 V.

lincalcl.wk3
Calibration Curve.
A calibration curve is used to obtain the chloride concentration of your unknown sample. This curve is generated by determining the potential of the Ag/AgCl electrode in solutions of varying [Cl-], and plotting the potential vs Log[Cl-], as shown in Figure 3. The resultant plot should be linear (as predicted by the Nernst Equation above). The potential of the unknown sample can then be determined and the unknown chloride concentration read directly from the calibration curve. For this experiment it is recommended that linear regression analysis (least squares fitting) be used to obtain quantitative data from the calibration curve.
Assuming a linear calibration curve, with the general behavior of:
y = a + bx, then (4)
from a least squares analysis, the slope and intercept can be calculated as:
b = sxy/sxx and a = Y(avg) - bx(avg), where (5)
sxx =
S[xi - x(avg)]2 = Sxi2 - (Sxi)2/N (6)
sxy =
S[(xi - x(avg)][(yi - y(avg)] = Sxiyi - (SxiSyi)/N (7)
N = total number of calibration data points
The determination of an unknown with concentration = xu can be obtained from the measured value of yu, using the regression slope and intercept:
xu = (yu - a)/b (8)
Because the experimental procedure here calls for the determinations of unknown chloride concentrations for nine different chloride solutions prepared from the original unknown sample, the recommended way of reporting the results is to calculate the percent chloride in the original solid sample from each determination, and then average these results. You should report the average and the 95% confidence interval, using
m = X(avg) ± t(su/Ö ¯n ), where (9)
X(avg) = average of computed % chloride in solid sample from 9 separate determinations
su = standard deviation of the 9 values used to obtain X(avg)
n = the number of separate determinations of % chloride (n = 9)
t = tabulated value from t-Table at 95% confidence level for n-1 degs of freedom
Sample Calculations
If the unknown sample weight is 0.1512g., and the [Cl-] obtained from the calibration curve for the undiluted solution of the unknown is 0.00835M, what is the % chloride and what are the fractions of NaCl and KCl in the unknown?
G = 0.1512g.; A.Wt Cl = 35.45; F.Wt NaCl = 58.44; F.Wt KCl = 74.56
%Cl- = 100% x (A.Wt. Cl)(# moles Cl)/(Sample wt., G)
%Cl- = 100% x (35.45)(0.00835M)(0.250L)/(0.1512g.) = 48.94%
To determine fNa, re-arrange Equation 1 from p. 65:
%Cl-/(A.Wt. Cl)(100%) - 1/(FWt KCl) = fNa[1/(FWt NaCl) - 1/(FWt KCl)]
fNa = [(48.94%)/(35.45)(100%) - 1/(74.56)] = 0.106
[1/(58.44) - 1/(74.56)]
Thus, fNa = 0.106 and (1 - fNa) = 0.894
That is, the original sample was a binary mixture of 10.6% NaCl and 89.4% KCl.
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