"CH 429 Fluoride"
Jarrett Long Determination of Fluoride Ion Potentiometrically INTRODUCTION Fluoride happens to be a good candidate for ion electrode potentiometric measurements due to certain characteristics of the fluoride ion. Namely, it is a small ion which is capable of traveling through certain membranes. These membranes allow for the migration of the small, negatively charged fluorine ions between the calibration solution and the solution in question until a charge equilibrium is reached due to the difference in potential of the solutions. When there are concentration of fluoride anions in solution is greater then more ions will flow through the membrane, leading to a larger potential difference. This laboratory experiment will demonstrate this relation by determining the concentration of fluoride ions in an unknown sample by mathematical extrapolation. METHOD Since the relationship between the concentration and the voltage is generally logarithmic, it is possible to graph the correlation in such a way that allows for the determination of the original concentration. Due to fluoride etching, all containers except for weighing containers are plastic. The first step is making up 100 ml “Known Add” solution of 10mg/l of F by mixing 221 mg NaF to 1 liter of water. Then pour 10 ml of this solution and 10 ml of a pH buffer (TISAB in this case) to a 100 ml volumetric flask with water. Transfer from flask to plastic. This will be added in 1 ml aliquots to three various solutions. A 100 ml blank solution, with no original fluoride concentrations, a test solution (quality control) of 0.2 mg/l fluoride concentration in 100 ml (stored in plastic), and 100 ml of unknown in it’s own flask. 10 additions of the “Known Add” is added to each solution one solution at a time with measurements of voltage after each of the additions. The data is then plotted in two different fashions. Using a computer, a variation of the Nernst function (Vs+Va)exp(-nEF/RT) can be plotted against the concentration of fluoride added, which we know to be 0.01 mg of F per mL aliquot. Using a linear fit of this relationship, the slope and the y-intercept of this graph can be determined for each of the three solutions. The original concentration is found by taking the absolute value of the quotient (|fluoride intercept| = |fi|) of the y-intercept (b) by the slope (m) and dividing this by the solution volume. This is: | fi | | b / m | co Vo Vo where |m/b| is the least fit line of the corresponding linear fit. (See: FIGURES AND GRAPHS) The other form of the data is by using Gran’s paper. This is special logarithmic paper that allows the concentration-potential relationship to easily be hand-graphed and then through graphical analysis the same information can be gathered. The graphs are attached in the DATA section and will be the results will be discussed briefly in the next section. CALIBRATION The blank solution was analyzed so that through this relatively-analog process can have a zero calibration. The ten aliquots were added and graphed onto the attached Graph 1. With a |fi|/Vo of 1.15 x 10-3 4.37 x 10-3 mg/l then this can be assumed as determinate error of the experiment, and therefore subtracted from the masses of the other two trials. For instance, through the same procedure the original fluoride mass in the quality control was calculated to be 0.207 0.017 mg/l when the original known amount was to be 0.200 0.001 mg/l. That shows a relative error of 3.5%, a seemingly respectful mark of accuracy with 8.5% total uncertainty. When discussing the Gran’s Paper method, it must be noted that it uses no fancy least-squared fit systems, or even an automated linear fit process! It simply boils down to graphing the voltage vs. concentration and from there using a ruler to determine the fluoride intercept. This is done for all three on the attached Gran’s Paper, however, the results for the quality control were about 0.23mg/L (0.26-0.03 for the blank determinate error) and the results were 0.51 mg/l for the unknown. The 0.51is a good estimate, however, as indicated by the quality control’s 15% error, this method should be used only when electricity and computers have disappeared from Earth. ERROR ANALYSIS The error budget is presented as Table 1 in TABLES. A note on the lack of indeterminate error: As understood, any other source of indeterminate error would involve either the concentration of the “Known Add” or perhaps error on the Voltage detector. However, it seems that either one of very small error sources would be taken into account by the standard deviation calculations from the linear fit of the model which would account for random error from the machine and random error on aliquots. Also, the error for the linear fit is from the formula: <(WRITE) ERROR ON LINEAR EXTRAPOLATION FORMULA HERE> RESULTS By looking at Graph 3 and through the same procedure as for the other two solutions, it is determined that the original concentration of fluoride in the unknown sample is 0.49 0.02 mg/l. The calculation for this and the error are given in the SAMPLE CALCULATIONS. The 95% confidence interval lies within (0.45 < < 0.53). DISCUSSION A relatively simple experiment that just requires patience and time, along with a good hand at making weighing out compound and making solutions to ensure proper mathematical models. The Gran’s paper would be an excellent way of avoiding the mean-looking Nernst function in the days before heavy computational analysis. However, nowadays, it seems to just increase error likelihood, especially due to the technique of human interpolation. The 3.5% error on the quality control assures at least adequate reassurance of the final determination, though, as with most labs, more trials and quality controls would be appreciated - infinitely so would be ideal. TABLES Error Budget Error Source Indeterminate Standard Deviation Determinate Blank Solution -1.15 x 10-3 Correction 4.37 x 10-3 mg/l Machine Error - Random, See: linear - fit error Aliquot Fluctuations - Random, See: linear - fit error Linear Fit Error Blank: 1.15 x 10-2 mg/l QC: 1.71 x 10-2 mg/l Unknown: 1.65 x 10-2 mg/l FIGURES AND GRAPHS GRAPH 1 0.30 Blank Calibration Curve 0.25 (-EnF/RT) 0.20 (Vs+Va)e 0.15 y = 2.922x + 0.0127 0.10 |b/m| = |fluoride intercept| = |-0.0127/2.922| = 0.0486 0.05 0 20 40 60 80 -3 100x10 Total Fluoride Added (mg) GRAPH 2 0.35 Quality Control Calibration Curve 0.30 (-EnF/RT) 0.25 0.20 (Vs+Va)e y = 2.8683x + 0.072085 0.15 0.10 |b/m| = |fluoride intercept| = |-0.072085/2.8683| = 0.02513 0.05 0.00 0 20 40 60 80 -3 100x10 Total Flouride Added (mg) GRAPH 3 0.45 "Unknown" Calibration Curve 0.40 (-EnF/RT) 0.35 y = 3.0145x + 0.1598 (Vs+Va)e 0.30 |b/m| = |fluoride intercept| 0.25 = |-0.1598/3.0145| = |0.05301| 0.20 0 20 40 60 80 -3 100x10 Total Fluoride Added (mg) DATA Raw Uncharted Data BLANK TEST UNKNOWN E (V) Fluoride Added (mg) E (V) Fluoride Added (mg) E (V) Fluoride Added (mg) 0.213 0 0.183 0 0.163 0 0.199 0.01 0.176 0.01 0.159 0.01 0.187 0.02 0.17 0.02 0.156 0.02 0.177 0.03 0.165 0.03 0.153 0.03 0.17 0.04 0.16 0.04 0.15 0.04 0.165 0.05 0.157 0.05 0.148 0.05 0.16 0.06 0.154 0.06 0.146 0.06 0.157 0.07 0.151 0.07 0.144 0.07 0.155 0.08 0.149 0.08 0.142 0.08 0.152 0.09 0.147 0.09 0.14 0.09 0.149 0.1 0.146 0.1 0.139 0.1 SAMPLE CALCULATIONS <(WRITE) DO CALCULATION FOR Unknown Mass AND Error HERE> QUESTIONS 7.1| a) E = c - 59log[F] where c is a constant b) c = 159, so 0.92ppm after converting from M. c) Yes 7.2| a) Should be low b) high c) 4.7 d) 9.4 7.3| a) 1.9 mM b) 0.044 mM c) 0.057 mM 7.4| 7.5| 11ppm 2.7| About 4% Question 1 from Lab: 3.17 Question 2 from Lab: