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Experiment 1 and 2: Determination of Chloride Concentration in Water by Gravimetric Methods and Precipitation Titration Chem 374 Analytical Chemistry Laboratory



Principle and Background The vast majority of water sources such s groundwater and rainfall contain chloride, a mobile anion. Due to the plethora of chloride in various natural water sources, several methods of analysis for the determination of chloride are utilized. In these experiments, two methods are employed and compared in the analysis of the content of chloride in water. These two methods for the determination of chloride are by gravimetric methods and precipitation titration. In precipitation methods, “the analyte is converted to a slightly soluble precipitate which is filtered, washed free of impurities, and converted to a product of known composition by suitable heat treatment.” The product is weighed and then analyte content of the sample is then determined through stoichiometry. The mass of the sample and the mass of a product of known composition are necessary in order to compute the results of a gravimetric analysis. If the product is the analyte, the concentration is



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found as a percentage of the analyte by taking the ratio of the mass of analyte to the mass of sample. A product that is not the analyte is multiplied by the gravimetric factor (GF) which accounts for the stoichiometric relationship between the analyte and product weighed. Selective reagents react with a narrow amount of species so that a product with a high purity, low solubility, unreactive nature, and high molar mass is obtained. Gravimetric methods are absolute meaning that they don’t require any form of calibration, requiring a shorter amount of operator time per sample. Sensitivity and accuracy are typically dependent on the type of analytical instrument that is utilized. The chloride content of water can be determined by the precipitation of the anion as AgCl (s). The precipitate is collected in a weighed crucible, that is washed, and then dried to determine its mass. The solution is held somewhat acidic to eliminate any interference from anions of weak acids that may produce a soluble salt in a neutral medium. Excess silver ion is necessary to reduce the solubility of AgCl, but should also be avoided to prevent coprecipitation of silver nitrate. AgCl is initially formed as a colloid and then coagulated with heat. This coagulation is promoted by nitric acid and excess silver nitrate in order to provide a high electrolyte concentration. Nitric acid maintains this electrolyte concentration, and eliminates any possible peptization during washing. Titrimetric methods of analysis make up a large group of quantitative procedures. A measured quantity is equivalent to the volume of a standard solutions needed to react completely with the analyte. Neutralization titrations are primarily used for determining the concentrations of analytes that are acids/bases or convertible to a species by an acceptable treatment. These concentrations are expressed in molarity (M), which provides the number of moles of a reagent in a L of solution. An equivalent is a compound that can product a mole of positive or negative charge, but an amount of a substance in an equivalent varies from reaction reaction. The equivalence point of a titration is reached when the amount of an added titration is chemically equivalent to the amount of analyte within a sample. This experiment utilises a Mohr titration, which reaches its equivalence point through the presence of chromate which forms a reddish-orange Ag2CrO4 precipitate after most of the chloride has precipitated. The pH is



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buffered within a range of 7-10 as the chromate is converted to dichromate Cr2O72-, a soluble salt at a lower pH. At pH greater than 10, Ag2O may precipitate, forming an upper pH limit for all argentometric titrations. Silver chromate precipitation will ideally start at the equivalence point



2−¿ Cr O¿4 =¿ of silver chloride titration, [Ag+] = √❑ . The solubility of Ag2CrO4 we have + ¿¿ 2 ¿ A g¿ ¿ KspAg2CrO4 so that the onset of silver chromate precipitation occurs where [Ag+] = √ ❑ . Solving these two equations and solving for the chromate concentration provides [chromate]= K spAg 2CrO 4 / KspAgCl. By equating these two results we find KspAg2CrO4 = 1.1 x 10-12 M3 and KspAgCl = 1.81 x 10-10 M2, we



2−¿ have Cr O¿4 =6mM. A lower chromate concentration may be used, because chromate itself is colored. ¿ This will lead to a possible l “overshoot” of the equivalence point somewhat before silver chromate starts to precipitate. To correct for the formation some precipitate before it is a blank must be titrated solution, and corresponding volume should be subtracted from titrant volume Vt. In this second experiment, the concentration of chloride in water is determined using a weight based titration based on the Mohr method. The end point is apparent by the formation of a orange-brownish silver chromate precipitate. Experimental Experiment 1 Two crucibles are brought to a constant mass by heating at 110 ℃ for an hour. Two 100-mL water samples are mixed with about 2-3 mL of 6 M HNO3 to each sample. 0.2 M AgNO3 is added and mixed with



each solution until AgCl coagulates, followed by an additional 3 to 5 mL. Heat and digest the



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solids for about 10 min. Add a few drops of AgNO3 to confirm that precipitation occurred. Each beaker is covered and stored in a dark place for a week. The following week, the solutions are decanted through the weighed crucibles. The precipitates are washed several times with 2 to 5 mL of 6 M HNO3 per liter of deionized water, and are then decanted through the filters. Transfer the AgCl from the beakers to the individual crucibles with fine streams of wash solution. Continue washing until the filtrates are essentially free of Ag+ ion. The precipitate should be dried at 110 °C for at least 1 hr. Store the crucibles in a desiccator while they cool. Weight the mass of the crucibles and the samples. Repeat the cycle of heating, cooling, and weighing until two consecutive weighings.. Calculate the concentration of chloride ions in the water sample. After analysis has been completed remove the precipitates by tapping the crucibles over a piece of paper. Transfer the collected AgCl to the waste container. Remove any traces of AgCl by filling the crucibles with 6 M NH3 and allowing them to stand.



Experiment 2 In this second experiment three samples are prepared with 50 mL of the unknown chloride sample, an additional 50 mL of water to bring the total volume of 100 mL, and 1 mL of indicator (potassium chromate). Prior to this, a blank consisting of 50 mL of DI water, a small amount of calcium carbonate, and 1 mL of indicator 5% K2CrO4 is prepared. The addition of the silver nitrate titrant is added dropwise, until an orange-brownish color change is observe. The bottle of silver nitrate is weighed prior to its addition and after each use for each of the three samples. The unknown chloride samples should be prepared as previously mentioned, and then silver



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nitrate should be added dropwise until the color change is observed. The reagent weights should be corrected for the blank, and the concentration of chloride in the unknown is determined using the average of the three determinations. Results and Calculations For this experiment



Table 1: Weight of Crucibles Weighing #



Mass of Crucible 1 (g)



Crucible 2 (g)



1



33.9870



33.8356



2



33.9846



33.8351



3



33.9903



33.8359



4



33.9901



33.8357



5



33.9906



33.8364



Average



33.9894



33.8354



Table 2: Weight of Crucibles with Sample *Note for sample 2 that a different crucible was utilized than listed in Table 1, to account for that the average mass of the crucible utilized was 35.8266g, and it’s weight with the sample was 35.8683



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Sample #



Mass of Sample+Crucible (g)



Mass of Sample (Mass of Sample+crucible)-(Average Mass of Crucible from Table 1) (g)



1



34.0321



.0427



2



35.8683



.0417



After determining the mass of the sample, the concentration of chloride was determined as demonstrated in the sample calculation below: Ex. for Sample 1 −¿ −¿=105.62mg Cl¿ g Cl−¿ 35.45 =0.0105617 g Cl ¿ mol g 143.33 AgCl mol M Cl =0.0427 g AgCl x ¿ M AgCl Cl−¿ =m AgCl x ¿ m¿ −¿



Concentration of chloride=105.62 ppm



Chloride concentration of Sample 2: 103.14ppm Table 3: Average chloride concentration and Standard Deviation



Analyst



Calculated Chloride Concentration (ppm) 102.0 101.7 107.6 105.1 108.8 109.4 105.6 103.1



Average = 105.4 s = 2.8



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¿ 110−105.4∨ ¿ x 100=4.2 error 110 ¿ ¿ Expected conc .−Calculated conc .∨ x 100=¿ Expected conc . Percent error=¿ N = 8 data points → N-1 = 7 → at 95% confidence, t = 2.365



Confidence Interval : µ=x ±



ts √❑



Comparing the Experimental Mean to the True Value (110 ppm Cl-)



¿ µ−x∨ ¿ √❑ s t calc=¿ Experiment 2:



The mass of silver nitrate was determined by subtracting the differences in the mass of the dropper bottle from the difference of the mass of silver nitrate use with the blank (0.77g). The differences in mass of silver nitrate for trials 1, 2, and 3 were a respective 3.437g, 3.253g, and 2.645g. After accounting for the blank the masses were a respective, 2.667, 3.253, and 2.645g. The chloride concentration was then determined utilizing similar calculations to the one below:



Ex. for Sample 1:



0.4877 molAgNO 3 35.453 mg x 2.667 g x x 20=92.2 ppm kg solution mmol



Table 4: Calculated chloride concentration (ppm), average, concentration, and the standard deviation



Analyst



Calculated Chloride Concentration (ppm) 106 118.3 106.1 104.2 102.9 104.7 105.0 100.7 102.5



Average = 103.7 s = 6.80



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101.6 92.2 112.5 91.5 ¿ 110−103.71∨ ¿ x 100=5.72 error 110 ¿ ¿ Expected conc .−Calculated conc .∨ x 100=¿ Expected conc . Percent error=¿ N = 13 data points → N-1 = 12 → at 95% confidence, t = 2.179



Confidence Interval : µ=x ±



ts √❑



Comparing the Experimental Mean to the True Value of 110 ppm Cl -



¿ µ−x∨ ¿ √❑ s t calc=¿



Comparison of Two Experimental Means s pooled =√❑ t calc=



x 1−x 2 √❑ s pooled



Comparison of Precision: From Table 4-4 in the textbook, at df1=7, df2=12, Ftable = 2.91 Fcalc =



s1 s2



2



= 2



2



(2.8) =0.169< 2.91→ difference is significant 2 (6.8)



The following test demonstrates that the two standard deviations are significantly different from each other.



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Discussion and Results In these experiments, gravimetric and argentometric methods were conducted in order to determine the concentration of an unknown chloride solution. Several calculations were done to account for the differences in these methods, and to evaluate which method was more efficient for this determination. Utilizing the calculations done for the critical value 4 of the gravimetric precipitation values fell within the 95% confidence interval, whereas only 1 of the argentometric titration values fell within the 95% confidence interval. These results demonstrated that the gravimetric methods were more accurate in obtaining results within their independent range. When comparing the percent errors, the argentometric titration yielded a 5.72% error from the actual concentration of chloride (110 ppm), while the gravimetric precipitation yielded a 4.2% error. This further demonstrates that the gravimetric precipitation is slightly more accurate than the argentometric titration, and has slightly fewer errors. On the other hand, both tests failed the t-test which had compared the experimental means to the true value. This is significant as this would show that neither method held an advantage over the either in the determination of chloride concentration, if simply accounting for the t-test The argentometric method seemed to have yielded several outliers within the data including the second data point with a value of 118.3 when compared to Brandon’s other values, and the value of 112.5 for the second to last data point collected from my second trial. These results may have been due to varying errors which will be discussed below. Several systematic and random errors during the experimentation between the two methods. The t-test for both results was rejected, demonstrating that neither result had an advantage over the other. However, this may have been due to any errors during the process of experimentation or any systematic errors that may not have been accounted for. In order to



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account for systematic errors in the gravimetric titration a better system could be employed in drying, cooling, and weighing the crucibles. Keeping the time between each interval of drying and cooling consistent, would provide a more efficient manner of weighing the crucibles and would reduce potential deviations in the data. Argentometric titrations may have been affected by overshooting the equivalence point, which had been accounted for by the preparation of a blank. Argentometric titrations may be suspect to more random errors than gravimetric methods due to the potential of overshooting, inconsistent pipetting of the various chemicals which would lead to the varied concentrations, potential contaminants, and the addition of inconsistent amounts of the silver nitrate. To account for any inconsistencies in the mixing, withdrawal, or addition of certain chemicals an automated pipettor could have been used to account for any personal errors that may have occurred. This would lead to consistent amounts of each chemical for each trial/sample. Additionally, instead of simply adding the silver nitrate to the unknown chloride mixture until the color change is noted, an added step of counting the number of drops of silver nitrate added should be enforced to ensure that an equivalent amount is added while titrating for each sample. Any random errors with the gravimetric errors may have included potential contaminants within the sample, additional moisture within the crucibles, and incomplete washing. These random errors, as demonstrated by the percent errors of each individual method, were less significant for the gravimetric method when compared to the argentometric method. While neither the gravimetric nor the argentometric method demonstrated to have had a significant overall advantage in the determination of chloride, the gravimetric had a lower percent error when compared to that of the argentometric. Overall, the gravimetric precipitation method demonstrated a greater effectiveness for determining the concentration of chloride in an unknown solution when compared to that of the argentometric titration method due to a lower



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percent error in addition to a larger amount of values that fell within the critical value test. . Both of these methods demonstrated an ability to determine the concentration of chloride and are practical methods for environmental testing, however, the gravimetric method simply demonstrated a greater ability to do so. Questions for Experiment 1 1. What would happen if you carried out the precipitation of chloride ions in a neutral medium? If the precipitation of chloride ions were carried out in a neutral medium the anions of the weak acids would form precipitates with Ag+ that would interfere with precipitation of the analyte. The solution has to be kept acidic during precipitation to eliminate interference from anions of weak acids that form soluble salts in a neutral medium.



2. Why does the wash solution contain HNO3 and not NaNO3? Why is it not deionized water or AgNO3 solution? The wash solution contains HNO3 in order to keep the solution mildly acidic to maintain the electrolyte concentration and prevent interference from soluble salts that could form. It also eliminates the possibility of peptization during the washing step causing the acid to decompose into a volatile product when the precipitate is dried. . DI water isn’t used because it causes peptization, and the usage NaNO3 and AgNO3 would cause too much interference during the precipitation of the analyte. 3. How would the results of this analysis be affected if photodecomposition of silver chloride took place before filtration, and the additional reaction (shown below) occurred? 3Cl2(aq) + 3H2O + 5Ag+ 5AgCl(s) + ClO3- + 6H+



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The photodecomposition of silver chloride would lead to the chloride being used up at a faster rate than it’s being produced, leading to a chloride amount lower than the true value. 4. A 5.000-g sample of a pesticide was decomposed with metallic sodium in alcohol, and the liberated chloride ion was precipitated as AgCl. Express the results of this analysis in terms of the percentage of 1,1,1-trichloro-2,2-bis(parachlorophenyl)ethane (DDT) (C 14H9Cl5) based upon the recovery of 0.1606 g of AgCl. ?=determine amount of DDT produced in the pesticide utilizing a recovered 0.1606g of AgCl 354.34 g DDT =0.079 g DDT 1 mol Cl−¿ 1 mol DDT 1 mol x ¿ 1mol AgCl 1mol AgCl 0.1606 g AgCl x x¿ 143.35 g



5 mol Cl−¿ x



Percentage of DDT =



0.079 g DDT x 100=1.6 DDT 5 g sample



5. Addition of an excess AgNO3 to a 0.5012-g sample yielded a mixture of AgCl and AgI that weighed 0.4715 g. The precipitate was then heated up in a stream of Cl 2 to convert the AgI to AgCl. 2AgI(s) + Cl2(g)  2AgCl(s) + I2(g) The precipitate was found to weigh 0.3922 g after this treatment. Calculate the percentages of KI and NH4Cl in the sample. x = mass of iodine, and y = mass of chloride. x+ y=0.4715



M AgCl=143.35 ; M AgI =234.8



x=0.4715− y x+ y (



M AgCl )=0.3922 M AgI



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−¿ 0.4715− y+ 0.611 y=0.3922 y=0.204 g → 43.3 I ¿ 0.0793=0.389 y y=0.204 g=0.204 /0.4715 g of mixture=43.3 KI x=0.4715−0.204=0.2675 g /0.415 g of mixture → 56.8 NH 4 Cl



Questions for Experiment 2 1. Write balanced chemical equations for all the reactions involved in this experiment. Cl- + Ag+ → AgCl(s) CrO42- + 2Ag+ → Ag2CrO4(s) 2. Why is solid CaCO3 suspended in the indicator blank? It imitates the presence of AgCl within the sample used for the experiment. 3. Why was NaHCO3 added to your samples prior to titration? This chemical was added to test the pH of the solution. The solution will bubble if it isn’t basic enough, and it needs to be buffered in between a pH of 7-10.



4. Briefly describe how you would determine carbonate ion by the Volhard method. A carbonate ion would be determined by the Volhard method by adding a measured excess of AgNO 3 to the solution, then filter out the AgCO 3 under back-titration.



A back titration using a standard



solution of KSCN would then need to be used in order to determine the concentration of carbonate.



5. Titration of a 0.485-g sample by the Mohr method required 36.8 mL of standard 0.1060 M AgNO3 solution. Calculate the percentage of chloride in the sample.



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−¿ Cl−¿ 35.45 g 1 mol x =0.138 gCl ¿ 1 mol AgNO3 1 mol 0.1060 mol 0.0368 L AgNO3 x x¿ L −¿ ¿ ¿ 0.138 g Cl− x 100=28.5 Cl 0.485 g total m Cl x 100=¿ mtotal −¿=¿ ¿ Percentage of Cl −¿



6. Because the yellow color of CrO42- tends to obscure the first appearance of the red Ag 2CrO4, it is common practice to hold the CrO 42- concentration to about 2.5 x 10-3 M. Calculate the relative titration error (neglecting the volume of AgNO 3 needed to produce a detectable amount of Ag2CrO4) in the titration of 50.0 mL of 0.0500 M NaCl with 0.1000 M AgNO3. 50 mL x



1L 0.05 mol NaCl 1mol AgNO3 1L x x x =0.025 L=25 mL titrant 1000 mL 1L 1 mol NaCl 0.1mol



+¿ Ag ¿= √❑ M ¿



[Ag+] =



2.608 x 10−4 =[



(0.05)(0.05) 0.1V 1.7 x 10−10 − ]+ 0.05+V 0.05+V 2.608 x 10−4



Percent error=



V ¿ 25.211 mL



25.211−25 mL x 100=0.84 25.211



7. The Fajans method is to be used in the routine analysis of solids for their chloride content. It is desired that the volume of standard AgNO3 used in these titrations be numerically equal to the percent Cl- when 0.2500-g samples are used. What should the molarity of the AgNO3 solution be?



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AgNO3+Cl- ↔ AgCl(s)+NO3Mass of solid sample=0.2500g



Cl- %=(mass Cl-/mass sample) x 100



Volume of AgNO3 =(mass Cl-/mass sample) x 100= 100 (mass of Cl-)/0.25 1) AgNO3 mols/L=mol of AgCl in sample= mol of Cl-/volume of AgNO3 =(0.25g Cl-)/((35.45g/mol) x (0.25g/100(mass of chloride)(value from 1) =0.25g/(35.45g.mol)(100)=7.052x 1 0−5 M 8. (a) A 0.1752-g sample of primary standard AgNO3 was dissolved in 502.3 g of deionized water. Calculate the weight molarity of Ag+ in this solution. 0.1752 g AgNO3 x



1 mol =1.03 x 10−3 mol AgNO3 169.87 g



Mass of solution = 502.3 g + 0.1752 g = 502.48 g = 0.502 kg solution −3



Weight molarity =



1.03 x 10 mol AgNO3 =2.05 x 10−3 mols /kg 0.502 kg soln



(b) The standard solution described in part (a) was used to titrate a 25.171-g sample of a KSCN solution. An end point was obtained after adding 23.765 g of the AgNO3 solution. Calculate the weight molarity of the KSCN solution. +¿ x



23.765 g soln =1.938 x 10−3 m KSCN 25.171 g sample 2.053 x 10−3 m Ag¿



(c) The solutions described in parts (a) and (b) were used to determine the BaCl2×2H2O in a 0.7120-g sample. A 20.102-g portion of the AgNO3 was added to a solution of the sample and the excess AgNO3 was back-titrated with 7.543 g of the KSCN solution. Calculate the percent BaCl2×2H2O in the sample. 2.053 x 10−3 mol



Ag+¿ =¿ 1 kg



+¿ x ¿ 2.010 x 10−2 kg Ag¿



+¿−1.462 x 10−5 mol 4.127 x 10−5 mol Ag¿



244.265 g 1 mol BaCl2 ∙ 2 H 2 O 1mol BaCl2 ¿ 2.665 x 10−5 mol x ¿



2 mol Ag+¿ x



−3



¿ 3.26 x 10 g BaCl2 ∙2 H 2 O



15



−3



mass of BaCl2 ∙ 2 H 2 O 3.26 x 10 g BaCl2 ∙2 H 2 O BaCl2 ∙ 2 H 2 O= x 100= x 100=0.457 BaCl2 ∙ 2 H 2 O total mass of sample 0.7120 g sample