Gravimetric Analysis Lab Report by Sarah Uddin

March 30, 2019 | Author: Sarah Uddin | Category: Precipitation (Chemistry), Physical Chemistry, Physical Sciences, Science, Chemical Substances
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Lab Report for gravimetric analysis of an unknown chloride compound....

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Chloride determination of an unknown sample by gravimetric analysis Sarah Uddin

Lab partners Rosio and Pierre Dates performed: 1/30/2017, 2/1/2017, 2/6/2017, 2/8/2017, 2/15/2017 Date due: 2/22/2017

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Introduction:

Gravimetric analysis is a process often used to determine the mass pe rcent of an ion in an unknown salt. Determination is made by comparing the mass of two compounds, both containing the ion being analyzed, the analyte1. During gravimetric analysis, the unknown salt is dissolved and a solution containing an ion which reacts with the aqueous analyte is added to form a solid, pure precipitate with a known molecular weight. This known precipitate’s mass determines the mass of analyte present in the solution. This mass is then compared to the initial mass of the unknown sample to determine  percent of analyte it contains. Since the unknown sample can absorb moisture from air exposure, the sample is dried. If mass is obtained prior to drying, initial mass will include impurities in the salt and result in a low  percent mass. In chloride determination by gravimetric analysis, AgNO3 is often used to precipitate AgCl, see the following equations 1 and 2: AgNO3 (s)

+

 Ag (aq) +

NO3-(aq)

Ag+(aq) + Cl-(aq)  AgCl (s)

eq. 1 eq. 2

Attention to factors which affect the final mass of AgCl is of primary importance. Specifically, colloidal precipitate formation, coprecipitates, common ions contamination, a nd pH fluctuation, and the photodecomposition of silver compounds need to be considered to ensure a true and accurate AgCl mass is obtained.

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Slowly adding the AgNO3 solution to just excess concentration is important to prevent suspended colloids in supersaturated conditions2. When aqueous ions come together to form salts, there Figure 1

are two separate processes:

nucleation, the initial ionic bonding of small aggregates of ions, then particle growth, which adds more molecules to that nucleus, to form a larger crystal2. See figure 1a. In supersaturated solution, nucleation occurs faster than particle growth, resulting in small suspended co lloids, which pass easily through filters, resulting in loss of the analyte. S ee figure 1b. By adding the AgNO3 slowly, while mixing vigorously, in dilute conditions helps prevent this loss of analyte2. Also, digestion, or standing in heated solution, allows cr ystal size to increase because the salt slowly recrystallizes, promoting particle growth. An electrolytic solution is ideal for formation of salt solids. The increased ions in the solution lowers the kinetic energy required for ions to react through the adsorption layers surrounding the crystal in solution2. HNO3 is added to the solution with the dissolved unknown salt to maintain electrolyte concentration. This also lowers pH which is beneficial to prevent coprecipitation of AgOH, by the following equation: Ag+(aq) + OH-(aq)  AgOH (s)

eq. 3

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Though the electrolyte solution is ideal for salt formation, alternate ions present results in coprecipitations. Both adsorbed, surface bonded, and occluded, internal coprecipitation, are  possible2,3. See equations 4 and 5 for possible coprecipitates formed, even if solubility constants are not met2. Equation 5 accounts for dissolved CO2 (g) from the air. 2Ag+(aq) + NO3-(aq) + Cl-(aq)

 AgNO3 (s) +

CO2 (g) + 2H2O (l) + 2Ag+(aq) + Cl-(aq)



AgCl (s)

eq. 4

H3O+(aq) + HCO3-(aq) + 2Ag+(aq) + Cl-(aq)

AgHCO3 (s) + AgCl (s)



eq. 5

Adsorbed and/or occluded silver ions can also be present. Washing the precipitate with a gathering agent, like dilute HNO3 solution, removes coprecipitates, like silver nitrate and silver  bicarbonate, and silver ions. The dilute HNO3 wash also maintains electrolyte concentration and  prevents breaking down, or peptization, of the AgCl, by resolubilizing the AgCl when washing2,3. Peptization results in loss of analyte. Another loss of analyte occurs during photodecomposition of AgCl from UV light exposure. Photodecomposition results in reduction of silver ions and oxidation of chloride ions to form solid silver and gaseous chlorine, see equation 63. Therefore, limiting light exposure to the  product, AgCl, is important. 2AgCl (s) +  hv

 2Ag (s) +

Cl2 (g)

eq. 6

In this experiment, an unknown salt sample con taining chloride was dissolved in a dilute HNO3 solution. AgCl was precipitated by an AgNO3 solution, heated to digest, then filtered in a  purified/dry crucible. The AgCl precipitate was washed with a diluted HNO3 solution, then dried. The mass of AgCl was used to determine the percent mass of chloride in the unknown sample.

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Results: Table 1: Gravimetric results after precipitating Cl - with AgNO3 Sample

Mass Unknown

Mass Crucible

Mass Crucible

Mass AgCl

(g)

(g)

and AgCl (g)

(g)

1

0.5420 (±.0001)

29.5337 (±.0001) 30.7529 (±.0001)

1.2192 (±.0001)

2

0.5066 (±.0001)

30.5641 (±.0001) 31.6887 (±.0001)

1.1246 (±.0001)

3

0.5054 (±.0001)

29.6612 (±.0001) 30.8108 (±.0001)

1.1496 (±.0001)

Result Calculations:

Sample 1

Sample 2

Sample 3

− − 1   1   35. 4 5   1.2192 ±.0001   143.32   1    1  − =0.3016    .5420 ±.00011    100%= .% − − 1   1   35. 4 5   1.1246±.0001   143.32   1    1  − = 0.2782    .5066 ±.00011    100%=.% − − 1   1   35. 4 5   1.1496±.0001   143.32   1    1  − = 0.2843    .5054 ±.00011    100%=.%

Average Percent Chloride:

.+.+. =.±.%

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Statistical Analysis Calculations:

Propagation of error for Gravimetric analysis:

 .0001  .0001   . (.05001 173) (29.9197)  (31.0841) = 55.60 = .01075= . 95% Confidence interval for Gravimetric Method:

4.303∗.√ 36757 =1.679 = ± % 4.303∗1. 2 07 √ 3 =2.999 = ± %

95% Confidence interval for Fajan’s method:

T-test to compare true value to Fajans method mean:

= 56.556.7 1.√ 231 = .2862
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