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Representation of silver chloride colloidal particle

Representation of silver chloride colloidal particle and adsorptive layers when Cl - is in excess. Cl - adsorbs on the particles when in excess (primary layer).

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Representation of silver chloride colloidal particle

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  1. Representation of silver chloride colloidal particle and adsorptive layers when Cl- is in excess.

  2. Cl- adsorbs on the particles when in excess (primary layer). A counter layer of cations forms. The neutral double layer causes the colloidal particles to coagulate. Washing with water will dilute the counter layer and the primary layer charge causes the particles to revert to the colloidal state (peptization). So we wash with an electrolyte that can be volatilized on heating (HNO3).

  3. 5. Drying and Ignition: The purpose of drying (heating at about 120-150 oC in an oven) or ignition in a muffle furnace at temperatures ranging from 600-1200 oC is to get a material with exactly known chemical structure so that the amount of analyte can be accurately determined.

  4. 6. Calculations: The last step in the gravimetric procedure is performing the analytical calculations, as will be seen shortly.

  5. Precipitation from Homogeneous Solution In order to make Q minimum we can, in some situations, generate the precipitating agent in the precipitation medium rather then adding it. For example, in order to precipitate iron as the hydroxide, we dissolve urea in the sample. Heating of the solution generates hydroxide ions from the hydrolysis of urea. Hydroxide ions are generated at all points in solution and thus there are no sites of concentration. We can also adjust the rate of urea hydrolysis and thus control the hydroxide generation rate. This type of procedure can be very advantageous in case of colloidal precipitates.

  6. Impurities in Precipitates 1. Occlusion: Some constituents of the precipitation medium may be trapped in the crystal structure resulting in positive or negative errors. The trapped materials can be water, analyte ions, precipitating agent ions, or other constituents in the medium. Slow addition of precipitating agent and stirring may avoid occlusion but if it does occur, dissolution of precipitate and repracipitation may have to be done.

  7. 2. Inclusion: If the precipitation medium contains ions of the same charge and size as one forming the crystal structure of the precipitate, this extraneous ion can replace an ion from the precipitate in the crystal structure. For example, in the precipitation of NH4MgPO4 in presence of K+ ammonium leaves the crystal magnesium ammonium phosphate and is replaced by K+ since both have the same charge and size. However, the FW fro NH4+ is 18 while that of K+ is 39. In this case a positive error occurs as the weight of precipitate will be larger when K+ replaces NH4+. In other situations we may get a negative error when the FW of the included species is less than the original replaced species

  8. 3. Surface Adsorption: This always results in positive errors in gravimetric procedures. See previous discussion on colloidal precipitates. 4. Postprecipitation: In cases where there are ions other than analyte ions which form precipitates with the precipitating agent but at much slower rate then analyte, and if the precipitate of the analyte is left for a long time without filtration then the other ions start forming a precipitate over the original precipitate leading to positive error. Examples include precipitation of copper as the sulfide in presence of zinc. Copper sulfide is formed first but if not directly filtered, zinc sulfide starts to precipitate on the top of it. The same is observed in the precipitation of calcium as the oxalate in presence of magnesium.

  9. Gravimetric Calculations The point here is to find the weight of analyte from the weight of precipitate. We can use the concepts discussed previously in stoichiometric calculations but let us learn something else. Assume Cl2 is to be precipitated as AgCl, then we can write a stoichiometric factor reading as follows: one mole of Cl2 gives 2 moles of AgCl. We can define a quantity called the gravimetric factor (GF) where: GF(analyte) = (a/b)*(FW analyte/FW ppt) a,b are stoichiometric coefficients GF for Cl2 = (1 mol Cl2/2 mol AgCl) * (FW Cl2/FW AgCl) = x g analyte/y g precipitate

  10. Weight of substance sought = weight of precipitate x GF One can also consider the problem by looking at the number of mmoles of analyte in terms of the mmoles of the precipitate where for the precipitation of Cl2 as AgCl, we can write Cl2D 2 AgCl mmol Cl2 = 1/2 mmol AgCl (mg Cl2/FW Cl2) = 1/2 (mg AgCl/FW AgCl)

  11. Example Calculate the mg analyte to mg precipitate for the following: P (at wt =30.97) in Ag3PO4 (FW = 711.22), Bi2S3 (FW 514.15) in BaSO4 (FW = 233.40) Solution P D Ag3PO4 mmol P = mmol Ag3PO4 Mg P/30.97 = mg Ag3PO4/711.22 Mg P/mg Ag3PO4 = 30.97/711.22 = 0.04354

  12. Bi2S3D 3 BaSO4 mmol Bi2S3 = 1/3 mmol BaSO4 mg Bi2S3/FW Bi2S3 = 1/3 mg BaSO4/FW BaSO4 mg Bi2S3/514.15 = 1/3 mg BaSO4/233.40 mg Bi2S3/ BaSO4 = 1/3 (514.15/ 233.40) = 0.73429

  13. Phosphate in a 0.2711 g sample was precipitated giving 1.1682 g of (NH4)2PO4.12 MoO3 (FW = 1876.5). Find percentage P (at wt = 30.97) and percentage P2O5 (FW = 141.95) in the sample. Solution First we set the mol relationship between analyte and precipitate P D (NH4)2PO4.12 MoO3 mmol P = mmol (NH4)2PO4.12 MoO3 mg P/at wt P = mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3 mg P = at wt P x (mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3) mg P = 30.97 (1.1682x103 / 1876.5) = 19.280 mg % P = (19.280/271.1) x 100 = 7.111%

  14. The same procedure is applied for finding the percentage of P2O5 P2O5D 2 (NH4)2PO4.12 MoO3 mmol P2O5 = 1/2 (NH4)2PO4.12 MoO3 mg P2O5/FW P2O5 = 1/2 (mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3) mg P2O5 = 1/2 x FW P2O5 x (mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3) mg P2O5 = 1/2 x 141.95 (1.1682x103/1876.5) = 44.185 mg % P2O5 = (44.185/271.1) x 100 = 16.30%

  15. Manganese in a 1.52 g sample was precipitated as Mn3O4 (FW = 228.8) weighing 0.126 g. Find percentage Mn2O3 (FW = 157.9) and Mn (at wt = 54.94) in the sample. Solution 3 Mn2O3D 2 Mn3O4 mmol Mn2O3 = 3/2 mmol Mn3O4 mg Mn2O3 / FW Mn2O3 = 3/2 (mg Mn3O4/FW Mn3O4) mg Mn2O3 = 3/2 FW Mn2O3 (mg Mn3O4/FW Mn3O4) mg Mn2O3 = 3/2 x 157.9 ( 126/228.8) = 130 mg % Mn2O3 = (130/1520) x 100 = 8.58%

  16. The same idea is applied for the determination of Mn in the sample 3 Mn D Mn3O4 mmol Mn = 3 mmol Mn3O4 mg Mn/ at wt Mn = 3 (mg Mn3O4/FW Mn3O4) mg Mn = 3 at wt Mn (mg Mn3O4/FW Mn3O4) mg Mn = 3 x 54.94 ( 126/228.8) = 90.8 mg % Mn = (90.8/1520) x 100 = 5.97%

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