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Lecture 17

Lecture 17. Gravimetric Analysis Steps in Gravimetric Analysis. Gravimetric Analysis.

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Lecture 17

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  1. Lecture 17 Gravimetric Analysis Steps in Gravimetric Analysis

  2. Gravimetric Analysis In this technique, the analyte is converted to an insoluble form which can then be washed, dried, and weighed in order to determine the concentration of the analyte in the original solution. Gravimetry is applied to samples where a good precipitating agent is available. The precipitate should be quantitative, easily washed and filtered and of a suitable quantity for accurate weighing.

  3. Therefore, gravimetry is regarded as a macro analytical technique. However, it is considered, when appropriately done, one of the most accurate analytical techniques. Also, Gravimetry is one of a few analytical methods that do not require standard solutions as the weight of precipitate is the only important parameter in analyte determination.

  4. Properties of Precipitates 1. Low solubility No significant loss of the analyte occurs during filtration and washing 2. High purity 3. Known composition After drying and ignition, known chemical composition 4. Easy to separate from reaction mixture 5. Stable under atmospheric conditions 4

  5. Steps in a Gravimetric Analysis After appropriate dissolution of the sample, the following steps should be followed for a successful gravimetric procedure: 1. Preparation of the Solution: This may involve several steps including adjustment of the pH of the solution in order for the precipitate to occur quantitatively and get a precipitate of desired properties, removing interferences, adjusting the volume of the sample to suit the amount of precipitating agent to be added.

  6. 2. Precipitation: This requires addition of a precipitating agent solution to the sample solution. Upon addition of the first drops of the precipitating agent, supersaturation occurs, and then nucleation starts to occur where every few molecules of precipitate aggregate together forming a nucleous. At this point, addition of extra precipitating agent will either form new nuclei or will build up on existing nuclei to give a precipitate. This can be predicted by Von Weimarn ratio where, according to this relation the particle size is inversely proportional to a quantity called the relative supersaturation where: Relative Supersaturation = (Q – S)/S

  7. The Q is the concentration of reactants before precipitation, S is the solubility of precipitate in the medium from which it is being precipitated. Therefore, in order to get particle growth instead of further nucleation we need to make the relative supersaturation ratio as small as possible. The optimum conditions for precipitation which make the supersaturation low are:

  8. Precipitation using dilute solutions to decrease Q • Slow addition of precipitating agent to keep Q as low as possible • Stirring the solution during addition of precipitating agent to avoid concentration sites and keep Q low • Increase solubility by precipitation from hot solution • Adjust the pH in order to increase S but not a too much increase as we do not want to lose precipitate by dissolution • Usually add a little excess of the precipitating agent for quantitative precipitation and check for completeness of the precipitation

  9. 7. 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.

  10. 3. Digestion of the Precipitate: The precipitate is left hot (below boiling) for 30 min to 1 hour in order for the particles to be digested. Digestion involves dissolution of small particles and reprecipitation on larger ones resulting in particle growth and better precipitate characteristics. This process is called Ostwald ripening. An important advantage of digestion is observed for colloidal precipitates where large amounts of adsorbed ions cover the huge area of the precipitate. Digestion forces the small colloidal particles to agglomerate which decreases their surface area and thus adsorption.

  11. During digestion at elevated temperature: • Small particles tend to dissolve and reprecipitate on larger ones. • Individual particles agglomerate. • Adsorbed impurities tend to go into solution. Ostwald ripening

  12. Lecture 18 Gravimetric Analysis, Cont…. Gravimetric Calculations

  13. 4. Filtering and Washing the Precipitate: It is crucial to wash the precipitate very well in order to remove all adsorbed species which will add to weight of precipitate. One should be careful not to use too much water since part of the precipitate may be lost. Also, in case of colloidal precipitates we should not use water as a washing solution since peptization would occur. In such situations dilute nitric acid, ammonium nitrate, or dilute acetic acid may be used. Usually, coagulated particles return to the colloidal state if washed with water, a process called peptization.

  14. Washing of Precipitates

  15. Adsorption and Particle size You should know that adsorption is a major problem in gravimetry in case of colloidal precipitate since a precipitate tends to adsorb its own ions present in excess, forming what is called a primary ion layer which attracts ions from solution forming a secondary or a counter ion layer. Individual particles repel each other keeping the colloidal properties of the precipitate. Particle coagulation can be forced by either digestion or addition of a high concentration of a diverse ions strong electrolytic solution in order to shield the charges on colloidal particles and force agglomeration.

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

  17. Cl- or Ag+ adsorbs on the particles when in excess (primary ion layer). A counter layer of cations forms. Washing with water will dilute the counter ion 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).

  18. 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.

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

  20. 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 reprecipitation may have to be done.

  21. 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

  22. 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.

  23. Name Student No. In the reaction 2A + B D C Calculate the concentrations of A, B, and C at equilibrium when 0.2 mol of A are mixed with 0.6 mol of B. The equilibrium constant is 2.5*10-7.

  24. 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)

  25. 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)

  26. 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

  27. 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

  28. 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%

  29. 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%

  30. Lecture 19 Gravimetric Calculations Precipitation Equilibria

  31. 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%

  32. 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%

  33. What weight of sulfur (FW = 32.064) ore which should be taken so that the weight of BaSO4 (FW = 233.40) precipitate will be equal to half of the percentage sulfur in the sample. Solution S D BaSO4 mmol S = mmol BaSO4 mg S/at wt S = mg BaSO4/FW BaSO4 mg S = at wt S x ( mg BaSO4 / FW BaSO4) mg S = 32.064 x (1/2 %S/233.40) mg S = 0.068689 %S %S = (mg S/mg sample) x 100 %S = 0.068689 %S/mg sample) x 100 mg sample = 0.068689 %S x 100 /%S = 6.869 mg

  34. A mixture containing only FeCl3 (FW = 162.2) and AlCl3 (FW = 133.34) weighs 5.95 g. The chlorides are converted to hydroxides and ignited to Fe2O3 (FW = 159.7) and Al2O3 (FW = 101.96). The oxide mixture weighs 2.62 g. Calculate the percentage Fe (at wt = 55.85) and Al (at wt = 26.98) in the sample. Solution g FeCl3 + g AlCl3 = 5.95 g Fe2O3 + g Al2O3 = 2.62

  35. Fe D FeCl3 1 mol Fe = 1 mol FeCl3 g Fe/at wt Fe = g FeCl3/ FW FeCl3 Rearrangement gives g FeCl3 = g Fe (FW FeCl3/at wt Fe) In the same manner g AlCl3= g Al ( FW AlCl3/at wt Al)

  36. g FeCl3 + g AlCl3 = 5.95 g Fe (FW FeCl3/at wt Fe) + g Al ( FW AlCl3/at wt Al) = 5.95 assume g Fe = x, g Al = y then: x(FW FeCl3/at wt Fe) + y ( FW AlCl3/at wt Al) = 5.95 x (162.2/55.85) + y (133.34/26.98) = 5.95 2.90 x + 4.94 y = 5.95 (1)

  37. The same treatment with the oxides gives 2 Fe D Fe2O3 mol Fe = 2 mol Fe2O3 g Fe/at wt Fe = 2 (gFe2O3/FW Fe2O3) g Fe2O3 = 1/2 g Fe (FW Fe2O3/at wt Fe) In the same manner g Al2O3 = 1/2 g Al (FW Al2O3/at wt Al) g Fe2O3 + g Al2O3 = 2.62 1/2 g Fe (FW Fe2O3/at wt Fe) + 1/2 g Al (FW Al2O3/at wt Al) = 2.62 1/2 x (159.7/55.85) + 1/2 y (101.96/26.98) = 2.62 1.43 x + 1.89 y = 2.62 (2)

  38. 2.90 x + 4.94 y = 5.95 (1) 1.43 x + 1.89 y = 2.62 (2) from (1) and (2) we get x = 1.07 y = 0.58 % Fe = (1.07/5.95) x 100 = 18.0% % Al = (0.58/5.95) x 100 = 9.8%

  39. Problem Consider a 1.0000 g sample containing 75% potassium sulfate (FW 174.25) and 25% MSO4. The sample is dissolved and the sulfate is precipated as BaSO4 (FW 233.39). If the BaSO4 ppt weighs 1.4900g, what is the atomic weight of M2+ in MSO4? K2SO4g BaSO4 MSO4g BaSO4

  40. mmol BaSO4 = mmol K2SO4 + mmol MSO4 Rearranging and solving:

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