Solutions for Gravimetric Analysis Exercises

Solutions for Gravimetric Analysis Exercises 1. The terms in a reaction quotient are actually dimensionless ratios of actual concentrations (or pressures) divided by standard concentrations (or pressures). The standard state for solutes is a 1 M solution and for gases it is a pressure of 1 bar (~ 1 atm), so these are the units used. For liquids and solids the standard state is the pure substance, so these ratios are unity. The solvent in a dilute solution is approximated as a pure substance. 2. The equilibrium constant for a reaction is dimensionless because it equals the reaction quotient at equilibrium, and, as discussed in Problem 1, this ratio involves only dimensionless terms. 3. a) Adding the two equations together gives 4. K1 = 1.8 x 10-10 AgCl(s) (2) Ag+(aq) + Cl-(aq) º AgCl(aq) K2 = 2.0 x 103 ______________________________________________ (3) b) º Ag+(aq) + Cl-(aq) (1) AgCl(s) º AgCl(aq) K3 = K1@K2 = 3.6 x 10-7 At equilibrium K3 = [AgCl(aq)], so [AgCl(aq)] = 3.6 x 10-7 M. Determine the solubility of La(IO3)3 using Ksp for La(IO3)3 and a table of initial and equilibrium concentration terms. La(IO3)3(s) X La3+(aq) + 3lO3-(aq) init 0 0.050 equil x 0.050 + 3x where x equals the increase in the [La3+] as a result of the dissolving process. at equil, Ksp = 1.0 x 10-11 = [La3+][lO3-]3 = (x)(0.050 + 3x)3 • (x)(0.050)3 assumes 3x << 0.050 (assumption valid) Solutions for Gravimetric Analysis Exercises 5. MgCO3 should be more soluble because it has the larger Ksp and the stoichiometry of the two salts is the same. If the stoichiometry of the salts is different, one cannot simply compare values of Ksp. 6. If only 1% of 0.010 M Ce3+ remains in solution this means [Ce3+] = 0.00010 M. The concentration of oxalate in equilibrium with 0.00010 M Ce3+ is determined by Ksp = 3 x 10-29 = [Ce3+]2[C2O42-]3 = (0.00010)2[C2O42-]3 To see if this oxalate concentration will precipitate 0.010 M Ca2+, calculate Q for the dissolution of CaC2O4 and compare to Ksp. Q = [Ca2+][C2O42-] = (0.010)(1.4 x 10-7) = 14. X 10-9 Since Q < Ksp (=1.3 x 10-8) for CaC2O4, Ca2+ will not precipitate. 7. A high relative supersaturation results in a precipitate with very small particle size and one that is more difficult to recover quantitatively. 8. To lower the relative supersaturation, ‚ precipitate from dilute solution ‚ add precipitating agent slowly, with stirring ‚ precipitate from hot solutions ‚ adjust solvent to increase the precipitate solubility 9. An electrolyte in the wash solution preserves the electric double layer and prevents peptization (break up of the precipitate). 10. HNO3 will volatilize during the drying process, NaNO3 will not. Solutions for Gravimetric Analysis Exercises 11. Salt At equilibrium [Ag+] at equilibrium for 0.10 M anion AgCl Ksp = 1.8 x 10-10 = [Ag+][Cl-] 1.8 x 10-9 M AgBr Ksp = 5.0 x 10-13 = [Ag+][Br-] 5.0 x 10-12 M AgI Ksp = 8.3 x 10-17 = [Ag+][I-] 8.3 x 10-16 M Ag2CrO4 Ksp = 1.2 x 10-12 = [Ag+]2[CrO42-] 3.5 x 10-6 M Precipitation occurs when Q > Ksp. If the mole ratio of Ag+ to the anion is the same, then the anions precipitate in order of increasing Ksp. Since this is not the case, the salt that needs the smallest [Ag+] to reach Q > Ksp will precipiate first, followed by the others in order of increasing [Ag+]. Based on the information in the above table, I- will precipitate first followed by Br-, then Cl- and finally CrO42-. 12. 13. Let x = grams of K2SO4 and y = grams of (NH4)2SO4. Thus, x + y = 0.649 g. Since there is one mole of SO42- in each mole of reactant and product, mol K2SO4 + mol (NH4)2SO4 = mol BaSO4 Substituting y = 0.649 - x into the above equation and solving for x yields x = 0.397 g. Therefore,