Gravimetric Analysis: Solutions for Equilibrium Constants and Solubility, Exams of Stoichiometry

Solutions to exercises on gravimetric analysis, focusing on the concepts of equilibrium constants, solubility calculations, and the relationship between ksp and the solubility of salts. It covers topics such as the meaning of reaction quotients, the calculation of equilibrium constants for various reactions, and the determination of solubility using ksp values.

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2021/2022

Uploaded on 07/05/2022

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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
(1) AgCl(s) º Ag+(aq) + Cl-(aq) K1 = 1.8 x 10-10
(2) Ag+(aq) + Cl-(aq) º AgCl(aq) K2 = 2.0 x 103
______________________________________________
(3) AgCl(s) º AgCl(aq) K3 = K1@K2 = 3.6 x 10-7
b) At equilibrium K3 = [AgCl(aq)], so [AgCl(aq)] = 3.6 x 10-7 M.
4. 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)
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  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

(1) AgCl(s) º Ag+(aq) + Cl-^ (aq) K 1 = 1.8 x 10-

(2) Ag+(aq) + Cl-^ (aq) º AgCl(aq) K 2 = 2.0 x 10^3

______________________________________________

(3) AgCl(s) º AgCl(aq) K 3 = K 1 @K 2 = 3.6 x 10-

b) At equilibrium K 3 = [AgCl(aq)], so [AgCl(aq)] = 3.6 x 10-7^ M.

  1. Determine the solubility of La(IO 3 ) 3 using Ksp for La(IO 3 ) 3 and a table of initial and equilibrium concentration terms.

La(IO 3 ) 3 (s) X La3+(aq) + 3lO 3 -^ (aq)

init 0 0.

equil x 0.050 + 3x

where x equals the increase in the [La 3+] as a result of the dissolving process.

at equil, Ksp = 1.0 x 10-11^ = [La3+][lO 3 -^ ] 3 = (x)(0.050 + 3x)^3 • (x)(0.050)^3

assumes 3x << 0.

(assumption valid)

  1. MgCO 3 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.
  2. 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 [C 2 O 4 2-^ ] 3 = (0.00010) 2 [C 2 O 4 2-^ ] 3

To see if this oxalate concentration will precipitate 0.010 M Ca 2+, calculate Q for the dissolution of CaC 2 O 4 and compare to Ksp.

Q = [Ca2+][C 2 O 4 2-^ ] = (0.010)(1.4 x 10-7^ ) = 14. X 10-

Since Q < Ksp (=1.3 x 10 -8^ ) for CaC 2 O 4 , Ca2+^ will not precipitate.

  1. A high relative supersaturation results in a precipitate with very small particle size and one that is more difficult to recover quantitatively.
  2. 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

  1. An electrolyte in the wash solution preserves the electric double layer and prevents peptization (break up of the precipitate).
  2. HNO 3 will volatilize during the drying process, NaNO 3 will not.