Precipitation Titration

Precipitation Reactions [1] A precipitation reaction occurs when two or more soluble species/reagents combine to form an insoluble product that we call a precipitate. We usually write the balanced reaction as a net ionic equation, in which only the precipitate and those ions involved in the reaction are included: In the equilibrium treatment of precipitation, however, the reverse reaction describing the dissolution of the precipitate is more frequently encountered.

Precipitation Reactions [2] The equilibrium constant for this reaction is called the solubility product, Ksp. Note that the precipitate, which is a solid, does not appear in the Ksp expression. The equation is valid only if PbCl2(s) is present and in equilibrium with the dissolved Pb2+ and Cl–.

Problem: Solubility of Pb(IO3)2 in water [1] When an insoluble compound such as Pb(IO3)2 is added to a solution a small portion of the solid dissolves. Equilibrium is achieved when the concentrations of Pb2+ and IO3– are sufficient to satisfy the solubility product for Pb(IO3)2. At equilibrium the solution is saturated with Pb(IO3)2. How can we determine the concentrations of Pb2+ and IO3–, and the solubility of Pb(IO3)2 in a saturated solution prepared by adding Pb(IO3)2 to distilled water? The equilibrium reaction: (1) Its equilibrium constant: (2)

Problem: Solubility of Pb(IO3)2 in water [2] Substituting the equilibrium concentrations into eq. 2: (x) (2x)2 = 2.5 x 10-13 and solving gives: 4x3 = 2,5 x 10-13 → x = 4.0 x 10-5 The equilibrium concentrations of Pb2+ and IO3–, therefore, are Since one mole of Pb(IO3)2 contains one mole of Pb2+, the solubility of Pb(IO3)2 is the same as the concentration of Pb2+; thus, the solubility of Pb(IO3)2 is 4.0 x 10–5 M.

A more complex problem: The common ion effect [1] How is the solubility of Pb(IO3)2 affected if we add Pb(IO3)2 to a solution of 0.10 M Pb(NO3)2? Substituting the equilibrium concentrations into the solubility product expression (0.10 + x) (2x)2 = 2.5 x 10-13 4x3 + 0.40 x2 = 2,5 x 10-13 The following approximation for the equilibrium concentration of Pb2+ Substituting into Ksp equation:

A more complex problem: The common ion effect [2] Before accepting this answer, we check to see if our approximation was reasonable. In this case the approximation 0.10 + x » 0.10 seems reasonable since the difference between the two values is negligible. The equilibrium concentrations of Pb2+ and IO3–, therefore, are The solubility of Pb(IO3)2 is equal to the additional concentration of Pb2+ in solution, or 7.9 x 10–7 mol/L. As expected, the solubility of Pb(IO3)2 decreases in the presence of a solution that already contains one of its ions. This is known as the common ion effect.

Selecting and Evaluating the End Point: Mohr Method Mohr method: determination of Cl¯ using Ag+ as a titrant. By adding a small amount of K2CrO4 to the solution containing the analyte, the formation of a precipitate of reddish-brown Ag2CrO4 signals the end point. Because K2CrO4 imparts a yellow color to the solution, obscuring the end point, the amount of CrO42– added is small enough that the end point is always later than the equivalence point. To compensate for this positive determinate error an analyte-free reagent blank is analyzed to determine the volume of titrant needed to effect a change in the indicator’s color. The volume for the reagent blank is subsequently subtracted from the experimental end point to give the true end point.

Selecting and Evaluating the End Point: Volhard Method Volhard method: Ag+ is titrated with SCN- in the presence of Fe3+. The end point for the titration reaction is the formation of reddish colored Fe(SCN)2+ complex.: The titration must be carried out in a strongly acidic solution to achieve the desired end point.

Selecting and Evaluating the End Point: Fajan Method Fajans’ method, which uses an adsorption indicator whose color when adsorbed to the precipitate is different from that when it is in solution. For example, when titrating Cl– with Ag+ the anionic dye dichlorofluoroscein is used as the indicator. Before the end point, the precipitate of AgCl has a negative surface charge due to the adsorption of excess Cl–. The anionic indicator is repelled by the precipitate and remains in solution where it has a greenish yellow color. After the end point, the precipitate has a positive surface charge due to the adsorption of excess Ag+. The anionic indicator now adsorbs to the precipitate’s surface where its color is pink. This change in color signals the end point.

Argentometric Titration A precipitation titration in which Ag+ is the titrant. Analyte Titrant a End Point b AsO43- AgNO3, KSCN Volhard Br– AgNO3 Mohr or Fajans Cl– Volhard* CO32- C2O42- Analyte Titrant a End Point b CrO42- AgNO3, KSCN Volhard* I– AgNO3 Fajans Volhard PO43- S2- SCN– a When two reagents are listed, the analysis is by a back titration. The first reagent is added in excess, and the second reagent is used to back titrate the excess. b For Volhard methods identified by an asterisk (*) the precipitated silver salt must be removed before carrying out the back titration. back titration: A titration in which a reagent is added to a solution containing the analyte, and the excess reagent remaining after its reaction with the analyte is determined by a titration.

Problem 1 A mixture containing only KCl and NaBr is analyzed by the Mohr method. A 0.3172-g sample is dissolved in 50 mL of water and titrated to the Ag2CrO4 end point, requiring 36.85 mL of 0.1120 M AgNO3. A blank titration requires 0.71 mL of titrant to reach the same end point. Report the %w/w KCl and NaBr in the sample. MW of KCl＝74.551 g/mol; NaBr = 102.89 g/mol M(Ag+) x V(Ag+) = g KCl/MW KCl + g NaBr/MW NaBr (1) g NaBr = 0.3172 g – g KCl (2) g KCl = …. g NaBr = ….

Problem 2 The %w/w I– in a 0.6712-g sample was determined by a Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach the end point. Report the %w/w I– in the sample. AW I- = 126.9 g/mol mol Ag+ = mol I- + mol SCN- gram I- = …. %w/w I- = gram I- / 0.6712 gram x 100% = ……