1. A number of anions form slightly soluble precipitates with certain metal ions and can be titrated with the metal solutions. for example: Cl - titrated with Ag +, SO 4 2- titrated with Ba 2+.

2. Solubility of Slightly Soluble Salts: AgCl (s)  (AgCl) (aq)  Ag + + Cl - Solubility Product K SP = ion product K sp = [Ag + ][Cl - ] The molar solubility depends on the stoichiometry of the salt. Ag 2 CrO 4(s)  2 Ag + + CrO 4 2- K sp = [Ag + ] 2 [CrO 4 2- ] Because the K sp product always holds, precipitation will not take place unless the product of precipitate ions exceeds the K sp When compound is referred to as insoluble, it is not completely insoluble but is slightly soluble.

3. Example 5.6 (p 162): The K sp of AgCl at 25  C is 1.0  10 -10. Calculate the concentrations of Ag + and Cl - in a saturated solution of AgCl, and the molar solubility of AgCl. Solution AgCl (s)  Ag + + Cl - K sp = [Ag + ][Cl - ] [Ag + ] = [Cl - ] = S S 2 = 1.0  10 -10  S = 1.0  10 -5 M The solubility of AgCl is 1.0  10 -5 M. Example 5.8 (p 163): What must be the concentration of added Ag + to just start precipitation of AgCl in a 1.0  10 -3 M solution of NaCl. Solution [Ag + ][Cl - ] = K sp [Ag + ] (1.0  10 -3 ) = 1.0  10 -10 [Ag + ] = 1.0  10 -7 M The concentration of Ag + must exceed 10 -7 M to begin precipitation.

4. Example 5.9 (p 164): What is the solubility of PbI 2 in g/L, if the K sp is 7.1  10 -9 ? Solution PbI 2(s)  Pb 2+ + 2I - K sp = [Pb 2+ ] [I - ] 2 = 7.1  10 -9 The molar solubility of PbI 2 : [Pb 2+ ] = S and [I - ] = 2S (S)(2S) 2 = 7.1  10 -9 Therefore, the solubility, in g/L, is 1.2  10 -3 mol/L  461.0 g/mol = 0.55 g/L HOME WORK: 25, 26, 27, 30 P.169-170

5. Predicted effect of excess barium ion on solubility of BaSO 4. The common ion effect is used to decrease the solubility. Sulfate concentration is the amount in equilibrium and is equal to the BaSO 4 solubility. In absence of excess barium ion, solubility is 10 -5 M. The common ion effect is used to decrease the solubility. Sulfate concentration is the amount in equilibrium and is equal to the BaSO 4 solubility. In absence of excess barium ion, solubility is 10 -5 M.

6. Precipitation titration, which is based on reactions that yield ionic compounds of limited solubility, is one of the oldest analytical techniques. The slow rate of formation of most precipitates, however, limits the number of precipitating agents that can be used in titrations to a handful. The most widely used and important precipitating reagent, silver nitrate, which is used for the determination of the halogens, the halogen-like anions. Titration Curves: Consider the titration of Cl - with standard solution of AgNO 3. A titration curve prepared by plotting pCl (-log[Cl - ]) againest the volume of AgNO 3. At the beginning of the titration: [Cl - ] = 0.1 M  pCl = 1 As the titration continues: pCl determined by [Cl - ] remaining (part of Cl - is removed as AgCl) At the equivalent point: [Cl - ] =  K sp = 10 -5 M  pCl = 5 Beyond the equivalent point: [Cl - ] determined from the concentration of excess Ag + and K sp

7. AgI is least soluble Sharpest change at equiv. point Least sharp, but steep enough for Equiv. point location The smaller K sp, the larger the break at the equivalent point.

8. 1. The precipitate formation is stoichiometric. 2. To allow the titrant to be added quickly, the equilibrium between the precipitate and its ions in solution much be attained rapidly. 3. The precipitate must be of low solubility in the solution. This is indicated by small equilibrium constant (K sp ). 4. A method to detect the stoichiometric point of the titration must be available. Although a number of indicators are available, in general, the best method for detecting the end point in precipitation titration is by an instrumental technique.

9. The end point can be detected by: I.measuring either pCl or pAg with an electrode and a potentiometer. II.using an indicator to produce a color change at or near the equivalence point.

11. The Mohr titration must be performed at a pH 8 (neutral or slightly basic medium ). Why? to prevent silver hydroxide formation (at pH > 8). 2Ag + + 2OH   2AgOH (s)  Ag 2 O (s) + H 2 O precipitate To prevent the formation of chromic acid (at pH  6). CrO 4 2  + H 3 O +  HCrO 4  + H 2 O [CrO 4 2  ] become lower, more Ag + to be added to reach end point, which cause error. The Mohr titration is useful for determining chloride in natural solutions, such as drink water.

13. The endpoint is routinely used for halide determinations where a known excess of silver ion is added to precipitate the halide ion, and the excess silver ion is back titrated using the thiocyanate/iron(III) as the indicator. The silver chloride precipitate is filtered, and the excess silver ion is titrated with thiocyanate producing a white precipitate of AgSCN. Once the silver is consumed, the excess thiocyante reacts with the iron(III) ion producing a red FeSCN 2+ complex. Thus, the appearance of the red color at the endpoint. The red FeSCN 2+ complex color is detectable at 6.4 x 10 -6 M concentrations and above. The titration is usually done in acidic pH medium to prevent precipitation of iron hydroxides, Fe(OH) 3.

14. The AgCl precipitate must be separated from the thiocyanate, why?? to prevent the reaction: AgCl + SCN -  AgSCN + Cl - Since AgSCN is less soluble than AgCl, equilibrium will shift to the right causing a negative error for the chloride analysis. To eliminate this error, AgCl must be filtered or add nitrobenzene before titrating with thiocyanate; nitrobenzene will form an oily layer on the surface of the AgCl precipitate, thus preventing its reaction with thiocyanate.

16. Before the end point (Cl  in excess). 1  Layer 2  Layer (AgCl) Cl  Na + When the equivalence point is reached, there is no longer an excess of analyte Cl , and the surface of the colloidal particles are largely neutral.

17. (AgCl) Ag + X  1  Layer 2  Layer After the end point (Ag + in excess). After the equivalence point, there will be an excess of titrant Ag +, some of these will adsorb to the AgCl particles, which will now be surrounded by a diffuse negative counterion layer. Now, negatively charged fluorescein can penetrate the counter ion layer and adsorb onto the AgCl lattice due to its affinity to Ag +.