Factors affecting solubility Common ion depress the solubility

Factors affecting solubility Common ion: depress the solubility and enhances precipitation Diverse ion: increase solubility; due to interionic attraction whic the ions of the precipitate in their ionic (i. e. not molecular) form Temperature: usually slightly increases the solubility (except P

Complex formation: usually increases the solubility Ag. Cl + 2 NH 3 [Ag(NH 3)2]+ + Cl- p. H: salts of weak acids dissolve at low p. H Solvent: inorganic salts more soluble in water than organic s

Precipitation titration • Precipitation titration is a titration method based on the formation of precipitate, which is slightly soluble • Titration with precipitating agents is useful for determining certain analytes, provided the equilibrium will take place rapidly and suitable means of detecting equivalent points is available

• Titration curve – plot of the changes in analyte concentration againts titrant volumes.

• Consider titration of Cl- with a standard solution of Ag. NO 3. • Before titration started – only have Cl-. p. Cl = - log[Cl-] • Titration proceed – part of Cl- is removed from solution by precipitation as Ag. Cl. p. Cl = -log [remaining Cl-]

At equivalence point - we have solution a saturated solution of Ag. Cl. Excess Ag. NO 3 added – excess Ag+. [Cl-] is determine from the concentration of Ag+ and Ksp.

Example Calculate p. Cl for the titration of 100. 0 ml 0. 100 M Na. Cl with 0. 100 M Ag. NO 3 for the addition of 0. 0, 20. 0, 100. 0 and 110. 0 ml Ag. NO 3. Ksp Ag. Cl is 1. 0 x 10 -10

a) Addition of 0. 0 ml Ag+ [Cl-] = 0. 100 M p. Cl = -log [Cl-] = -log 0. 100 =1

b) Addition of 20. 0 ml Ag+ + Cl– Ag. Cl Initial mmol Cl- = 100. 0 ml x 0. 100 M = 10. 0 mmol added Ag+= 20. 0 ml x 0. 100 M = 2. 0 mmol Cl- left = 8. 0 mmol [Cl-] left = 8. 0 = 0. 0667 M (100+20) ml p. Cl = -log [Cl-] = -log 0. 0667 = 1. 18

c) Addition of 99. 0 ml Ag Initial mmol Cl- = 100. 0 ml x 0. 100 M = 10. 0 mmol added Ag+= 99. 0 ml x 0. 100 M = 9. 9 mmol Cl- left = 0. 1 mmol [Cl-] left = 0. 1 = 5. 01 x 10 -4 M (100+99)ml p. Cl = -log [Cl-] = -log 5. 01 x 10 -4 = 3. 3