Precipitation Reactions and Titrations Dr AKM Shafiqul Islam & Dr Zarina Zakaria

Precipitation Reactions A precipitation reaction occurs when water solutions of two different ionic compounds are mixed and an insoluble solid separates out of solution. The precipitate is itself ionic; the cation comes from one solution and the anion from another. To predict the occurrence of these reactions, we must know which ionic substances are insoluble in water.

Precipitation Reaction A reaction which forms a solid (precipitate) AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) -AgCl is classified as an insoluble substance Solubility – amount of substance that can be dissolved in a specific amount of water (g/L or mg/L) - Any substance with a solubility less than 0.01mol/L is considered insoluble. Metathesis Reactions

1.Write the balanced molecular equation. 2.Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 3.Cancel the spectator ions on both sides of the ionic equation Write the net ionic equation for the reaction of silver nitrate with sodium chloride. AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) Ag + + NO Na + + Cl - AgCl (s) + Na + + NO 3 - Ag + + Cl - AgCl (s) Writing Net Ionic Equations

Precipitation Reactions Precipitate – insoluble solid that separates from solution molecular equation ionic equation net ionic equation Pb NO Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - Na + and NO 3 - are spectator ions Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb I - PbI 2 (s)

Solubility Equilibria  Solubility Product Constant K sp  The equilibrium constant for the dissolution of a solid.  M m X x (s)  mM n+ (aq) + xX y- (aq)  Ksp = [M n+ ] m [X y- ] x  Solution must be saturated in order to use or determine the constant.

Measuring K sp and Calculating Solubilty for K sp  Table from CRC handbook.  Discussion.

Factors That Affect Solubility  The Common-Ion Effect  Here it is again.  The pH of the Solution  Formation of Complex Ions  Amphoterism  All can be classified under the Le Châtelier principle.

Factors That Affect Solubility  Common Ion.  MgF 2 (s)  Mg 2+ (aq) + 2F - (aq)  Adding NaF to the solution will reduce the solubility of MgF 2  Ksp = [Mg 2+ ] [F - ] 2  Since the equilibrium constant stays the same increasing [F - ] reduces [Mg 2+ ].

 The shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction. AgCl(s)  Ag + (aq) + Cl  (aq) The “Common Ion Effect” Concept test: What happens if you add NaF(s) into a solution of HF?

Factors That Affect Solubility  The pH of the Solution  As you’ve often been told there are very few neutral solutions. Dissolved ions impart some acidic or basic character to the solution.  Adding an acid or base to a specific solution will certainly effect the solubility. (depending on the ions basic or acidic nature)  CaCO 3 (s)  Ca 2+ (aq) + CO 3 2- (aq)

Factors That Affect Solubility  Formation of Complex Ions  If you add another matterial which would form a more soluble species in solution (complex ion) the solubility of the initial material will increase.  AgCl  Ag + (aq) + Cl - (aq)  If you add something that reacts with Silver ions you are effectively removing Ag + from the right side of the equation.  Almost the reverse of the common ion effect.  Ksp for AgCl = 1.8 x  Assuming the complex ion is much more soluble.

Precipitation of Ionic Compounds  CaF 2 (s)  Ca 2+ (aq) + 2 F - (aq)  K sp = [Ca 2+ ] [F - ] 2 at equilibrium  IP = [Ca 2+ ] [F - ] 2 initially.  If IP is greater than K sp a precipitate will form. Equilibrium K sp

Separation of Ions by Selective Precipitation  Addition of certain counter ions to a solution can (if chosen correctly) precipitate one of the initial ions selectively.  E.g.  Since AgNO 3 is “relatively soluble” and AgCl is “relatively insoluble” one can add HCl to a solution of AgNO 3 and precipitate the Ag + ion from solution.

Separation of Ions by Selective Precipitation  Sulfide Precipitations  In this case the solubility of a metal sulfide is related to the hydronium ion concentration in solution.  MS(s) + 2 H 3 O + (aq)  M 2+ (aq) + H 2 S(aq) + 2 H 2 O(l)  Ksp = [M 2+ ][H 2 S] / [H 3 O + ]  By adding acid to the solution ….  Bring back Q c Equilibrium K sp

Precipitate Formation  crystallization –nucleation: particles join to produce aggregates –crystal growth aggregate grows and 'fall out' of solution –crystal growth aggregate grows and 'fall out' of solution –We want a few big chunks of precipitate! supersaturation: more solute than should be present in solution relative supersaturation: a measure of supersaturation, (Q-S)/S Q = actual solute concentration S = equilibrium solute concentration

Controlling Precipitation  Increase S –Increase temperature  Decrease Q –Dilute solution –Well mixed (stirring)

pH control of precipitation Ca 2+ + C 2 O 4 2-  CaC 2 O 4 (s) H 2 C 2 O 4  2 H + + C 2 O 4 2- Feeder Reaction:

Homogeneous Precipitation  Fe HCO 2 -  Fe(HCO 2 ) 3  nH 2 O (NH 2 )CO + 3 H 2 O + heat  HCOOH + OH - + CO NH 4 + Feeder Reaction:

High Electrolyte Concentration to Aid Precipitation Excess charge on colloid creates ionic atmosphere around particle

Qualitative Analysis  This is not quantitative. Simply identification of ions present.  Methods (some can be quantitative)  Precipitation  Flame Test (color)  Solution (color)  Sodium Hydroxide  Ammonia  Sulfuric and Hydrochloric acid tests.  Chromatography  Flame ionization emission spectroscopy.

Qualitative Analysis  Group I: Ag +, Hg 2+, Pb 2+ Precipitated in 1 M HCl  Group II: Bi 3+, Cd 2+, Cu 2+, Hg 2+, (Pb 2+ ), Sb 3+ and Sb 5+, Sn 2+ and Sn 4+ Precipitated in 0.1 M H 2 S solution at pH 0.5  Group III: Al 3+, (Cd 2+ ), Co 2+, Cr 3+, Fe 2+ and Fe 3+, Mn 2+, Ni 2+, Zn 2+ Precipitated in 0.1 M H 2 S solution at pH 9  Group IV: Ba 2+, Ca 2+, K +, Mg 2+, Na +, NH 4 + Ba 2+, Ca 2+, and Mg 2+  Precipitated in 0.2 M (NH 4 ) 2 CO 3 solution at pH 10

Precipitation Titration Curve Equations  At Equivalence Point [A][T] = K sp  After the Equivalence Point [T] = (Original Concentration of Titrant)(Volume of excess Titrant / Total Volume of Solution) pT = -log[T]

Precipitation Titration Curve Example  mL of M I - was titrated with M Ag +. Ag + + I -  AgI(s) The solubility product for AgI is 8.3 x Calculate the concentration of I - ion in solution (a) after addition of mL of Ag + ; (b) after addition of mL of Ag + ; (c) at the equivalence point.

Precipitation Titration Curve Example  mL of M Hg 2 (NO 3 ) 2 was titrated with M KIO 3. Hg IO 3 -  Hg 2 (IO 3 ) 2 (s) The solubility product for Hg 2 (IO 3 ) 2 is 1.3 x Calculate the concentration of Hg 2 2+ ion in solution (a) after addition of mL of KIO3; (b) after addition of mL of KIO3; (c) at the equivalence point.

Precipitation Titration Curve Example  Consider the titration of mL of M Hg(NO 3 ) 2 with KSCN. Calculate the value of pHg 2 2+ at each of the following points and sketch the titration curve: 0.25V T 0.5V T 0.75V T 1.05V T 1.25V T

Titration Curve Shape  1:1 Stoichiometry of Reagents –Equivalence point is the steepest point of a curve  Maximum slope  An inflection point  Other Stoichiometries –The curve is not symmetric about the equivalence point –The equivalence point is not at the center of the steepest section of the curve  The less soluble the product, the sharper the curve around the equivalence point

Titration Curve Shape

Titration of a Mixture  The less soluble product forms first  If there is sufficient difference in solubility of products –First precipitation is nearly complete before the second one begins  Separation by precipitation –Coprecipitation  Alters the expected endpoints

Titration of a Mixture Example

Detection of End Point  Methods for determining the end point for precipitation titrations –Indicators –Electrodes –Light scattering  We’ll discuss indicators used for the Ag + Cl - titration equilibrium –Titrations with Ag + are called argentometric titrations

Indicators  Volhard titration –Uses the formation of a soluble colored complex at the end point  Fajans titration –A colored indicator is adsorbed onto the precipitate at the end point

 Used as a procedure for titrating Ag + ; determination of Cl - requires a back-titration –First, Cl - is precipitated by excess AgNO 3 –Excess Ag + is titrated with KSCN in the presence of Fe 3+ –When Ag + has been consumed, a red complex forms  The Volhard titration can be used for any anion that forms an insoluble salt with silver Volhard Titration red

Fajans Titration The technique uses an adsorption indicator. Prior to the equivalence point, there is excess Cl- in solution. Some is adsorbed on the surface of the crystal, giving a partial negative charge. After the equivalence point, there is excess Ag+ in solution. Some adsorbs to the surface imparting a partial positive charge to the precipitate. Choosing an indicator with a partial negative charge will cause it to adsorb to the surface. Dichlorofluorescein is green in solution but pink when absorbed on AgCl