Precipitation Formation Titrations I. General Principles A. Involves formation of a precipitate B. Must determine the volume of a standardized titrant needed to just precipitate all of the ion. C. Need an indicator or electrode to determine when the precipitation is complete II. For Method to Work A. Must produce a very insoluble precipitate
B. Must have an indicator
III. Factors Affecting Solubility of A Precipitate A. Temperature
B. Solvent
C. Common Ion Effect
D. Diverse (Neutral Salt, Activity) Effect Molarity KNO3
0.000 M 0.001 M 0.005 M 0.010 M
Molar Solub. AgCl
1.00 x 10-5 M 1.04 x 10-5 M 1.08 x 10-5 M 1.12 x 10-5 M
Molar Solub. BaSO4
1.00 x 10-5 M 1.21 x 10-5 M 1.48 x 10-5 M 1.70 x 10-5 M
E. Effect of pH (“high” acidity) 1. Solubility of some salts (salts of weak acids) depends on pH.
2. Example CaC2O4 º Ca +2 + C2O4-2 Ksp = [Ca+2] [ C2O4-2] &2
C2O4-2 + H+ º HC2O4-
Ka2 '
HC2O4- + H+ º H2C2O4
CaC2O4 9
H+
&
[HC2O4 ] &
Ka1 '
Ca+2
[H %][C2O4 ]
[H %][HC2O4 ] [H2C2O4]
H+
º C O º HC O º H C O -2
2
4
F. Effect of Hydrolysis
-
2
4
2
2
4
1. Insoluble Salt MA 9 MA º M+ + A2. Suppose A- is a salt of a very weak Acid A- + H2O º HA + OH 3. Effect:
G. Effect of Complexation 1. Insoluble Salt e.g. AgCl AgCl º Ag+ + Cl 2. Suppose Ag+ forms a complex Ag+ + 2 NH3 º Ag (NH3)2+ 3. Effect-
IV.
Regents Forming Organic Precipitates
A. Reagents Forming Chelate Compounds
1.
N
O H N
O
2.
H
H
Al+3
N
:N
H
B. Reagents Forming Salt-like Precipitate 1. O O O O || || || || C - C + Ca+2 –> C- C / \ / \ O O O O \ / Ca
2. B (C6H5)4- Tetraphenylboron (-)
B
K+ + B(C6H5)4 - -----> KB (C6H5)4 9 3. Benzidine To Precipitate SO4-2 H
H N:
:N
+ H2SO4
H
H
9 H
H
:N
N-H
H
H
+ H
+
SO4-2
V. Properties Desired of a Rxn For ppt’n Titration A. Very rapid rxn
B. Single stoichiometric rxn
C. Marked change in conc. of one of the reactants or products at the endpoint
D. Indicator must exist to show the change in concentration of the reactant or product at the endpoint.
VI.
Titration Curve A. For Ag+ + Cl- º AgCl B. Titrate Cl- unknown with standardized Ag+ C. Monitor [Cl-]. Calculate pCl pCl= -log [Cl-] D. Plot pCl versus ml Ag+ added Until EP [Cl-] relatively high After EP [Cl-] relatively low See large change in [Cl-] (pCl) at EP
PCl
ml Ag+
VII.
Theoretical Calculation of Titration Curve A. For known [Cl-] & known volume of Cl- -titrate with standardized Ag+ sol’n. Calculate pCl as a function of ml of Ag+ added B. Major Points 1. Start of Titration 0.00 ml [Cl-] = p Cl = - log [Cl-]
2. Prior To Equivalence Point Calc. # moles Cl# moles Ag+ That are mixed together Calc. # moles Cl- remaining number of moles Cl & remaining [Cl ] ' number liters of solution &
3. At Equivalence Point Calc. solubility of AgCl in “distilled” water Solubility = [Cl-]
4. Beyond the equiv point Calc. excess [Ag+] [Cl -] = Ksp, AgCl ---------[Ag+] excess pCl = -log [Cl-]
Calculation of Titration Curve for 50 ml of 0.100 M NaCl with 0.100 M of AgNO3. Plot pCl vs ml of AgNO3 added at 1) start of titration 2) after addition of 10.0 ml of AgNO3 3) after addition of 49.9 ml AgNO3, 4) Equiv. pt; 5) After addition of 60 ml AgNO3 1. At start of Titration
ie 0 ml AgNO3
[Cl-] = 0.100 M
-log [Cl-] = -(-1) = 1.0
2. After 10 ml AgNO3 Ag+ + Cl- º AgCl [Cl-]i = 0.100 m moles Cl & '
.100 mmoles (50 ml) ml
= 5.0 mmoles mmoles Ag+ added = .100 mmoles/ml (10.0 ml) = 1.0 mmoles mmoles Cl- left = 5.0 mmoles -1.0 mmoles = = 4.0 mmoles Total Vol = 50 ml + 10 ml = 60 ml [Cl-] = 4 mmoles ' 0.067 M 60 ml
pCl = 1.17
3. After Add of 49.9 ml m moles of Cl- in solution = . 0.100 mmoles (50 ml) = 5.00 mmoles ml
m moles of Ag+ added= 0.100 mmoles (49.9ml) = 4.99 mmoles ml
_____________________________ mmoles Cl- left = 0.01 mmoles Total Vol= 50 ml + 49.9 ml [Cl-] = .01 mmoles ' 1.00x 10&4 M 99.9 ml
pCl = 4.00
4. After Add 50.0 ml-Equiv. pt a) Neither an excess of Cl- or Ag+ b) Only source of Cl- is from AgClº Ag+ + ClKsp = [ Ag + ] [ Cl - ]= 1.0 x 10-10 [ Ag + ] = [ Cl - ] = x pCl = 5.00
5. After Addition of 60.0 ml AgNO3 Excess in Ag+ mmoles of Cl & i n solution ' .100
mmoles of Ag% in solution ' .100
mmoles ml
50ml'5.0 mmoles
mmoles 60 ml ' 6.0 mmoles ml
___________________ Excess Ag+ =1.0 mmoles
Total Vol = 50 ml + 60 ml = 110 ml [Ag+] = 1.0 mmoles/ 110 ml = 9.1 x 10 -3 M Ksp = 1 x 10-10 = [Ag+][Cl-] [Cl &] '
