Precipitimetry
By
Mahmoud A. Tantawy
(Associate Professor of Analytical Chemistry)
• Volumetric analysis involving ppt. formation.
1- The precipitate must be practically insoluble.
2- Precipitation should be rapid.
3- The end point must be easily detected.
2
Law of mass action
For a reversible chemical reaction between A and B to form products C and D:
At the beginning [C][D]
Q=
A + B [A][B]
A + B Forward reaction C + D Q < Keqm
A + B Backward reaction C + D Q > Keqm
A + B C + D Q = Keqm
3
Important terms
Solubility of a substance (S)
• Maximum amount of solute that dissolves in a given volume of solvent at a
given temperature.
• Solubility (S), is an equilibrium position
Solubility product constant (Ksp)
• Is the equilibrium constant representing the dissolving of an ionic solid in water.
• i.e. compound AnBm nA + mB therefore, Ksp = [A]n [B]m
• Solubility product (Ksp), is an equilibrium constant 4
AgCl AgCl Ag+ + Cl-
Solid Soluble
• For simplicity
AgCl Ag+ + Cl-
Solid
Ksp = [Ag+] [Cl-]
• Only applied for sparingly soluble salts, at a specified temperature.
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• For a precipitate to be formed, the ionic product of its ions (Q)
should exceed its solubility product (Ksp).
Q < Ksp Soluble Solution
• When do a salt precipitate: Q = Ksp Saturated solution
Q > Ksp Precipitation occurs.
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Comparing relative solubilities
1. Salts having the same formula
• Ksp can be used to compare solubility of different salts.
• The smaller the Ksp of a cpd the faster it will ppt.
• eg : Ksp of AgI < AgBr < AgCl , therfore AgI will ppt first.
solubility
Ag+
AgCl
Ksp
Cl- AgBr
I-
Br-
AgI
pptn 7
2. Salts having different formulae AB, A2B, AB2 and A2B3
• Example: AgCl, (Ag)2SO4 , Cu(Cl)2 and Bi2S3
• Ksp cannot be used to compare relative solubilities of these salts.
Bi2S3 Ksp = 1.1x10-73
Order of Ksp: CuS > Ag2S > Bi2S3
Ag2S Ksp = 1.6x10-49
CuS Ksp = 8.5 x 10-45
• Solubility (S) should be calculated and used to compare relative solubilities.
CuS Solubility = 9.2x10-23
Order of solubility: Bi2S3 > Ag2S > CuS
Ag2S Solubility = 3.4x10-17
Bi2S3 Solubility = 1.0x10-15 8
Molar Solubility Solubility product
Solubility
S Ksp
Salt concentration at
Molar concentration of a For a saturated soln of compound
saturation AnBm Ksp equals the product of the
Salt at saturation
(at specified temp.) molar concentration of its [ions] each
(at specified temp.)
raised to a power equal the number of
g/L ions furnished on ionization.
M (mol/L)
Ksp=[A]n [B]m
AgCl Ag + Cl For Agcl Ksp = [Ag+] [Cl-]
S S S
Pb3(PO4)2 3Pb2+ + 2PO42- Pb3(PO4)2 Ksp = [Pb2+]3 [PO43-]2
S 3S 2S
= Solublity (g/L)
= Molar solubility (S) x M.Wt
M.Wt
9
mole
x 3x 2x
Pb3(PO4)2 3Pb+2 + 2PO4-3
1 3 2
S 3S 2S 10
Calculation of the Solubility product from the solubility
Problems :-
1- If the solubility of AgCl is 0.0015 g/L. Calculate solubility product.
Mol.wt of AgCl = 143
Soln.
a) Molar solubility [S] = 0.0015/143 = 1.05 x 10-5 mol/L
AgCl Ag+ + Cl-
1mole 1mole 1mole
S S S
b) Molar concentration of ions 1.05 x 10-5M ? ?
Ag+ = S = 1.05 x 10-5 mol/L
Cl- = S = 1.05 x 10-5 mol/L
c) Solubility product Ksp = [Ag+] [Cl-] = 1.05 x 10-5 x 1.05 x 10-5 = 1.1 x 10-10
or Ksp = [S] [S] = S2 = (1.05 x 10-5)2= 1.1 x 10-10
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2- Calculate solubility product of Pb3(PO4)2, solubility = 0.00014 g/L, mol.wt of Pb3(PO4)2
= 811.7
Soln.
a) Molar solubility [S] = 0.00014/811.7 = 1.7 x 10-7 mol/L
Pb3(PO4)2 3Pb+2 + 2PO4-3
1 3 2
S 3S 2S
1.7 x 10-7 M ? ?
b) Molar concentration of ions
Pb2+ = 3S = 3 x (1.7 x 10-7) mol/L
PO4-3 = 2S = 2 x (1.7 x 10-7) mol/L
c) Solubility product Ksp = [Pb2+]3 [PO4-3]2 =[3(1.7 x 10-7)]3 x [2(1.7 x 10-7)]2 = 1.5 x 10-32
or Ksp = [3S]3 [2S]2 = 108 S5 = 108 x (1.7 x 10-7)5 = 1.5 x 10-32 12
Calculation of the Solubility from the solubility product
3- Calculate solubility of Ag2S in pure water Ksp (Ag2S) = 6 x 10-51
Soln.
Ag2S 2Ag+ + S-2
S 2S S Where S is the molar solubility of Ag2S
a) Ksp = [Ag+]2 [S2-] = [2S]2 [S] = 4S3
b) Molar solubility S can be calculated
4S3 = 6 x 10-51
S3 = 1.5 x 10-51
S = 1.14 x 10-17 mol/L
C) Solubility = Molar Solubility x Mwt = 1.14 x 10-17 x 248 = 2.82 x 10-15 g/L
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Prediction on precipitation
Q < Ksp Soluble solution
Q = Ksp Saturated solution, at equilibrium
Q > Ksp Precipitation occurs.
Q (reaction Quotient) = Product of the molar concentration of involved
[ions] each raised to a power equal the number of ions furnished on
ionization.
Q (reaction Quotient) calculated same way as Ksp.
Q (reaction Quotient) at saturation = Ksp 14
Example 1:
Determine whether a precipitate will be formed when 100 mL of 0.1 Pb2+ solution
mixed with 100 mL of 0.3 Cl- solution. (Ksp of PbCl2=1.7x10-4).
Soln.
C1V1 = C2V2 For Pb2+ C1=0.1M, V1=100ml, C 2= ? and V2= 200ml
The lead and chloride ion concentrations would be 0.05M and 0.15M, respectively.
PbCl2 Pb2+ + 2Cl-
But when we add the two solution the product of ionic concentrations would be :
[Pb2+] [Cl-]2 = (0.05) (0.15)2 =1.1x10-3 which is > Ksp
i.e. exceeds the Ksp, thus precipitation occurs.
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Example 2:
What is the maximum pH for a solution of 0.1 Mg2+ from which Mg(OH)2 will not
precipitate, given that Ksp of Mg(OH)2 is 1.2x10-11.
Mg(OH)2 Mg2+ +2 OH-
Ksp = [Mg2+] [OH-]2 = 1.2x10-11
i.e. precipitation occurs at pH above 9.04, at lower pH values no precipitation will occur.
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1. Effect of common ion on solubility:
+ +
AgCl Ag AgCl Ag + Cl
+ Cl +
+ AgNO3 Ag + NO3
NaCl Na + Cl
Common ion Common ion
-The common ion depresses the solubility. We can calculate the extent of this depression.
Example ;-
- Calculate the solubility of AgCl in 0.001M, 0.01M, 0.1M KCl
Soln.
- In 0.001M KCl [Cl-] = 10-3 M Similarily,
Ksp of AgCl = [Ag+] [Cl-] = [Ag+] [10-3] - In 0.01M KCl [Cl-] = 10-2 Therefore Ag+ = 10-8 M
[Ag+] = 1.1 x 10-10 / 10-3 = 10-7 - In 0.1M KCl [Cl-] = 10-1 Therefore Ag+ = 10-9 M
i.e Molar solubility of [Ag+] decreased from 10-5 in H2O to 10-7 M in 0.001M KCl.
i.e. As [Cl-] increase Solubility of AgCl decreases
- Cl- (precipitating agent) can be used to suppress the solubility of AgCl. 17
AgCl Ag2+ +Cl-
Ksp = [Ag+] [Cl-] = 10-10
[Cl-] = 10-3 [Cl-] = 10-1
+
AgCl Ag + Cl
+
NaCl Na + Cl
[Cl-] = S Common ion
Ksp = [Ag+] [Cl-] Ksp = [Ag+] [Cl-]
Ksp = S2
Ksp Ksp
S2 = 10-10 [Ag+] = [Ag+] =
[Cl-] [Cl-]
10 -10 10 -10
S = 10-5 M [Ag+] = [Ag+] =
10-3 10-1
[Ag+] = 10-7 [Ag+] = 10-9 18
2. Effect of temperature
- Increasing temp. will increase solubility.
3. Effect of complex formation
- Complex formation increase solubility when soluble complex is formed with one ion of
ppt. components.
Eg 1 : AgCl is ppt . when we add excess NH3 it will dissolve the ppt forming silver amine
complex [Ag(NH3)2]+.
[Ag(NH3)2]+ Ag+ + 2NH3 Kinstab (small)
When this complex dissociate it yields Ag+ of very low conc. which cannot exceed Ksp
of AgCl
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Complex Precipitate
Ag+ Ag+
Dissociation
Dissociation
Ag+ = Solubility
Ag+ AgCl
Ag+
Ksp
Ag+ Ag+
≈ AgBr
Kinst
Ag+ Ag+
[Ag(NH3)2]+ Ag+ Ag+
Ag+ Ag+
Ag+
Ag+ AgI
Ag+
Ag+
[Ag(CN)2]- Ag+
Formation
formation =pptn
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Example : Ag+ + Cl- AgCl
[Ag(NH3)2]+ +
2NH3
Ag+ + Br- AgBr
[Ag(NH3)2]+ +
2NH3
Ag+ + I- AgI
[Ag(NH3)2]+ +
2NH3 21
Since Ksp of AgI < AgBr < AgCl
Therefore, when we add NH3 on ppt of AgI or AgBr or AgCl. It will dissolve AgCl while
AgBr is partially soluble and AgI will remain insoluble. Give reason
Ans.; Because silver amine complex produces silver ions insufficient to exceed
Ksp of AgCl because it is large (1.1 x 10-10) but approach Ksp of AgBr (1.1 x 10-14)
and exceeds that of AgI (1 x 10-16)
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Fractional precipitation :-
• When a pptating agent like Ag+ is added to a soln. containing equal conc. (0.1M) of
both Cl- and I- . Which will ppt 1st AgCl or AgI?
• How completely will the 1st salt be ppt. before the 2nd ion begins to react with Ag+ ?
Answer:
Since both have the same formulae; The salt of lower Ksp will ppt 1st i.e . AgI (1.7 x 10-16)
i.e If iodide conc. is 0.1M therefore Ag+ needs to ppt AgI will be
[Ag+] = Ksp AgI / [I-] = 10-16/10-1 = 10-15 M
• If chloride conc. is 0.1M therefore Ag+ needs to ppt AgCl will be
[Ag+] = Ksp AgCl / Cl- = 10-10/10-1 = 10-9 M
N.B. 10-15 < 10-9
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• When [Ag+] conc. reaches 10-15 iodide will ppt as AgI
• while Cl- will not ppt till Ag+ conc. reaches 10-9.
• Therefore, when the conc. of Ag+ reaches 10-9 all iodide will almost be
precipitated as AgI and then Cl- will start precipitation as AgCl.
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Question:
• State a case at which both AgCl and AgI could precipitate
simultaneously (at the same time)?
Answer:
In fact both AgI & AgCl will ppt. simultaneously when :
[Ag+] = Ksp AgI / [I-] = Ksp AgCl / [Cl-]
therefore [Cl-] / [I-] = Ksp AgCl / Ksp AgI = 10-10 /10-16 = 106
i.e. When Cl- : I- ratio = 106 : 1
This type of successive precipitation, using the same precipitating agent, is
called fractional pptn.
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Titration curve
- Useful in:
1. Determining sharpness of the end point, this is indicated by characteristic
inflection of the sigmoid curve.
2. To choose the suitable indicator.
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- Sharpness of end point depends on:
1- The initial conc. of both the titrant and ion to be determined, the higher this conc. the
sharper will be the end point.
2- The solubility of the formed salt : the lower the Ksp the more sharper will be the end
point.
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