Chapter six
Gravimetric Analysis
Gravimetric Analysis :- Is one of the most accurate and
precise methods of quantative analysis .It is based on the
measurements of the mass of solid product of a chemical or
electrochemical reaction.
In gravimetric Analysis :Precipitate is separated by filtration , washing ,drying and
finally is accurately weighed .
Calculation proceduses , some common examples of gravimetric
Analysis are given in this chapter.
Solubility:The solubility of compound is the maximum amount of the
compound which can dissolve in a specific volume of solution at
equilibrium.
It is often measured in units of gm/l or gm/100ml.
Saturated solution :- When a solution at equilibrium contains
the maximum allowable quantity of a substance.
Super Saturated:- A solution which contains more than the
maximum allowable equilibrium concentration of a compound.
When supersaturated solution is allowed to settle until
equilibrium is established, a precipate is obtained and the
solution becomes saturated.
Solubility product
When a substance has limited solubility. i.e it is not completely
insoluble, but it is slightly soluble,the ions of the dissolved
portion exist in equilibrium with the solid material.
For e.g AgCl is added to water, small portion of it will dissolved
AgCl(s)
( AgCl ) aq
Ag+ + ClA small a mount of undissociated compound exist in
equilibrium in the aqueous phase (e.g about 0.1%) and its
concentration is constant, the prerence of undisociated
molecules can be generally neglected .
The equilibrium constand for such dissociation is called
Solubility product (Ksp) for AgCl
Ksp = [Ag+] [Cl-]
In general for slightly soluble salt AxBy
Ksp = [An+]x [Bm-]y
n, m = charges for cation and anion respectively
Solubility product for particular salt is constant at fixed temp. in
specific volume
Note = Precipitation will not accur unless
[Ag+] [Cl-] > Ksp
[Ag+] and [Cl-] remain in solution
If [Ag+] [Cl-] = Ksp
Example (1)
Calculate the concentration of Ag+ and Cl- in a saturated
solution of AgCl at 25oC and calculate the molar solubility of
AgCl, Given Ksp = 1.8x10-10?
AgCl
Ag+ + Cl-
Ksp = 1.8x10-10 [Ag+] [Cl-]
[Ag+] = [Cl-] = √1.8x10-10 = 1.34x10-5 M
Let S = molar solubility 0f AgCl
Since each mole of AgCl that dissolves will give (S) moles of
Ag+ and Cl- then SxS =1.8x10-10
S = 1.34x10-5 M the solubility of AgCl
Example (2)
Calculate the solubility of lead iodide (pbI2) in water at 250c in
the units (g/liter) ?
pbI2 = pb+2 +2I-
Ksp = 1×10-9 = [Pb+] [I-]2
1x10-9 = [S] [2S]-2 = 4S3
S = √0.25 × 10-9 = 6.3× 10-4M
= 6.3× 10-4 mol/l × 46.1 g/mole = 0.29 g/liter
Example (3)
Calculate the concentration of Cl- remainin in solution at
equilibrium and the solubility of AgCl when 10ml of 0.2M
AgNO3 in a solution is added to 10 ml of 0.10 M NaCl
Mmoles of Ag+ =10 x 0.2 = 2mmoles
Mmoles of Cl- = 10 x 0.1 = 1mmole
excess of Ag+ = 2-1 =1 mmole
Total vol. of solution = 10+10 = 20ml
Ag+ 1.0/20 = 0.05 M
Ksp = 1.8x 10-10 = [Ag+][Cl-]
[Cl-] =
= 3.6 x 10-9 M
The Cl- conc. Equals the solubility of the AgCl ,
hence the solubility =3.6 x 10-9 M
Example (4)
What must be the concentration of added Ag+ to 1.0x10-3M
NaCl solution to just start precipitation of AgCl?
Conc. of Cl- = 1x 10-3 M
[Ag+][1x10-3] =Ksp = 1.8x10-10
[Ag+] = 1.8x10-10 / 10-3 = 1.8x10-7 M
Hence the conc.of Ag+ must just exceeds 1.8x10-7 M to begin
precipitation
Factors affecting solubility
1- Temp. Ksp and the solubility increase with temp. but for
Sod.Sulfate decrease with above 340C .
2- Solvent Polar solutes = are more soluble in polar solvents
Non polar solutes = more soluble in non polar solvent
Hence
The solubility of AgCl is greater in polar solvent and it is better
to performe titration of chloride with silver ions by using water
as the solvent
Where as the precipitation of AgCl is more complete in a less
polar solvent( e.g hexane )
3- Concentration
Change in the concentration of one of the ions in the solubility
product equation has an effect upon the solubility of the slightly
soluble substance Solubility of the a salt is considerably
decrease by the addition of one of the salt to the solution this is
known as common-ion effect (example3)
4- PH the pH of a solution has effect upon solubility if one of
the ions of the salt is either an acid or base
Cu+2 + 2OH-
Cu(OH)2
Changing the [OH-] = will change the solubility
Example (5)
Calculate the solubility of Zn(OH)2 in a solution at pH=6, pH=9
Zn(OH)2
Zn+2 + 2OH-
Ksp = [Zn+2][OH-]2 =2x 10-17
Since the only source of Zn+2 is Zn(OH)2, solubility of Zn(OH)2
S = [Zn+2] at PH = 6 [H+] = 10-6 M [OH-]=10-8 M
SpH = 6
[Zn+2] = 2x10-17/ (10-8)2 = 0.2 M
b) at PH=9
SpH = 9
[H+]= 1x10-9 M [OH-] = 1x10-5M
[Zn+2] = 2x10-17/ (10-5)2 = 2x10-7 M
Typical Gravimetric analysis calculation
Gravimetric factor =
a/b : stoichiometric ratio between element and the precipitate
for example
Cl-
AgCl
1Cl2
2AgCl
Cl2 → PbCl2
Desire element
PPt
SO4
BaSO4
Fe
Fe2O3
CO32-
CaCO3
P
M2P2O7
G.F
Example (6)
An ore is analyzed for the manganese content (%Mn) by
converting the manganese to Mn3O4 and weighing it.If the
sample with mass 1.52gm of Mn3O4,What is the %Mn in the
sample ?
= 0.720 gm Mn / gm Mn3O4
Mass of Mn = 0.126 (0.720) = 0.0907 gm
Wt% Mn =
Example (7)
× 100 =
× 100 = 5.97%
Phosphate ion can be analyzed by converting it to ammonium
Phosphtmolybdate (NH4)3PO4.12MoO3 Calculate the %P and
P2O5 in the sample .If 1.1682gm of precipitate were obtained
from a sample with mass of 0.2711gm
= 0.0165 gm P / gm molybdate
Wt% P =
× 100 = 7.11%
Similarly
Mass of P2O5 = 0.0193 P (
Wt% P2O5 =
) ( ) = 0.0442 gm
× 100 =
× 100 = 16.31%
Example (8)
A sample of fertilizer mixture is analyzed and found to contain
45wt% nitrogen . Calculate the wt% of urea ithe sample ?
Sample = 100 gm
mass of nitrogen 45 urea= NH2CONH2
= 0.4667 gm N2 / gm urea
Mass of urea = 45 gm N2 ×
Wt% urea =
= 96.43gm
× 100 =
Gravimetric Analysis Steps
× 100 = 96.43%
1- Preparation of solution
It is necessary to separate any interfering materials by adjusting
conditions before precipitation
e.g Calcium oxalate is insoluble in basic medium but at low pH
oxalate ion carbines [H+] to form oxalic acid, aluminum ion can
be precipitated pH=4 higher pH is required to ppted magnesium.
2- Precipitation
precipitation agent solution is added to the test sample to form
precipitate e.g addition of AgNO3 to chloride a solution to
precipitate AgCl
3-Aging or Digestion
When ppted allow to stand in preance of original solution large
crystals grow
4-Washing and Filtration
Impurities can be removed by washing the ppted throughout
filtration the filtrate should be tested continuously to detect
completion of washing .
5-Drying or Ignition
Drying of the collected ppt can usually be done to remove water
by heating at (110-120)0C for 1-2 hours in an oven, Ignition at a
much higher temp. is usually required if a ppt must be converted
to more suitable from for weighing such as :Ignition of hydrous ferric oxide (Fe2O3.xH2O) to the anhydrous
Fe2O3= Ignition may be done in afurnace or by heating with a
burner using a porcelain or ptatinum crucible.
6-Weighing
Dry ppted is weighed using an analytical balance. A constant
weight, of difference about (0.3-0.4mg) should be obtained from
successive weighings.
7-Calculations
Are usually made on a wt% according to the typical procedure
discussed in the previous section of this chapter.
Problems
1- Calculate the concentrations of Ag+ and CrO4-2in a saturated
solution of Ag2CrO4 ? Ans:1.3x10-4M , 6.5x10-5M
2-An excess of silver nitrate solution was added to 50ml of a
sample solution containing Chloride ion. The mass of AgCl
precipitate observed was 0.5872gm. Calculate the Conc. of
chloride in the original sample solution
Ans:8.194×10-2M
3- Calculate the Conc. of a saturated solution of Calcium sulfate
at 250C if Ksp for CaSO4 =1.2×10-6
Ans:1.1×10-3M
4-A 0.100gm sample of impure silver was dissolved in nitric
acid and gravimetrically analyzed by the addition of an excess
of NaCl solution . The mass of AgCl collected was 0.1248gm
Calculate the wt% of silver in the sample
Ans: 93.93%
5-A saturated solution of PbF2 was analyzed and was found to
contain 4.2×10-3M of fluoride . Calculate the solubility product
of PbF2 .
Ans: 3.7×10-8
6-A saturated solution of Al(OH)3 with pH=4 was found to
contain 0.02M of Al+3 calculate the solubility product of
Al(OH)3
Ans:2×10-32