Application of Statistical and mathematical equations in Chemistry Part 4 Solubility Theory Precipitation Percentage calculation

1. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
Application of Statistical and mathematical
equations in Chemistry
Part 4
Solubility Theory
Precipitation
Percentage calculation

2. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
Solubility Theory
Solubility is the property of a solid, liquid, or gaseous chemical
substance called solute to dissolve in a solid, liquid, or gaseous solvent to form a
homogeneous solution of the solute in the solvent. The solubility of a substance
fundamentally depends on the physical and chemical properties of the solute and
solvent as well as on temperature, pressure and the pH of the solution. The extent of
the solubility of a substance in a specific solvent is measured as
the saturation concentration, where adding more solute does not increase the
concentration of the solution and begin to precipitate the excess amount of solute.
Some ionic compounds (salts) dissolve in water, which arises because of the
attraction between positive and negative charges (see: solvation). For example, the
salt's positive ions (e.g. Ag+) attract the partially negative oxygens in H2O. Likewise,
the salt's negative ions (e.g. Cl−) attract the partially positive hydrogens in H2O. Note:
oxygen is partially negative because it is moreelectronegative than hydrogen, and
vice-versa (see: chemical polarity).
AgCl(s) Ag+(aq) + Cl−(aq)
However, there is a limit to how much salt can be dissolved in a given volume of
water. This amount is given by the solubility product, Ksp. This value depends on
the type of salt (AgCl vs. NaCl, for example), temperature, and the common ion
effect.
One can calculate the amount of AgCl that will dissolve in 1 liter of water, some
algebra is required.
Ksp = [Ag+] × [Cl−] (definition of solubility product)
Ksp = 1.8 × 10−10 (from a table of solubility products)
[Ag+] = [Cl−], in the absence of other silver or chloride salts,

3. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
[Ag+]2 = 1.8 × 10−10
[Ag+] = 1.34 × 10−5
The result: 1 liter of water can dissolve 1.34 × 10−5 moles of AgCl(s) at room
temperature. Compared with other types of salts, AgCl is poorly soluble in water. In
contrast, table salt (NaCl) has a higher Ksp and is, therefore, more soluble.
Furmala
example
Example 1
The solubility of barium sulphate at 298 K is 1.05 x 10-5 mol dm-3.
Calculate the solubility product.
The equilibrium is:
Notice that each mole of barium sulphate dissolves to give 1 mole
of barium ions and 1 mole of sulphate ions in solution.
That means that:
[Ba2+] = 1.05 x 10-5 mol dm-3

4. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
[SO42-] = 1.05 x 10-5 mol dm-3
All you need to do now is to put these values into the solubility
product expression, and do the simple sum.
Don't forget to work the units out.
Important: Get your calculator and work this out!
Students frequently mis-enter numbers like 1.05 x
10-5. If you try this sum, and get a different answer,
then you are probably misusing the EXP button. To
enter this number, you would enter 1.05, press the
EXP button, and then enter -5 (probably by entering
5 and then pressing the +/- button). People often try
to enter x 10 in the middle of this process as well.
The EXP button includes this.
Example 2
These calculations are very simple if you have a compound in
which the numbers of positive and negative ions are 1 : 1. This next
example shows you how to cope if the ratio is different.
The solubility of magnesium hydroxide at 298 K is 1.71 x 10-4 mol
dm-3. Calculate the solubility product.

5. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
The equilibrium is:
For every mole of magnesium hydroxide that dissolves, you will
get one mole of magnesium ions, but twice that number of
hydroxide ions.
So the concentration of the dissolved magnesium ions is the same
as the dissolved magnesium hydroxide:
[Mg2+] = 1.71 x 10-4 mol dm-3
The concentration of dissolved hydroxide ions is twice that:
[OH-] = 2 x 1.71 x 10-4 = 3.42 x 10-4 mol dm-3
Now put these numbers into the solubility product expression and
do the sum.
Calculating solubilities from solubility products
Reversing the sums we have been doing isn't difficult as long as
you know how to start. We will take the magnesium hydroxide
example as above, but this time start from the solubility product
and work back to the solubility.

6. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
If the solubility product of magnesium hydroxide is 2.00 x 10-
11mol3 dm-9 at 298 K, calculate its solubility in mol dm-3 at that
temperature.
The trick this time is to give the unknown solubility a symbol like x
or s. I'm going to choose s, because an x looks too much like a
multiplication sign.
If the concentration of dissolved magnesium hydroxide is s mol
dm-3, then:
[Mg2+] = s mol dm-3
[OH-] = 2s mol dm-3
Put these values into the solubility product expression, and do the
sum.
Note: This is where you might need to find your
calculator instruction book! With a bit of luck, you

7. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
will find a button for cube roots, and this will enable
you to do sums with compounds like magnesium
hydroxide with a formula AB2or A2B.
Anything more complicated than this would need
you to be able to find 4th or 5th roots. It would be a
good idea to find out how your calculator does this.
Mine has an x1/y button. You would need to practice
using this or something similar. On my calculator, to
find fourth root of 16 using this button, you would
have the number 16 in the display, press the
x1/y button, enter 4 (for 4th root) and then press
equals. If you have done it right, you should get an
answer of 2.
Precipitation is the formation of a solid in a solution or inside another solid during
a chemical reaction or by diffusion in a solid. When the reaction occurs in a liquid
solution, the solid formed is called the precipitate. The chemical that causes the solid
to form is called the precipitant
n solids, precipitation occurs if the concentration of one solid is above the solubility
limit in the host solid, due to e.g. rapid quenching or ion implantation, and the
temperature is high enough that diffusion can lead to segregation into precipitates.
Precipitation in solids is routinely used to synthesize nanoclusters
Percentage calculation

8. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
Gravimetric methods of analysis are based on the measurement of mass. The two
gravimetric methods are precipitation methods and volatilization methods. In
precipitation methods the analyte is converted to an insoluble product, filtered,
washed and heated. The mass of the resulting residue is determined. In volatilization
methods the analyte is heated and the analyte or its decomposition product is
collected. The resulting loss of mass is determined.
General form for calculations
The calculations for gravimetric analyses are fairly straight-forward.
Gravimetric calculations are based on the fundamental stoichiometric
calculations. (Note: You may wish to review these calculations before
continuing.) The basic form of the calculation is:
The gravimetric factor (GF) comes from a combination of the mole ratios and the
formula weights used in the stoichiometric calculation.
For example, if you were looking for SO3 and your precipitate was BaSO4, the
gravimetric factor would be:

9. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
The numbers, 80.064 and 233.391, are the formula weights of
SO3 andBaSO4, respectively.
The main question is how to determine the mole ratio without knowing the entire
reaction. This is actually quite easy. Simply balance the common element. Most of the
time oxygen is not considered. In the above example, sulfur appears in both terms.
There is only one sulfur in each term and the sulfurs are balanced. In other words, the
mole ratio is 1.
Consider the following GF:
The common element is silver, Ag.
However, there are two silver atoms represented in the upper term and only one in
the lower term.
To "balance" the silver atoms, a 2 is placed in front of the substance in the lower term.

10. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
The calculation set-up for this gravimetric factor would be: